Q
stringlengths 4
3.96k
| A
stringlengths 1
3k
| Result
stringclasses 4
values |
|---|---|---|
Theorem 1. Let \( X \) be an arbitrary set, and \( \mathcal{C} \) a collection of subsets of \( X \), countably many of which cover \( X \) . Let \( \beta \) be a function from \( \mathcal{C} \) to \( {\mathbb{R}}^{ * } \) such that\n\n\[ \inf \{ \beta \left( C\right) : C \in \mathcal{C}\} = 0 \]\n\nThen the equation\n\n\[ \mu \left( A\right) = \inf \left\{ {\mathop{\sum }\limits_{{i = 1}}^{\infty }\beta \left( {C}_{i}\right) : A \subset \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{C}_{i},{C}_{i} \in \mathcal{C}}\right\} \]\n\ndefines an outer measure on \( X \) .
|
Proof. Assume all the hypotheses. There are now three postulates for an outer measure to be verified. Our assumption about \( \beta \) implies that \( \beta \left( C\right) \geq 0 \) for all \( C \in \mathcal{C} \) . Therefore, \( \mu \left( A\right) \geq 0 \) for all \( A \) . Since \( \varnothing \subset C \) for all \( C \in \mathcal{C},\mu \left( \varnothing \right) \leq \beta \left( C\right) \) for all \( C \) . Taking an infimum yields \( \mu \left( \varnothing \right) \leq 0 \).\n\nIf \( A \subset B \) and \( B \subset \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{C}_{i} \), then \( A \subset \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{C}_{i} \) and \( \mu \left( A\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\beta \left( {C}_{i}\right) \) . Taking an infimum over all countable covers of \( B \), we have \( \mu \left( A\right) \leq \mu \left( B\right) \).\n\nLet \( {A}_{i} \subset X\left( {i \in \mathbb{N}}\right) \) and let \( \varepsilon > 0 \) . By the definition of \( \mu \left( {A}_{i}\right) \) there exist \( {C}_{ij} \in \mathcal{C} \) such that \( {A}_{i} \subset \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{C}_{ij} \) and \( \mathop{\sum }\limits_{{j = 1}}^{\infty }\beta \left( {C}_{ij}\right) \leq \mu \left( {A}_{i}\right) + \varepsilon /{2}^{i} \) . Since \( \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i} \subset \mathop{\bigcup }\limits_{{i, j = 1}}^{\infty }{C}_{ij} \), we obtain\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \leq \mathop{\sum }\limits_{{i, j}}\beta \left( {C}_{ij}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\left\lbrack {\mu \left( {A}_{i}\right) + \varepsilon /{2}^{i}}\right\rbrack = \varepsilon + \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \]\n\nSince this is true for each positive \( \varepsilon \), we obtain \( \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \).
|
Yes
|
Lemma 1. If \( \left( {X,\mathcal{A}}\right) \) is a measurable space, then \( X \) and \( \varnothing \) belong to \( \mathcal{A} \) . Furthermore, \( \mathcal{A} \) is closed under countable intersections and set difference.
|
Proof. This is left to the reader as a problem.
|
No
|
Lemma 2. For any subset \( \mathcal{F} \) of \( {2}^{X} \) there is a smallest \( \sigma \) -algebra containing \( \mathcal{F} \) .
|
Proof. As Example 10 shows, there is certainly one \( \sigma \) -algebra containing \( \mathcal{F} \) . The smallest one will be the intersection of all the \( \sigma \) -algebras \( {\mathcal{A}}_{\nu } \) containing \( \mathcal{F} \) . It is only necessary to verify that \( \bigcap {\mathcal{A}}_{\nu } \) is a \( \sigma \) -algebra. If \( {A}_{i} \in \bigcap {\mathcal{A}}_{\nu } \), then \( {A}_{i} \in {\mathcal{A}}_{\nu } \) for all \( \nu \) . Since \( {\mathcal{A}}_{\nu } \) is a \( \sigma \) -algebra, \( \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i} \in {\mathcal{A}}_{\nu } \) . Since this is true for all \( \nu ,\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i} \in \bigcap {\mathcal{A}}_{\nu } \) . A similar proof is needed for the other axiom.
|
Yes
|
Lemma 1. If \( \left( {X,\mathcal{A},\mu }\right) \) is a measure space, then:\n\n(1) \( \mu \left( A\right) \leq \mu \left( B\right) \) if \( A \in \mathcal{A}, B \in \mathcal{A} \), and \( A \subset B \) .\n\n(2) \( \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \) if \( {A}_{i} \in \mathcal{A} \) .
|
Proof. For (1), write\n\n\[ \mu \left( B\right) = \mu \left\lbrack {A \cup \left( {B \smallsetminus A}\right) }\right\rbrack = \mu \left( A\right) + \mu \left( {B \smallsetminus A}\right) \geq \mu \left( A\right) \]\n\nFor (2), we create a disjoint sequence of sets \( {B}_{i} \) by writing \( {B}_{1} = {A}_{1},{B}_{2} = \) \( {A}_{2} \smallsetminus {A}_{1},{B}_{3} = {A}_{3} \smallsetminus \left( {{A}_{1} \cup {A}_{2}}\right) \), and so on. (Try to remember this little trick.) These sets are in \( \mathcal{A} \) by Lemma 1 in the preceding section. Also, \( {B}_{i} \subset {A}_{i} \) and\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{B}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {B}_{i}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \]
|
Yes
|
Lemma 1. If \( \left( {X,\mathcal{A},\mu }\right) \) is a measure space, then:\n\n(1) \( \mu \left( A\right) \leq \mu \left( B\right) \) if \( A \in \mathcal{A}, B \in \mathcal{A} \), and \( A \subset B \) .\n\n(2) \( \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \) if \( {A}_{i} \in \mathcal{A} \) .
|
Proof. For (1), write\n\n\[ \mu \left( B\right) = \mu \left\lbrack {A \cup \left( {B \smallsetminus A}\right) }\right\rbrack = \mu \left( A\right) + \mu \left( {B \smallsetminus A}\right) \geq \mu \left( A\right) \]\n\nFor (2), we create a disjoint sequence of sets \( {B}_{i} \) by writing \( {B}_{1} = {A}_{1},{B}_{2} = \) \( {A}_{2} \smallsetminus {A}_{1},{B}_{3} = {A}_{3} \smallsetminus \left( {{A}_{1} \cup {A}_{2}}\right) \), and so on. (Try to remember this little trick.) These sets are in \( \mathcal{A} \) by Lemma 1 in the preceding section. Also, \( {B}_{i} \subset {A}_{i} \) and\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{B}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {B}_{i}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \]
|
Yes
|
Lemma 1. If \( \left( {X,\mathcal{A},\mu }\right) \) is a measure space, then:\n\n(1) \( \mu \left( A\right) \leq \mu \left( B\right) \) if \( A \in \mathcal{A}, B \in \mathcal{A} \), and \( A \subset B \) .\n\n(2) \( \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \) if \( {A}_{i} \in \mathcal{A} \) .
|
Proof. For (1), write\n\n\[ \mu \left( B\right) = \mu \left\lbrack {A \cup \left( {B \smallsetminus A}\right) }\right\rbrack = \mu \left( A\right) + \mu \left( {B \smallsetminus A}\right) \geq \mu \left( A\right) \]\n\nFor (2), we create a disjoint sequence of sets \( {B}_{i} \) by writing \( {B}_{1} = {A}_{1},{B}_{2} = \) \( {A}_{2} \smallsetminus {A}_{1},{B}_{3} = {A}_{3} \smallsetminus \left( {{A}_{1} \cup {A}_{2}}\right) \), and so on. (Try to remember this little trick.) These sets are in \( \mathcal{A} \) by Lemma 1 in the preceding section. Also, \( {B}_{i} \subset {A}_{i} \) and\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{B}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {B}_{i}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \]
|
Yes
|
Lemma 1. If \( \left( {X,\mathcal{A},\mu }\right) \) is a measure space, then:\n\n(1) \( \mu \left( A\right) \leq \mu \left( B\right) \) if \( A \in \mathcal{A}, B \in \mathcal{A} \), and \( A \subset B \) .\n\n(2) \( \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \) if \( {A}_{i} \in \mathcal{A} \) .
|
Proof. For (1), write\n\n\[ \mu \left( B\right) = \mu \left\lbrack {A \cup \left( {B \smallsetminus A}\right) }\right\rbrack = \mu \left( A\right) + \mu \left( {B \smallsetminus A}\right) \geq \mu \left( A\right) \]\n\nFor (2), we create a disjoint sequence of sets \( {B}_{i} \) by writing \( {B}_{1} = {A}_{1},{B}_{2} = \) \( {A}_{2} \smallsetminus {A}_{1},{B}_{3} = {A}_{3} \smallsetminus \left( {{A}_{1} \cup {A}_{2}}\right) \), and so on. (Try to remember this little trick.) These sets are in \( \mathcal{A} \) by Lemma 1 in the preceding section. Also, \( {B}_{i} \subset {A}_{i} \) and\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{B}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {B}_{i}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \]
|
Yes
|
Lemma 1. If \( \left( {X,\mathcal{A},\mu }\right) \) is a measure space, then:\n\n(1) \( \mu \left( A\right) \leq \mu \left( B\right) \) if \( A \in \mathcal{A}, B \in \mathcal{A} \), and \( A \subset B \) .\n\n(2) \( \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \) if \( {A}_{i} \in \mathcal{A} \) .
|
Proof. For (1), write\n\n\[ \mu \left( B\right) = \mu \left\lbrack {A \cup \left( {B \smallsetminus A}\right) }\right\rbrack = \mu \left( A\right) + \mu \left( {B \smallsetminus A}\right) \geq \mu \left( A\right) \]\n\nFor (2), we create a disjoint sequence of sets \( {B}_{i} \) by writing \( {B}_{1} = {A}_{1},{B}_{2} = \) \( {A}_{2} \smallsetminus {A}_{1},{B}_{3} = {A}_{3} \smallsetminus \left( {{A}_{1} \cup {A}_{2}}\right) \), and so on. (Try to remember this little trick.) These sets are in \( \mathcal{A} \) by Lemma 1 in the preceding section. Also, \( {B}_{i} \subset {A}_{i} \) and\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{B}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {B}_{i}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \]
|
Yes
|
Lemma 1. If \( \\left( {X,\\mathcal{A},\\mu }\\right) \) is a measure space, then:\n\n(1) \( \\mu \\left( A\\right) \\leq \\mu \\left( B\\right) \) if \( A \\in \\mathcal{A}, B \\in \\mathcal{A} \), and \( A \\subset B \) .\n\n(2) \( \\mu \\left( {\\mathop{\\bigcup }\\limits_{{i = 1}}^{\\infty }{A}_{i}}\\right) \\leq \\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }\\mu \\left( {A}_{i}\\right) \) if \( {A}_{i} \\in \\mathcal{A} \) .
|
Proof. For (1), write\n\n\[ \n\\mu \\left( B\\right) = \\mu \\left\\lbrack {A \\cup \\left( {B \\smallsetminus A}\\right) }\\right\\rbrack = \\mu \\left( A\\right) + \\mu \\left( {B \\smallsetminus A}\\right) \\geq \\mu \\left( A\\right) \n\]\n\nFor (2), we create a disjoint sequence of sets \( {B}_{i} \) by writing \( {B}_{1} = {A}_{1},{B}_{2} = \) \( {A}_{2} \\smallsetminus {A}_{1},{B}_{3} = {A}_{3} \\smallsetminus \\left( {{A}_{1} \\cup {A}_{2}}\\right) \), and so on. (Try to remember this little trick.) These sets are in \( \\mathcal{A} \) by Lemma 1 in the preceding section. Also, \( {B}_{i} \\subset {A}_{i} \) and\n\n\[ \n\\mu \\left( {\\mathop{\\bigcup }\\limits_{{i = 1}}^{\\infty }{A}_{i}}\\right) = \\mu \\left( {\\mathop{\\bigcup }\\limits_{{i = 1}}^{\\infty }{B}_{i}}\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }\\mu \\left( {B}_{i}\\right) \\leq \\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }\\mu \\left( {A}_{i}\\right) \n\]
|
Yes
|
Lemma 2. Let \( \left( {X,\mathcal{A},\mu }\right) \) be a measure space. If \( {A}_{i} \in \mathcal{A} \) for \( i \in \mathbb{N} \) and \( {A}_{1} \subset {A}_{2} \subset \cdots \), then \( \mu \left( {A}_{n}\right) \uparrow \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \) .
|
Proof. Let \( {A}_{0} = \varnothing \) and observe that \( {A}_{n} = \mathop{\bigcup }\limits_{{i = 1}}^{n}\left( {{A}_{i} \smallsetminus {A}_{i - 1}}\right) \) . It follows from the disjoint nature of the sets \( {A}_{i} \smallsetminus {A}_{i - 1} \) that \( \mu \left( {A}_{n}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\mu \left( {{A}_{i} \smallsetminus {A}_{i - 1}}\right) \) . Hence\n\n\[ \mu \left( {A}_{n}\right) \uparrow \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {{A}_{i} \smallsetminus {A}_{i - 1}}\right) = \mu \left\lbrack {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{A}_{i} \smallsetminus {A}_{i - 1}}\right) }\right\rbrack = \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \]
|
Yes
|
Theorem 2. Let \( \left( {X,\mathcal{A},\mu }\right) \) be a measure space. For \( S \in {2}^{X} \) define\n\n\[ \n{\mu }^{ * }\left( S\right) = \inf \{ \mu \left( A\right) : S \subset A \in \mathcal{A}\}\n\]\n\nThen \( {\mu }^{ * } \) is an outer measure whose restriction to \( \mathcal{A} \) is \( \mu \) . Furthermore, each set in \( \mathcal{A} \) is \( \mu \) -measurable.
|
Proof. I. Since \( \mu \) is nonnegative, so is \( {\mu }^{ * } \) . Since \( \varnothing \in \mathcal{A} \), we have \( 0 \leq {\mu }^{ * }\left( \varnothing \right) \leq \) \( \mu \left( \varnothing \right) = 0 \).\n\nII. If \( S \subset T \), then \( \{ A : S \subset A \in \mathcal{A}\} \) contains \( \{ A : T \subset A \in \mathcal{A}\} \) . Hence\n\n\[ \n{\mu }^{ * }\left( S\right) = \inf \{ \mu \left( A\right) : S \subset A \in \mathcal{A}\} \leq \inf \{ \mu \left( A\right) : T \subset A \in \mathcal{A}\} = {\mu }^{ * }\left( T\right)\n\]\n\nIII. If \( {S}_{i} \in {2}^{X} \) and \( \varepsilon > 0 \), select \( {A}_{i} \in \mathcal{A} \) so that \( {S}_{i} \subset {A}_{i} \) and \( {\mu }^{ * }\left( {S}_{i}\right) \geq \) \( \mu \left( {A}_{i}\right) - \varepsilon /{2}^{i} \) . Then \( \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{S}_{i} \subset \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i} \in \mathcal{A} \) . Consequently,\n\n\[ \n{\mu }^{ * }\left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{S}_{i}}\right) \leq \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\left\lbrack {{\mu }^{ * }\left( {S}_{i}\right) + \frac{\varepsilon }{{2}^{i}}}\right\rbrack\n\]\n\nSince \( \varepsilon \) can be any positive number, \( {\mu }^{ * }\left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{S}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {S}_{i}\right) \).\n\nIV. If \( S \in \mathcal{A} \) and \( S \subset A \in \mathcal{A} \), then \( {\mu }^{ * }\left( S\right) \leq \mu \left( S\right) \leq \mu \left( A\right) \) . Taking an infimum for all choices of \( A \), we get \( {\mu }^{ * }\left( S\right) \leq \mu \left( S\right) \leq {\mu }^{ * }\left( S\right) \) . This proves that \( {\mu }^{ * } \) is an extension of \( \mu \).\n\nV. To prove that each \( A \) in \( \mathcal{A} \) is \( {\mu }^{ * } \) -measurable, let \( S \) be any subset of \( X \) . Given \( \varepsilon > 0 \), we find \( B \in \mathcal{A} \) such that \( {\mu }^{ * }\left( S\right) \geq \mu \left( B\right) - \varepsilon \) and \( S \subset B \) . Then\n\n\[ \n{\mu }^{ * }\left( S\right) + \varepsilon \geq \mu \left( B\right) = \mu \left( {B \cap A}\right) + \mu \left( {B \smallsetminus A}\right) = {\mu }^{ * }\left( {B \cap A}\right) + {\mu }^{ * }\left( {B \smallsetminus A}\right)\n\]\n\n\[ \n\geq {\mu }^{ * }\left( {S \cap A}\right) + {\mu }^{ * }\left( {S \smallsetminus A}\right) \geq {\mu }^{ * }\left( S\right)\n\]\n\nThis calculation used Parts III and IV of the present proof. Since \( \varepsilon \) was arbitrary, \( {\mu }^{ * }\left( S\right) = {\mu }^{ * }\left( {S \cap A}\right) + {\mu }^{ * }\left( {S \smallsetminus A}\right) \) for all \( S \) . Hence \( A \) is \( {\mu }^{ * } \) -measurable.
|
Yes
|
Theorem 3. Under the same hypotheses as in Theorem 2, the outer measure \( {\mu }^{ * } \) is regular.
|
Proof. Let \( S \) be any subset of \( X \) . For each \( n \in \mathbb{N} \) select \( {A}_{n} \in \mathcal{A} \) so that \( S \subset {A}_{n} \) and \( {\mu }^{ * }\left( S\right) \geq \mu \left( {A}_{n}\right) - 1/n \) . Put \( A = \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{A}_{n} \) . Since \( \mathcal{A} \) is a \( \sigma \) -algebra, \( A \in \mathcal{A} \) . (See Lemma 1 in the preceding section, page 384.) From the inclusion \( S \subset A \subset {A}_{n} \) we get\n\n\[ \n{\mu }^{ * }\left( S\right) \leq {\mu }^{ * }\left( A\right) = \mu \left( A\right) \leq \mu \left( {A}_{n}\right) < {\mu }^{ * }\left( S\right) + 1/n \n\]\n\nSince this is true for all \( n,{\mu }^{ * }\left( S\right) = {\mu }^{ * }\left( A\right) \) . By the preceding theorem, \( A \) is \( {\mu }^{ * } \) -measurable.
|
Yes
|
Theorem 1. The Lebesgue outer measure of an interval is its length.
|
Proof. Consider first a compact interval \( \left\lbrack {a, b}\right\rbrack \) . Since \( \left\lbrack {a, b}\right\rbrack \subset \left( {a - \varepsilon, b + \varepsilon }\right) \), we conclude from the definition (1) that \( \mu \left( \left\lbrack {a, b}\right\rbrack \right) \leq b - a + {2\varepsilon } \) for every positive \( \varepsilon \) . Hence \( \mu \left( \left\lbrack {a, b}\right\rbrack \right) \leq b - a \) . Suppose now that \( \mu \left( \left\lbrack {a, b}\right\rbrack \right) < b - a \) . Find intervals \( \left( {{a}_{i},{b}_{i}}\right) \) such that \( \left\lbrack {a, b}\right\rbrack \subset \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{a}_{i},{b}_{i}}\right) \) and \( \mathop{\sum }\limits_{{i = 1}}^{\infty }\left| {{b}_{i} - {a}_{i}}\right| < b - a \) . We can assume \( {a}_{i} < {b}_{i} \) for all \( i \) . By compactness and renumbering we can get \( \left\lbrack {a, b}\right\rbrack \subset \mathop{\bigcup }\limits_{{i = 1}}^{n}\left( {{a}_{i},{b}_{i}}\right) \) . It follows that \( \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{b}_{i} - {a}_{i}}\right) < b - a \) . By renumbering again we can assume \( a \in \left( {{a}_{1},{b}_{1}}\right) ,{b}_{1} \in \left( {{a}_{2},{b}_{2}}\right) ,{b}_{2} \in \left( {{a}_{3},{b}_{3}}\right) \), and so on. There must exist an index \( k \leq n \) such that \( b < {b}_{k} \) . Then we reach a contradiction:\n\n\[ b - a > \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{b}_{i} - {a}_{i}}\right) \geq \mathop{\sum }\limits_{{i = 1}}^{k}\left( {{b}_{i} - {a}_{i}}\right) = {b}_{k} - {a}_{1} + \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}\left( {{b}_{i} - {a}_{i + 1}}\right) > {b}_{k} - {a}_{1} > b - a \]\n\nIf \( J \) is a bounded interval of the type \( \left( {a, b}\right) ,(a, b\rbrack \), or \( \lbrack a, b) \), then from the inclusions\n\n\[ \left\lbrack {a + \varepsilon, b - \varepsilon }\right\rbrack \subset J \subset \left\lbrack {a - \varepsilon, b + \varepsilon }\right\rbrack \]\n\nwe obtain \( b - a - {2\varepsilon } \leq \mu \left( J\right) \leq b - a + {2\varepsilon } \) and \( \mu \left( J\right) = b - a \) .\n\nFinally, if \( J \) is an unbounded interval, then it contains intervals \( \left\lbrack {a, b}\right\rbrack \) of arbitrarily great length. Hence \( \mu \left( J\right) = \infty \) .
|
Yes
|
Theorem 2. Every Borel set in \( \mathbb{R} \) is Lebesgue measurable.
|
Proof. (S.J. Bernau) The family of Borel sets is the smallest \( \sigma \) -algebra containing all the open sets. The Lebesgue measurable sets form a \( \sigma \) -algebra. Hence it suffices to prove that every open set is Lebesgue measurable.\n\nRecall that every open set in \( \mathbb{R} \) can be expressed as a countable union of open intervals \( \left( {a, b}\right) \) . Thus it suffices to prove that each interval \( \left( {a, b}\right) \) is Lebesgue measurable. We begin with an interval of the form \( \left( {a,\infty }\right) \), where \( a \in \mathbb{R} \) .\n\nTo prove that the open interval \( \left( {a,\infty }\right) \) is measurable, we must prove, for any set \( S \) in \( \mathbb{R} \), that\n\n(2)\n\n\[ \mu \left( S\right) \geq \mu \left\lbrack {S \cap \left( {a,\infty }\right) }\right\rbrack + \mu \left\lbrack {S \smallsetminus \left( {a,\infty }\right) }\right\rbrack \]\n\nLet us use the notation \( \left| I\right| \) for the length of an interval \( I \) . Given \( \varepsilon > 0 \), select open intervals \( {I}_{n} \) such that \( S \subset \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{I}_{n} \) and \( \sum \left| {I}_{n}\right| < \mu \left( S\right) + \varepsilon \) . Define \( {J}_{n} = \) \( {I}_{n} \cap \left( {a,\infty }\right) ,{K}_{n} = {I}_{n} \cap \left( {-\infty, a}\right) \), and \( {K}_{0} = \left( {a - \varepsilon, a + \varepsilon }\right) \) . Then we have\n\n\[ S \cap \left( {a,\infty }\right) \subset \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{J}_{n} \]\n\n\[ S \smallsetminus \left( {a,\infty }\right) = S \cap ( - \infty, a\rbrack \subset \mathop{\bigcup }\limits_{{n = 0}}^{\infty }{K}_{n} \]\n\n\[ {J}_{n} \cup {K}_{n} \subset {I}_{n}\;\text{ and }\;{J}_{n} \cap {K}_{n} = \varnothing \]\n\nConsequently,\n\n\[ \mu \left\lbrack {S \cap \left( {a,\infty }\right) }\right\rbrack + \mu \left\lbrack {S \smallsetminus \left( {a,\infty }\right) }\right\rbrack \leq \mathop{\sum }\limits_{{n = 1}}^{\infty }\left\{ {\left| {J}_{n}\right| + \left| {K}_{n}\right| }\right\} + \left| {K}_{0}\right| \]\n\n\[ \leq \mathop{\sum }\limits_{{n = 1}}^{\infty }\left| {I}_{n}\right| + {2\varepsilon } < \mu \left( S\right) + {3\varepsilon } \]\n\nBecause \( \varepsilon \) was arbitrary, this establishes Equation (2). Since the measurable sets make up a \( \sigma \) -algebra, each set of the form \( ( - \infty, b\rbrack = \mathbb{R} \smallsetminus \left( {b,\infty }\right) \) is measurable. Hence the set \( \left( {-\infty, b}\right) = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }( - \infty, b - \frac{1}{n}\rbrack \) is measurable and so is \( \left( {a, b}\right) = \) \( \left( {-\infty, b}\right) \cap \left( {a,\infty }\right) \) .
|
Yes
|
Theorem 3. Lebesgue outer measure is invariant on the group \( \left( {\mathbb{R}, + }\right) \) .
|
Proof. The statement means that \( \mu \left( S\right) = \mu \left( {v + S}\right) \) for all \( S \in {2}^{\mathbb{R}} \) and all \( v \in \mathbb{R} \) . The translate \( v + S \) is defined to be \( \{ v + x : x \in S\} \) . Notice that the condition \( S \subset \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{a}_{i},{b}_{i}}\right) \) is equivalent to the condition \( x + S \subset \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {x + {a}_{i}, x + {b}_{i}}\right) \) . Since the length of \( \left( {x + {a}_{i}, x + {b}_{i}}\right) \) is the same as the length of \( \left( {{a}_{i},{b}_{i}}\right) \), the definition of \( \mu \) gives equal values for \( \mu \left( S\right) \) and \( \mu \left( {x + S}\right) \) .
|
Yes
|
Theorem 4. There exists no translation-invariant measure \( \nu \) defined on \( {2}^{\mathbb{R}} \) such that \( 0 < \nu \left( \left\lbrack {0,1}\right\rbrack \right) < \infty \) . Consequently, there exist subsets of \( \mathbb{R} \) that are not Lebesgue measurable.
|
Proof. The second assertion follows from the first because if every set of reals were Lebesgue measurable, then Lebesgue measure would contradict the first assertion.\n\nTo prove the first assertion, suppose that a measure \( \nu \) exists as described. By the preceding lemma, the set \( P \) given there has the property\n\n\[ \left\lbrack {0,1}\right\rbrack \subset \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{r}_{i} + P}\right) \subset \left\lbrack {-1,2}\right\rbrack \]\n\nAlso by the lemma, the sequence of sets \( {r}_{i} + P \) is disjoint. Consequently,\n\n\[ 0 < \nu \left( \left\lbrack {0,1}\right\rbrack \right) \leq \nu \left\lbrack {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{r}_{i} + P}\right) }\right\rbrack = \mathop{\sum }\limits_{{i = 1}}^{\infty }\nu \left( {{r}_{i} + P}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\nu \left( P\right) \]\n\nTherefore, \( \nu \left( P\right) > 0,\sum \nu \left( P\right) = \infty \), and we have the contradiction\n\n\[ \infty = \nu \left\lbrack {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{r}_{i} + P}\right) }\right\rbrack \leq \nu \left( \left\lbrack {-1,2}\right\rbrack \right) \leq {3\nu }\left( \left\lbrack {0,1}\right\rbrack \right) \]
|
Yes
|
Theorem 1. Let \( \left( {X,\mathcal{A}}\right) \) be a measurable space. A function \( f \) from \( X \) to the extended reals \( {\mathbb{R}}^{ * } \) is measurable if it has any one of the following properties:\na. \( {f}^{-1}(\left( {a,\infty \rbrack }\right) \in \mathcal{A} \) for each \( a \in {\mathbb{R}}^{ * }\)\nb. \( {f}^{-1}\left( \left\lbrack {a,\infty }\right\rbrack \right) \in \mathcal{A} \) for each \( a \in {\mathbb{R}}^{ * }\)\nc. \( {f}^{-1}\left( {\lbrack - \infty, a}\right) ) \in \mathcal{A} \) for each \( a \in {\mathbb{R}}^{ * }\)\nd. \( {f}^{-1}\left( \left\lbrack {-\infty, a}\right\rbrack \right) \in \mathcal{A} \) for each \( a \in {\mathbb{R}}^{ * }\)\ne. \( {f}^{-1}\left( \left( {a, b}\right) \right) \in \mathcal{A} \) for all \( a \) and \( b \) in \( {\mathbb{R}}^{ * }\)\nf. \( {f}^{-1}\left( O\right) \in \mathcal{A} \) for each open set \( O \) in \( {\mathbb{R}}^{ * } \)
|
Proof. We shall prove that each condition implies the one following it, and that \( \mathbf{f} \) implies that \( f \) is measurable. That \( \mathbf{a} \) implies \( \mathbf{b} \) follows from the equation \( {f}^{-1}\left( \left\lbrack {a,\infty }\right\rbrack \right) = \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{f}^{-1}\left( \left( {a - \frac{1}{n},\infty }\right\rbrack \right) \) and from the properties of a \( \sigma \) -algebra. That \( \mathbf{b} \) implies \( \mathbf{c} \) follows in the same manner from the equation \( {f}^{-1}\left( {\lbrack - \infty, a}\right) ) = \) \( X \smallsetminus {f}^{-1}\left( \left\lbrack {a,\infty }\right\rbrack \right) \) . That \( \mathbf{c} \) implies \( \mathbf{d} \) follows from the equation \( {f}^{-1}\left( \left\lbrack {-\infty, a}\right\rbrack \right) = \) \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{f}^{-1}\left( {\lbrack - \infty, a + \frac{1}{n}}\right) ) \) . That \( \mathbf{d} \) implies \( \mathbf{e} \) follows from writing \( {f}^{-1}\left( \left( {a, b}\right) \right) = \) \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{f}^{-1}\left( {\lbrack - \infty, b - \frac{1}{n}}\right) ) \smallsetminus {f}^{-1}\left( \left\lbrack {-\infty, a}\right\rbrack \right) \) . That e implies \( \mathbf{f} \) is a consequence of the theorem that each open set in \( {\mathbb{R}}^{ * } \) is a countable union of intervals of the form \( \left( {a, b}\right) \), where \( a \) and \( b \) are in \( {\mathbb{R}}^{ * } \) . To complete the proof, assume condition f. Let \( \mathcal{S} \) be the family of all sets \( S \) contained in \( {\mathbb{R}}^{ * } \) such that \( {f}^{-1}\left( S\right) \in \mathcal{A} \) . It is straightforward to verify that \( \mathcal{S} \) is a \( \sigma \) -algebra. By hypothesis, each open set in \( {\mathbb{R}}^{ * } \) belongs to \( \mathcal{S} \) . Hence \( \mathcal{S} \) contains the \( \sigma \) -algebra of Borel sets. Consequently, \( {f}^{-1}\left( B\right) \in \mathcal{A} \) for each Borel set \( B \), and \( f \) is measurable.
|
Yes
|
Theorem 3. Let \( \\left( {X,\\mathcal{A},\\mu }\\right) \) be a complete measure space, as defined in Section 8.2, page 387. If \( f \) is a measurable function and if \( f\\left( x\\right) = g\\left( x\\right) \) almost everywhere, then \( g \) is measurable.
|
Proof. Define \( A = \\{ x : f\\left( x\\right) \\neq g\\left( x\\right) \\} \) . Then \( A \) is measurable and \( \\mu \\left( A\\right) = 0 \) . Also, \( X \\smallsetminus A \) is measurable. For \( a \\in {\\mathbb{R}}^{ * } \) we write\n\n\[ \n{g}^{-1}(\\left( {a,\\infty \\rbrack }\\right) = \\left\\{ {{g}^{-1}(\\left( {a,\\infty \\rbrack }\\right) \\cap \\left( {X \\smallsetminus A}\\right) }\\right\\} \\cup \\left\\{ {{g}^{-1}(\\left( {a,\\infty \\rbrack }\\right) \\cap A}\\right\\} \n\]\n\nOn the right side of this equation we see the union of two sets. The first of these is measurable because it is \( {f}^{-1}\\left( \\left( {a,\\infty }\\right) \\right) \\smallsetminus A \) . The second set is measurable because it is a subset of a set of measure 0 , and the measure space is complete.
|
Yes
|
Theorem 4 (Egorov’s Theorem). Let \( \left( {X,\mathcal{A},\mu }\right) \) be a measure space such that \( \mu \left( X\right) < \infty \) . For a sequence of finite-valued measurable functions \( f,{f}_{1},{f}_{2},\ldots \) these properties are equivalent:\n\na. \( {f}_{n} \rightarrow f \) almost everywhere\n\nb. \( {f}_{n} \rightarrow f \) almost uniformly
|
Proof. Assume that \( \mathbf{b} \) is true. For each \( m \) in \( \mathbb{N} \) there is a measurable set \( {A}_{m} \) such that \( \mu \left( {A}_{m}\right) < 1/m \), and on \( X \smallsetminus {A}_{m},{f}_{n}\left( x\right) \rightarrow f\left( x\right) \) uniformly. Define \( A = \mathop{\bigcap }\limits_{{m = 1}}^{\infty }{A}_{m} \) . Then \( \mu \left( A\right) = 0 \) because \( A \subset {A}_{m} \) for all \( m \) . Also, \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) on \( X \smallsetminus A \) because \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) for \( x \in X \smallsetminus {A}_{m} \) and \( X \smallsetminus A = X \smallsetminus \bigcap {A}_{m} = \) \( \bigcup \left( {X \smallsetminus {A}_{m}}\right) \) . Thus \( \mathbf{a} \) is true.\n\nNow assume that \( \mathbf{a} \) is true. Let \( {g}_{n} = f - {f}_{n} \) . By altering \( {g}_{n} \) on a set of measure 0, we can assume that \( {g}_{n}\left( x\right) \rightarrow 0 \) everywhere. Next, we define \( {A}_{n}^{m} = \left\{ {x : \left| {{g}_{i}\left( x\right) }\right| \leq 1/m\text{for}i \geq n}\right\} \) . Thus \( {A}_{1}^{m} \subset {A}_{2}^{m} \subset \cdots \) For each \( x \) there is an index \( n \) such that \( x \in {A}_{n}^{m} \) ; in other words, \( x \in \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{A}_{n}^{m} \) and \( X \subset \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{A}_{n}^{m} \) . Since \( X \) has finite measure, \( \mu \left( {X \smallsetminus {A}_{n}^{m}}\right) \rightarrow 0 \) as \( n \rightarrow \infty \) . (See Lemma 2 in Section 8.2, page 387.) Let \( \varepsilon > 0 \) . For each \( m \), let \( {n}_{m} \) be an integer such that \( \mu \left( {X \smallsetminus {A}_{{n}_{m}}^{m}}\right) < \varepsilon /{2}^{m} \) . Define \( A = \mathop{\bigcap }\limits_{{m = 1}}^{\infty }{A}_{{n}_{m}}^{m} \) . Then\n\n\[ \mu \left( {X \smallsetminus A}\right) = \mu \left( {X \smallsetminus \mathop{\bigcap }\limits_{{m = 1}}^{\infty }{A}_{{n}_{m}}^{m}}\right) = \mu \left( {\mathop{\bigcup }\limits_{{m = 1}}^{\infty }\left( {X \smallsetminus {A}_{{n}_{m}}^{m}}\right) }\right) \leq \mathop{\sum }\limits_{{m = 1}}^{\infty }\mu \left( {X \smallsetminus {A}_{{m}_{n}}^{m}}\right) < \varepsilon \]\n\nOn \( A,{g}_{i}\left( x\right) \rightarrow 0 \) uniformly. Indeed, for \( x \in A \) we have (for all \( m \) )\n\n\[ i \geq {n}_{m} \Rightarrow \left| {{g}_{i}\left( x\right) }\right| \leq \frac{1}{m} \]
|
Yes
|
Theorem 5. Let \( \\left( {X,\\mathcal{A}}\\right) \) be a measurable space, and \( f \) any nonnegative measurable function. Then there exists a sequence of nonnegative simple functions \( {g}_{n} \) such that \( {g}_{n}\\left( x\\right) \\uparrow f\\left( x\\right) \) for each \( x \) . If \( f \) is bounded, this sequence can be constructed so that \( {g}_{n} \\uparrow f \) uniformly.
|
Proof. ([HewS], page 159.) Define\n\n\[ \n{A}_{i}^{n} = \\left\\{ {x \\in X : \\frac{i}{{2}^{n}} \\leq f\\left( x\\right) < \\frac{i + 1}{{2}^{n}}}\\right\\} \\;\\left( {0 \\leq i < n{2}^{n}}\\right) \n\]\n\n\[ \n{B}^{n} = \\{ x \\in X : f\\left( x\\right) \\geq n\\} \n\]\n\n\[ \n{g}_{n} = \\mathop{\\sum }\\limits_{i}\\frac{i}{{2}^{n}}{x}_{{A}_{i}^{n}} + n{x}_{{B}^{n}} \n\]\n\nThe sets \( {A}_{i}^{n} \) and \( {B}^{n} \) are measurable, by Theorem 1. Hence \( {g}_{n} \) is a simple function. The definition of \( {g}_{n} \) shows directly that \( {g}_{n} \\leq f \) . In order to verify that \( {g}_{n}\\left( x\\right) \) converges to \( f\\left( x\\right) \) for each \( x \), consider first the case when \( f\\left( x\\right) \\neq \\infty \) . For large \( n \) and a suitable \( i, x \\in {A}_{i}^{n} \) . Then \( f\\left( x\\right) - {g}_{n}\\left( x\\right) < \\frac{i + 1}{{2}^{n}} - \\frac{i}{{2}^{n}} = \\frac{1}{{2}^{n}} \) . On the other hand, if \( f\\left( x\\right) = + \\infty \), then \( {g}_{n}\\left( x\\right) = n \\rightarrow f\\left( x\\right) \) .\n\nFor the monotonicity of \( {g}_{n}\\left( x\\right) \) as a function of \( n \) for \( x \) fixed, first verify (Problem 3) that (for \( i < n{2}^{n} \) ) \( {A}_{i}^{n} = {A}_{2i}^{n + 1} \\cup {A}_{{2i} + 1}^{n + 1} \) . If \( x \\in {A}_{2i}^{n + 1} \) , then \( {g}_{n + 1}\\left( x\\right) = {2i}/{2}^{n + 1} = i/{2}^{n} = {g}_{n}\\left( x\\right) \) . If \( x \\in {A}_{{2i} + 1}^{n + 1} \), then \( {g}_{n + 1}\\left( x\\right) = \) \( \\left( {{2i} + 1}\\right) /{2}^{n + 1} \\geq {2i}/{2}^{n + 1} = {g}_{n}\\left( x\\right) \) . If \( x \\in {B}^{n} \), then \( f\\left( x\\right) \\geq n \), and therefore \( x \\in \\mathop{\\bigcup }\\limits_{{i \\geq n{2}^{n + 1}}}{A}_{i}^{n + 1} \\cup {B}^{n + 1} \) . It follows that \( {g}_{n + 1}\\left( x\\right) \\geq n = {g}_{n}\\left( x\\right) \) .\n\nFinally, if \( f \) is bounded by \( m \), then for \( n \\geq m \) we have \( 0 \\leq f\\left( x\\right) - {g}_{n}\\left( x\\right) \\leq \) \( {2}^{-n} \) . In this case the convergence is uniform.
|
Yes
|
Lemma 1. Let \( f = \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{x}_{{A}_{i}} \), where we assume only that the sets \( {A}_{i} \) are mutually disjoint measurable sets. Then \( \int f = \) \( \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}\mu \left( {A}_{i}\right) . \)
|
Proof. The function \( f \) is simple, and its range contains at most \( n \) elements. Let \( \left\{ {{\beta }_{1},\ldots ,{\beta }_{k}}\right\} \) be the range of \( f \), and let \( {B}_{i} = {f}^{-1}\left( \left\{ {\beta }_{i}\right\} \right) \) . Then \( f = \mathop{\sum }\limits_{{i = 1}}^{k}{\beta }_{i}{x}_{{B}_{i}} \) , and this representation is canonical; i.e., it conforms to the requirements of Equation (2). Putting \( {J}_{i} = \left\{ {j : {\alpha }_{j} = {\beta }_{i}}\right\} \), we have\n\n\[ \int f = \mathop{\sum }\limits_{{i = 1}}^{k}{\beta }_{i}\mu \left( {B}_{i}\right) = \mathop{\sum }\limits_{{i = 1}}^{k}{\beta }_{i}\mu \left( {\mathop{\bigcup }\limits_{{j \in {J}_{i}}}{A}_{j}}\right) = \mathop{\sum }\limits_{{i = 1}}^{k}\mathop{\sum }\limits_{{j \in {J}_{i}}}{\beta }_{i}\mu \left( {A}_{j}\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 1}}^{k}\mathop{\sum }\limits_{{j \in {J}_{i}}}{\alpha }_{j}\mu \left( {A}_{j}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{\alpha }_{j}\mu \left( {A}_{j}\right) \]
|
Yes
|
Lemma 2. If \( g \) and \( f \) are simple functions such that \( g \leq f \), then \( \int g \leq \int f \) .
|
Proof. Start with canonical representations, as described following Equation (2):\n\n\[ g = \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{x}_{{A}_{i}}\;f = \mathop{\sum }\limits_{{j = 1}}^{k}{\beta }_{j}{x}_{{B}_{j}} \]\n\nThen we have (non-canonical) representations conforming to Lemma 1:\n\n\[ g = \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{j = 1}}^{k}{\alpha }_{i}{X}_{{A}_{i} \cap {B}_{j}}\;f = \mathop{\sum }\limits_{{j = 1}}^{k}\mathop{\sum }\limits_{{i = 1}}^{n}{\beta }_{j}{X}_{{A}_{i} \cap {B}_{j}} \]\n\nSince \( g \leq f \), we have \( {\alpha }_{i} \leq {\beta }_{j} \) whenever \( {A}_{i} \cap {B}_{j} \neq \varnothing \) . By Lemma 1\n\n\[ \int g = \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{j = 1}}^{k}{\alpha }_{i}\mu \left( {{A}_{i} \cap {B}_{j}}\right) \;\int f = \mathop{\sum }\limits_{{j = 1}}^{k}\mathop{\sum }\limits_{{i = 1}}^{n}{\beta }_{j}\mu \left( {{A}_{j} \cap {B}_{j}}\right) \]\n\nHence\n\n\[ \int f - \int g = \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{j = 1}}^{k}\left( {{\beta }_{j} - {\alpha }_{i}}\right) \mu \left( {{A}_{i} \cap {B}_{j}}\right) \geq 0 \]
|
Yes
|
Lemma 3. If \( f \) is a nonnegative simple function, then its integral as given in Equation (2) equals its integral as given in Equation (3).
|
Proof. Since \( f \) itself is simple, the expression on the right of Equation (3) is at least \( \int f \) . On the other hand, if \( g \) is simple and if \( g \leq f \), then by Lemma 2, \( \int g \leq \int f \) . By taking a supremum, we see that the right side of Equation (3) is at most \( \int f \) .
|
Yes
|
Lemma 4. If \( f \) and \( g \) are nonnegative simple functions, then \( f\left( {f + g}\right) = \int f + \int g \) .
|
Proof. Proceed exactly as in the proof of Lemma 2. Then\n\n\[ g + f = \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{j = 1}}^{k}\left( {{\alpha }_{i} + {\beta }_{j}}\right) {X}_{{A}_{i} \cap {B}_{j}} \]\n\nBy the disjoint nature of the family \( \left\{ {{A}_{i} \cap {B}_{j}}\right\} \) we have\n\n\[ \int \left( {g + f}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{j = 1}}^{k}\left( {{\alpha }_{i} + {\beta }_{j}}\right) \mu \left( {{A}_{i} \cap {B}_{j}}\right) \]\n\nThis is the same as \( \int g + \int f \), as we see from an equation in the proof of Lemma 2.
|
Yes
|
Lemma 5. For two measurable functions \( f \) and \( g \), the condition \( 0 \leq f \leq g \) implies \( 0 \leq \int f \leq \int g \) .
|
Proof. Since 0 is a simple function, the definition of \( \int f \) in Equation (3) gives \( \int f \geq \int 0 = 0 \) . If \( h \) is a simple function such that \( h \leq f \), then \( h \leq g \) and \( \int h \leq \int g \) by the definition of \( \int g \) . In this last inequality, take the supremum in \( h \) to get \( \int f \leq \int g \) .
|
Yes
|
Theorem 1. Monotone Convergence Theorem. is a sequence of measurable functions such that \( 0 \leq {f}_{n} \uparrow f \), then \( 0 \leq \int {f}_{n} \uparrow \int f \)
|
Proof. (Rudin) Since \( 0 \leq {f}_{n} \leq {f}_{n + 1} \leq f \), we have \( 0 \leq \int {f}_{n} \leq \int {f}_{n + 1} \leq \int f \) by Lemma 5. Hence \( \lim \int {f}_{n} \) exists and is no greater than \( \int f \) . For the reverse inequality, let \( 0 < \theta < 1 \) and let \( g \) be a simple function satisfying \( 0 \leq g \leq f \) . Put \( {A}_{n} = \left\{ {x : {f}_{n}\left( x\right) \geq {\theta g}\left( x\right) }\right\} \) . If \( f\left( x\right) = 0 \), then \( g\left( x\right) = {f}_{n}\left( x\right) = 0 \), and \( x \in {A}_{n} \) for all \( n \) . If \( f\left( x\right) > 0 \), then eventually \( {f}_{n}\left( x\right) \geq {\theta g}\left( x\right) \) . Hence \( x \in \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{A}_{n} \) and \( X = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{A}_{n} \) . Also, we have \( {A}_{n} \subset {A}_{n + 1} \) for all \( n \) . By Lemma 2 in Section 8.2, page 387, we have, for any measurable set \( E \) ,\n\n(4)\n\n\[ \mu \left( {{A}_{n} \cap E}\right) \uparrow \mu \left( E\right) \]\n\nFrom this it is easy to prove that \( \int g{x}_{{A}_{n}} \uparrow \int g \) . Indeed, we write \( g = \) \( \mathop{\sum }\limits_{{i = 1}}^{m}{\lambda }_{i}{x}_{{E}_{i}} \) ( \( {E}_{i} \) being mutually disjoint) and observe that\n\n\[ \int g{X}_{{A}_{n}} = \int \mathop{\sum }\limits_{{i = 1}}^{m}{\lambda }_{i}{X}_{{A}_{n}}{X}_{{E}_{i}} = \int \mathop{\sum }\limits_{{i = 1}}^{m}{\lambda }_{i}{X}_{{A}_{n} \cap {E}_{i}} = \mathop{\sum }\limits_{{i = 1}}^{m}{\lambda }_{i}\mu \left( {{A}_{n} \cap {E}_{i}}\right) \]\n\nAs \( n \uparrow \infty \), we have \( \mu \left( {{A}_{n} \cap {E}_{i}}\right) \uparrow \mu \left( {E}_{i}\right) \) by Equation (4). Since the coefficients \( {\lambda }_{i} \) are nonnegative, \( \int g{\mathbf{X}}_{n} \uparrow \mathop{\sum }\limits_{{i = 1}}^{m}{\lambda }_{i}\mu \left( {E}_{i}\right) = \int g \) . We have proved that\n\n\[ \theta \int g = \mathop{\lim }\limits_{n}\int {\theta g}{x}_{{A}_{n}} \leq \mathop{\lim }\limits_{n}\int {f}_{n} \]\n\nSince this is true for any \( \theta \) in \( \left( {0,1}\right) \), one concludes that \( \int g \leq \mathop{\lim }\limits_{n}\int {f}_{n} \) . In this inequality take a supremum over all simple \( g \) for which \( 0 \leq g \leq f \), arriving at \( \int f \leq \mathop{\lim }\limits_{n}\int {f}_{n} \)
|
Yes
|
Theorem 2. For nonnegative measurable functions \( f \) and \( g \) we have \( \int \left( {f + g}\right) = \int f + \int g \) .
|
Proof. By Theorem 5 in Section 8.4, page 397, there exist nonnegative simple functions \( {f}_{n} \uparrow f \) and \( {g}_{n} \uparrow g \) . Then \( {f}_{n} + {g}_{n} \uparrow f + g \) . By Theorem 1 (the Monotone Convergence Theorem) and Lemma 4 above, we have \[ \int \left( {f + g}\right) = \mathop{\lim }\limits_{n}\int \left( {{f}_{n} + {g}_{n}}\right) = \lim \left\lbrack {\int {f}_{n}+\int {g}_{n}}\right\rbrack = \int f + \int g \]
|
Yes
|
Theorem 3. Let \( f \) be nonnegative and measurable. The conditions\n\n\( \int f = 0 \) and \( f\left( x\right) = 0 \) almost everywhere are equivalent.
|
Proof. Let \( A = \{ x : f\left( x\right) > 0\} \) and \( B = X \smallsetminus A \) . If \( f\left( x\right) = 0 \) almost everywhere, then \( \mu \left( A\right) = 0 \) . Hence\n\n\[ \n\int f = \int \left( {f{x}_{A} + f{x}_{B}}\right) = \int f{x}_{A} + \int f{x}_{B} \n\]\n\n\[ \n\leq \int \infty {x}_{A} + \int 0{x}_{B} = \infty \mu \left( A\right) + {0\mu }\left( B\right) = 0 \n\]\n\nFor the other implication, assume \( \int f = 0.\; \) Define \( \;{A}_{n} = \{ x : f\left( x\right) > \frac{1}{n}\} . \) Then \( A = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{A}_{n} \) . Since \( \frac{1}{n}{x}_{{A}_{n}} \) is a simple function bounded above by \( f \) we have\n\n\[ \n0 = \int f \geq \int \frac{1}{n}{x}_{{A}_{n}} = \frac{1}{n}\mu \left( {A}_{n}\right) \n\]\n\nThus \( \mu \left( {A}_{n}\right) = 0 \) for all \( n \) and \( \mu \left( A\right) = 0 \) by Lemma 1 in Section 8.2, page 386. ∎
|
Yes
|
Theorem 4. Fatou's Lemma. For a sequence of nonnegative measurable functions, \( \int \left( {\liminf {f}_{n}}\right) \leq \liminf \int {f}_{n} \) .
|
Proof. Recall that the limit infimum of a sequence of real numbers \( \left\lbrack {x}_{n}\right\rbrack \) is defined to be \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\inf }\limits_{{i \geq n}}{x}_{i} \) . The limit infimum of a sequence of real-valued functions is defined pointwise: \( \left( {\lim \inf {f}_{n}}\right) \left( x\right) = \lim \inf {f}_{n}\left( x\right) = \lim {g}_{n}\left( x\right) \), where \( {g}_{n}\left( x\right) = \mathop{\inf }\limits_{{i \geq n}}{f}_{i}\left( x\right) \) . Observe that \( {g}_{n - 1}\left( x\right) \leq {g}_{n}\left( x\right) \leq {f}_{n}\left( x\right) \) and that \( {g}_{n} \uparrow \) lim inf \( {f}_{n} \) . Hence by Theorem 1 (The Monotone Convergence Theorem)\n\n\[ \int \left( {\lim \inf {f}_{n}}\right) = \int \lim {g}_{n} = \lim \int {g}_{n} = \lim \inf \int {g}_{n} \leq \lim \inf \int {f}_{n} \]
|
Yes
|
Theorem 5. If \( f \) and \( g \) are nonnegative measurable functions that are equal almost everywhere, then \( \int f = \int g \) .
|
Proof. Let \( A = \{ x : f\left( x\right) = g\left( x\right) \} \) and \( B = X \smallsetminus A \) . Then\n\n\[ 0 \leq \int f{x}_{B} \leq \int \infty {x}_{B} = \infty \mu \left( B\right) = \infty 0 = 0 \]\n\nSimilarly, \( \int g{\mathbf{X}}_{B} = 0 \) . Hence\n\n\[ \int f = \int (f{X}_{X} + f{X}_{B}) = \int f{X}_{A} = \int g{X}_{A} = \int (g{X}_{A} + g{X}_{B}) = \int g \]
|
Yes
|
Lemma 1. A function \( f \) is integrable if and only if its positive and negative parts, \( {f}^{ + } \) and \( {f}^{ - } \), are integrable.
|
Proof. Assume that \( f \) is integrable. Then it is measurable, and the measurability of \( {f}^{ + } \) follows from the fact that \( \left\{ {x : {f}^{ + }\left( x\right) \geq a}\right\} \) is \( X \) when \( a \leq 0 \) and is \( \{ x : f\left( x\right) \geq a\} \) when \( a > 0 \) . The finiteness of the integral of \( \left| {f}^{ + }\right| \) is immediate from the inequality \( \left| {{f}^{ + }\left( x\right) }\right| \leq \left| {f\left( x\right) }\right| \) . The remainder of the proof involves similar elementary ideas.
|
No
|
Theorem 1. The set \( {L}^{1}\left( {X,\mathcal{A},\mu }\right) \) is a linear space, and the integral is a linear functional on it.
|
Proof. Let \( f \) and \( g \) be members of \( {L}^{1} \). To show that \( f + g \in {L}^{1} \), write \( h = f + g \), and \[ {h}^{ + } - {h}^{ - } = h = {f}^{ + } - {f}^{ - } + {g}^{ + } - {g}^{ - } \] From this it follows that \[ {h}^{ + } + {f}^{ - } + {g}^{ - } = {h}^{ - } + {f}^{ + } + {g}^{ + } \] Since these are all nonnegative functions, Theorem 2 of Section 8.5, page 402, is applicable, and \[ \int {h}^{ + } + \int {f}^{ - } + \int {g}^{ - } = \int {h}^{ - } + \int {f}^{ + } + \int {g}^{ + } \] Therefore, by Lemma 1, \[ \int (f + g) = \int h = \int {h}^{ + } - \int {h}^{ - } = \int {f}^{ + } - \int {f}^{ - } + \int {g}^{ + } - \int {g}^{ - } = \int f + \int g \] With this equation now established, we use Lemma 5 in Section 8.5 (page 401) to write \[ \int \left| {f + g}\right| \leq \int \left( {\left| f\right| + \left| g\right| }\right) = \int \left| f\right| + \int \left| g\right| < \infty \] For scalar multiplication, observe first that if \( \lambda \geq 0 \) and \( f \geq 0 \), then the definition of the integral in Equation (3) of Section 8.5 (page 400) gives \( \int {\lambda f} = \lambda \int f \). If \( f \geq 0 \) and \( \lambda < 0 \), then \[ \int {\lambda f} = \int {\left( \lambda f\right) }^{ + } - \int {\left( \lambda f\right) }^{ - } = - \int {\left( \lambda f\right) }^{ - } = - \int \left( {-\lambda {f}^{ + }}\right) = \lambda \int {f}^{ + } = \lambda \int f \] In the general case, we use what has already been proved: \[ \int {\lambda f} = \int \left\lbrack {\lambda {f}^{ + } + \left( {-\lambda }\right) {f}^{ - }}\right\rbrack = \int \lambda {f}^{ + } + \int - \lambda {f}^{ - } = \lambda \int {f}^{ + } - \lambda \int {f}^{ - } \] \[ = \lambda \left\lbrack {\int {f}^{ + }-\int {f}^{ - }}\right\rbrack = \lambda \int f \] The finiteness of the integral is now trivial: \[ \int \left| {\lambda f}\right| = \int \left| \lambda \right| \left| f\right| = \left| \lambda \right| \int \left| f\right| < \infty \]
|
Yes
|
Theorem 2. Dominated Convergence Theorem. \( g,{f}_{1},{f}_{2},\ldots \) be functions in \( {L}^{1}\left( {X,\mathcal{A},\mu }\right) \) such that \( \left| {f}_{n}\right| \leq g \) . If the sequence \( \left\lbrack {f}_{n}\right\rbrack \) converges pointwise to a function \( f \), then \( f \in {L}^{1} \) and \( \int {f}_{n} \rightarrow \int f \) .
|
Proof. The functions \( {f}_{n} + g \) are nonnegative. By Fatou’s Lemma (Theorem 4 in Section 8.5, page 403) and by the preceding theorem,\n\n\[ \int g + \int f = \int \left( {g + f}\right) = \int \liminf \left( {g + {f}_{n}}\right) \leq \liminf \int \left( {g + {f}_{n}}\right) \]\n\n\[ = \lim \inf \left\lbrack {\int g+\int {f}_{n}}\right\rbrack = \int g + \lim \inf \int {f}_{n} \]\n\nSince \( \int g < \infty \), we conclude that \( \int f \leq \liminf \int {f}_{n} \) . Since \( - f \) and \( - {f}_{n} \) satisfy the hypotheses of our theorem, the same conclusion can be drawn for them: \( \int - f \leq \liminf \int - {f}_{n} \) . This is equivalent to \( - \int f \leq - \limsup \int {f}_{n} \) and to \( \int f \geq \) lim sup \( \int {f}_{n} \) . Putting this all together produces\n\n\[ \lim \inf \int {f}_{n} \leq \lim \sup \int {f}_{n} \leq \int f \leq \lim \inf \int {f}_{n} \]
|
Yes
|
Theorem 3. Let \( f \) be Lebesgue integrable on the real line. For any positive \( \varepsilon \) there exist a simple function \( g \), a step function \( h \) and a continuous function \( k \) having compact support such that\n\n\[ \int \left| {f - g}\right| < \varepsilon \;\int \left| {f - h}\right| < \varepsilon \;\int \left| {f - k}\right| < \varepsilon \]
|
Proof. By Lemma \( 1,{f}^{ + } \) and \( {f}^{ - } \) are integrable. By the definition of the integral, Equation (3) in Section 8.5, page 400, there exist simple functions \( {g}_{1} \) and \( {g}_{2} \) such that \( {g}_{1} \leq {f}^{ + },{g}_{2} \leq {f}^{ - },\int {f}^{ + } < \int {g}_{1} + \varepsilon \), and \( \int {f}^{ - } < \int {g}_{2} + \varepsilon \) . Then \( {g}_{1} - {g}_{2} \) is a simple function such that\n\n\[ \int \left| {f - {g}_{1} + {g}_{2}}\right| \leq \int \left| {{f}^{ + } - {g}_{1}}\right| + \int \left| {{f}^{ - } - {g}_{2}}\right| = \int \left( {{f}^{ + } - {g}_{1}}\right) + \int \left( {{f}^{ - } - {g}_{2}}\right) < {2\varepsilon } \]\n\nIn order to establish the second part of the theorem, it now suffices to prove it in the special case that \( f \) is an integrable simple function. It is therefore a linear combination of characteristic functions of measurable sets of finite measure. It then suffices to prove this part of the theorem when \( f = {x}_{A} \) for some measurable set \( A \) having finite measure. By the definition of Lebesgue measure, there is a countable family of open intervals \( \left\{ {I}_{n}\right\} \) that cover \( A \) and satisfy \( \mu \left( A\right) \leq \mathop{\sum }\limits_{{n = 1}}^{\infty }\mu \left( {I}_{n}\right) < \mu \left( A\right) + \varepsilon \) . There is no loss of generality in assuming that the family \( \left\{ {I}_{n}\right\} \) is disjoint, because if two of these intervals have a point in common, their union is a single open interval. Since the series \( \sum \mu \left( {I}_{n}\right) \) converges, there is an index \( m \) such that \( \mathop{\sum }\limits_{{n = m + 1}}^{\infty }\mu \left( {I}_{n}\right) < \varepsilon \) . Put \( B = \mathop{\bigcup }\limits_{{n = 1}}^{m}{I}_{n} \) , \( E = \mathop{\bigcup }\limits_{{n = m + 1}}^{\infty }{I}_{n}, h = {\mathbf{X}}_{B} \), and \( \varphi = {\mathbf{X}}_{E} \) . Then \( h \) is a step function. Since \( A \subset B \cup E \), we have \( f \leq h + \varphi \) . Then\n\n\[ \left| {h - f}\right| \leq \left| {h + \varphi - f}\right| + \left| \varphi \right| = \left( {h + \varphi - f}\right) - \varphi \]\n\nConsequently,\n\n\[ \int \left| {h - f}\right| \leq \int \left( {h + \varphi - f}\right) + \int \varphi \]\n\n\[ = \mu \left( B\right) + \mu \left( E\right) - \mu \left( A\right) + \mu \left( E\right) \leq {2\varepsilon } \]\n\nFor the third part of the proof it suffices to consider an \( f \) that is an integrable step function. For this, in turn, it is enough to prove that the characteristic function of a single compact interval can be approximated in \( {L}^{1} \) by a continuous function that vanishes outside that interval. This can certainly be done with a piecewise linear function.
|
Yes
|
Theorem 1. Hölder’s Inequality. Let \( 1 \leq p \leq \infty ,\frac{1}{p} + \frac{1}{q} = 1 \) , \( f \in {L}^{p} \), and \( g \in {L}^{q} \) . Then \( {fg} \in {L}^{1} \) and \[ \int \left| {fg}\right| = \parallel {fg}{\parallel }_{1} \leq \parallel f{\parallel }_{p}\parallel g{\parallel }_{q} \]
|
Proof. The seminorms involved here are homogeneous: \( \parallel {\lambda f}\parallel = \left| \lambda \right| \parallel f\parallel \) . Consequently, it will suffice to establish Equation (3) in the special case when \( \parallel f{\parallel }_{p} = \) \( \parallel g{\parallel }_{q} = 1 \) . At first, let \( p = 1 \) and \( q = \infty \) . Since \( g \in {L}^{\infty } \), we have \( \left| {g\left( x\right) }\right| \leq M \) a.e. for some \( M \) . From this it follows that \( \int \left| {fg}\right| \leq M\int \left| f\right| = M\parallel f{\parallel }_{1} \) . By taking the infimum for all \( M \), we obtain \( \parallel {fg}{\parallel }_{1} \leq \parallel f{\parallel }_{1}\parallel g{\parallel }_{\infty } \) . Suppose now that \( p > 1 \) . We prove first that if \( a > 0, b > 0 \), and \( 0 \leq t \leq 1 \) , then \( {a}^{t}{b}^{1 - t} \leq {ta} + \left( {1 - t}\right) b \) . The accompanying Figure 8.1 shows the functions of \( t \) on the two sides of this inequality (when \( a = 2 \) and \( b = {12} \) ). It is clear that we should prove convexity of the function \( \varphi \left( t\right) = {a}^{t}{b}^{1 - t} \) . This requires that we prove \( {\varphi }^{\prime \prime }\left( t\right) \geq 0 \) . Since \( \log \varphi \left( t\right) = t\log a + \left( {1 - t}\right) \log b \), we have \[ \frac{{\varphi }^{\prime }\left( t\right) }{\varphi \left( t\right) } = \log a - \log b = c \] whence \( {\varphi }^{\prime \prime }\left( t\right) = c{\varphi }^{\prime }\left( t\right) = {c}^{2}\varphi \left( t\right) \geq 0 \) . Now let \( a = {\left| f\left( x\right) \right| }^{p}, b = {\left| g\left( x\right) \right| }^{q}, t = 1/p,1 - t = 1/q \) . Our inequality yields then \( \left| {f\left( x\right) g\left( x\right) }\right| \leq \frac{1}{p}{\left| f\left( x\right) \right| }^{p} + \frac{1}{q}{\left| g\left( x\right) \right| }^{q} \) . By hypothesis, the functions on the right in this inequality belong to \( {L}^{1} \) . Hence by integrating we obtain \[ \parallel {fg}{\parallel }_{1} = \int \left| {fg}\right| \leq \frac{1}{p}\int {\left| f\right| }^{p} + \frac{1}{q}\int {\left| g\right| }^{q} = \frac{1}{p} + \frac{1}{q} = 1 = \parallel f{\parallel }_{p}\parallel g{\parallel }_{q} \]
|
Yes
|
Theorem 2. Minkowski’s Inequality Let \( 1 \leq p \leq \infty \) . If \( f \) and \( g \) belong to \( {L}^{p} \), then so does \( f + g \), and \[ \parallel f + g{\parallel }_{p} \leq \parallel f{\parallel }_{p} + \parallel g{\parallel }_{p} \]
|
Proof. The cases \( p = 1 \) and \( p = \infty \) are special. For the first of these cases, just write \[ \int \left| {f + g}\right| \leq \int \left( {\left| f\right| + \left| g\right| }\right) = \int \left| f\right| + \int \left| g\right| \] For \( p = \infty \), select constants \( M \) and \( N \) for which \( \left| {f\left( x\right) }\right| \leq M \) a.e. and \( \left| {g\left( x\right) }\right| \leq N \) a.e. Then \( \left| {f\left( x\right) + g\left( x\right) }\right| \leq M + N \) a.e. This proves that \( f + g \in {L}^{\infty } \) and that \( \parallel f + g{\parallel }_{\infty } \leq M + N \) . By taking infima we get \( \parallel f + g{\parallel }_{\infty } \leq \parallel f{\parallel }_{\infty } + \parallel g{\parallel }_{\infty } \) . Now let \( 1 < p < \infty \) . From the observation that \( \left| {f + g}\right| \leq 2\max \{ \left| f\right| ,\left| g\right| \} \), we have \[ {\left| f + g\right| }^{p} \leq {2}^{p}\max \left\{ {{\left| f\right| }^{p},{\left| g\right| }^{p}}\right\} \leq {2}^{p}\left( {{\left| f\right| }^{p} + {\left| g\right| }^{p}}\right) \] This establishes that \( f + g \in {L}^{p} \) . Next, write \[ {\left| f + g\right| }^{p} = \left| {f + g}\right| {\left| f + g\right| }^{p - 1} \leq \left| f\right| {\left| f + g\right| }^{p - 1} + \left| g\right| {\left| f + g\right| }^{p - 1} \] Since \( \left| {f + g}\right| \in {L}^{p} \), we can infer that \( {\left| f + g\right| }^{p - 1} \in {L}^{q} \) (where \( \frac{1}{p} + \frac{1}{q} = 1 \) ) because \[ \int {\left| f + g\right| }^{\left( {p - 1}\right) q} = \int {\left| f + g\right| }^{p} < \infty \] By the homogeneity of Minkowski’s inequality, we may assume that \( \parallel f + g{\parallel }_{p} = 1 \) . Observe now that Hölder’s Inequality is applicable to the product \( \left| f\right| {\left| f + g\right| }^{p - 1} \) and to the product \( \left| g\right| {\left| f + g\right| }^{p - 1} \) . Consequently, \[ 1 = \int {\left| f + g\right| }^{p} \leq \int \left| f\right| {\left| f + g\right| }^{p - 1} + \int \left| g\right| {\left| f + g\right| }^{p - 1} \] \[ \leq {\begin{Vmatrix}f\end{Vmatrix}}_{p}{\begin{Vmatrix}{\left| f + g\right| }^{p - 1}\end{Vmatrix}}_{q} + {\begin{Vmatrix}g\end{Vmatrix}}_{p}{\begin{Vmatrix}{\left| f + g\right| }^{p - 1}\end{Vmatrix}}_{q} \] This is equivalent to \[ 1 \leq \left\{ {\parallel f{\parallel }_{p} + \parallel g{\parallel }_{p}}\right\} \parallel f + g{\parallel }_{p}^{p/q} = \parallel f{\parallel }_{p} + \parallel g{\parallel }_{p} \]
|
Yes
|
Theorem 3. The Riesz-Fischer Theorem. \( E \)\n\n\( {L}^{p}\left( {X,\mathcal{A},\mu }\right) \), where \( 1 \leq p \leq \infty \), is complete.
|
Proof. The case \( p = \infty \) is special and is addressed first. Let \( \left\lbrack {f}_{n}\right\rbrack \) be a Cauchy sequence in \( {L}^{\infty } \) . Define\n\n\[ \n{E}_{nm} = \left\{ {x : \left| {{f}_{n}\left( x\right) - {f}_{m}\left( x\right) }\right| > {\begin{Vmatrix}{f}_{n} - {f}_{m}\end{Vmatrix}}_{\infty }}\right\} \n\]\n\nBy Problem 1, these sets all have measure 0 . Hence the same is true of their union, \( E \) . If \( x \in X \smallsetminus E \), then \( \left| {{f}_{n}\left( x\right) - {f}_{m}\left( x\right) }\right| \leq {\begin{Vmatrix}{f}_{n} - {f}_{m}\end{Vmatrix}}_{\infty } \), and thus \( \left\lbrack {{f}_{n}\left( x\right) }\right\rbrack \) is a Cauchy sequence in \( \mathbb{R} \) for each \( x \in X \smallsetminus E \) . This sequence converges to a number that we may denote by \( f\left( x\right) \) . Define \( f\left( x\right) = 0 \) for \( x \in E \) . On \( X \smallsetminus E \) , \( \left| {f\left( x\right) }\right| = \lim \left| {{f}_{n}\left( x\right) }\right| \leq \lim {\begin{Vmatrix}{f}_{n}\end{Vmatrix}}_{\infty } < \infty \) . (Use the fact that a Cauchy sequence in a metric space is bounded.) Thus, \( f \in {L}^{\infty } \) . To prove that \( {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{\infty } \rightarrow 0 \) , let \( \varepsilon > 0 \) and select \( N \) so that \( {\begin{Vmatrix}{f}_{n} - {f}_{m}\end{Vmatrix}}_{\infty } < \varepsilon \) when \( n > m > N \) . Then \( \left| {{f}_{n}\left( x\right) - {f}_{m}\left( x\right) }\right| < \varepsilon \) on \( X \smallsetminus E \), and \( \left| {f\left( x\right) - {f}_{m}\left( x\right) }\right| \leq \varepsilon \) for \( m > N.\)
|
No
|
Theorem 1. If \( \\left( {X,\\mathcal{A},\\mu }\\right) \) is a measure space, and if \( f \) is a nonnegative measurable function, then the equation\n\n(1)\n\n\[ \n\\nu \\left( A\\right) = {\\int }_{A}{fd\\mu }\\;\\left( {A \\in \\mathcal{A}}\\right) \n\]\n\ndefines a measure \( \\nu \) that is absolutely continuous with respect to \( \\mu \) .
|
Proof. The postulates for a measure are quickly verified.\n\n(a) \( \\nu \\left( \\varnothing \\right) = {\\int }_{\\varnothing }f = \\int f{x}_{\\varnothing } = \\int 0 = 0 \)\n\n(b) \( \\nu \\left( A\\right) \\geq 0 \) because \( f \\geq 0 \)\n\n(c) If \( \\left\\lbrack {A}_{i}\\right\\rbrack \) is a disjoint sequence of measurable sets, then\n\n\[ \n\\nu \\left( {\\mathop{\\bigcup }\\limits_{{i = 1}}^{\\infty }{A}_{i}}\\right) = {\\int }_{\\cup {A}_{i}}f = \\int f{x}_{\\cup {A}_{i}} = \\int f\\sum {x}_{{A}_{i}} = \\int \\sum f{x}_{{A}_{i}} \n\]\n\n\[ \n= \\int \\mathop{\\lim }\\limits_{n}\\mathop{\\sum }\\limits_{{i = 1}}^{n}f{x}_{{A}_{i}} = \\mathop{\\lim }\\limits_{n}\\int \\mathop{\\sum }\\limits_{{i = 1}}^{n}f{x}_{{A}_{i}} = \\mathop{\\lim }\\limits_{n}\\mathop{\\sum }\\limits_{{i = 1}}^{n}\\int f{x}_{{A}_{i}} \n\]\n\n\[ \n= \\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }\\nu \\left( {A}_{i}\\right) \n\]\nThis calculation used the Monotone Convergence Theorem (Section 8.5, page 401). The absolute continuity of \( \\nu \) is clear: if \( \\mu \\left( A\\right) = 0 \), then \( \\nu \\left( A\\right) = {\\int }_{A}f = 0 \) .
|
Yes
|
Theorem 2. Radon-Nikodym Theorem. Let \( \mu \) and \( \nu \) be \( \sigma \) -finite measures on a measurable space \( \left( {X,\mathcal{A}}\right) \) . If \( \nu \) is absolutely continuous with respect to \( \mu \), then there exists a nonnegative measurable function \( h \), determined uniquely up to a set of \( \mu \) -measure 0, such that \( \nu \left( A\right) = \) \( {\int }_{A}{hd\mu } \) for all \( A \in \mathcal{A} \) .
|
Proof. We prove the theorem first under the assumption that \( \mu \left( X\right) < \infty \) and \( \nu \left( X\right) < \infty \) . Consider the Hilbert space \( {L}^{2} = {L}^{2}\left( {X,\mathcal{A},\mu + \nu }\right) \) . For any \( f \) in \( {L}^{2} \), define \( \Phi \left( f\right) = \int {fd\mu } \) . It is easily verified that \( \Phi \) is a linear functional on \( {L}^{2} \) . Furthermore it is bounded (continuous) because by the Hölder Inequality (Theorem 1 in Section 8.7, page 409)\n\n\[ \left| {\Phi \left( f\right) }\right| = \left| {\int f \cdot {1d\mu }}\right| \leq \int \left| f\right| \cdot {1d}\left( {\mu + \nu }\right) \leq \parallel f{\parallel }_{2}\parallel 1{\parallel }_{2} \]\n\nBy the Riesz Representation Theorem for Hilbert space (Section 2.3, page 81) there exists an element \( {h}_{0} \) in \( {L}^{2} \) such that\n\n\[ \Phi \left( f\right) = \int f{h}_{0}d\left( {\mu + \nu }\right) \;f \in {L}^{2} \]\n\nThis means that \( \int {fd\mu } = \int f{h}_{0}d\left( {\mu + \nu }\right) \), whence\n\n\[ \int f\left( {1 - {h}_{0}}\right) {d\mu } = \int f{h}_{0}{d\nu } \]\n\nLet \( B = \left\{ {x : {h}_{0}\left( x\right) \leq 0}\right\} \) . Then \( 1 - {h}_{0} \geq 1 \) on \( B \), and consequently\n\n\[ 0 \leq \mu \left( B\right) \leq \int {x}_{B}\left( {1 - {h}_{0}}\right) {d\mu } = \int {x}_{B}{h}_{0}{d\nu } \leq 0 \]\n\nThus \( \mu \left( B\right) = 0 \) and \( {h}_{0}\left( x\right) > 0 \) a.e. (with respect to \( \mu \) ). Since \( \nu \ll \mu \), we have \( \nu \left( B\right) = 0 \) also. Hence for any \( A \in \mathcal{A} \),\n\n\[ \nu \left( A\right) = \int {\mathbf{X}}_{A}{d\nu } = \int {h}_{0}^{-1}{\mathbf{X}}_{A}{h}_{0}{d\nu } = \int {h}_{0}^{-1}{\mathbf{X}}_{A}\left( {1 - {h}_{0}}\right) {d\mu } \]\n\n\[ = {\int }_{A}{h}_{0}^{-1}\left( {1 - {h}_{0}}\right) {d\mu } = {\int }_{A}{hd\mu }\;\left( {h = {h}_{0}^{-1}\left( {1 - {h}_{0}}\right) }\right) \]\nTo see that \( h \geq 0 \) a.e., with respect to \( \mu \), write \( A = \{ x : h\left( x\right) < 0\} \), so that \( 0 \leq \nu \left( A\right) = {\int }_{A}{hd\mu } \leq 0 \), whence \( \mu \left( A\right) = 0 \) .
|
Yes
|
Jordan Decomposition. The difference of two measures (defined on the same \( \sigma \) -algebra), one of which is finite, is a signed measure. Conversely, every signed measure \( \mu \) is the difference of two measures \( {\mu }^{ + } \) and \( {\mu }^{ - } \), one of which is finite. Furthermore, we may require these two measures to be mutually singular, and in that case they are uniquely determined by \( \mu \) .
|
For the first assertion, let \( {\mu }_{1} \) and \( {\mu }_{2} \) be measures, and suppose that \( {\mu }_{1} \) is finite. Put \( \mu = {\mu }_{1} - {\mu }_{2} \) . To see that \( \mu \) is a signed measure, note first that \( \mu \) does not assume the value \( + \infty \) . Next, we have \( \mu \left( \varnothing \right) = 0 \) since \( {\mu }_{1} \) and \( {\mu }_{2} \) have this property. Finally, let \( \left\{ {A}_{i}\right\} \) be a disjoint sequence of measurable sets. Then\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = {\mu }_{1}\left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) - {\mu }_{2}\left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 1}}^{\infty }{\mu }_{1}\left( {A}_{i}\right) - \mathop{\sum }\limits_{{i = 1}}^{\infty }{\mu }_{2}\left( {A}_{i}\right) \]\n\n\[ = \mathop{\lim }\limits_{n}\mathop{\sum }\limits_{{i = 1}}^{n}{\mu }_{1}\left( {A}_{i}\right) - \mathop{\lim }\limits_{n}\mathop{\sum }\limits_{{i = 1}}^{n}{\mu }_{2}\left( {A}_{i}\right) \]\n\n\[ = \mathop{\lim }\limits_{n}\left\lbrack {\mathop{\sum }\limits_{{i = 1}}^{n}{\mu }_{1}\left( {A}_{i}\right) - \mathop{\sum }\limits_{{i = 1}}^{n}{\mu }_{2}\left( {A}_{i}\right) }\right\rbrack \]\n\n\[ = \mathop{\lim }\limits_{n}\mathop{\sum }\limits_{{i = 1}}^{n}\left\lbrack {{\mu }_{1}\left( {A}_{i}\right) - {\mu }_{2}\left( {A}_{i}\right) }\right\rbrack = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \]\n\nNotice that on the second line of this calculation the first sum is finite, although the second may be infinite.
|
Yes
|
Theorem 2. Radon-Nikodym Theorem for Signed Measures. Let \( \left( {X,\mathcal{A},\mu }\right) \) be a \( \sigma \) -finite measure space. If \( \nu \) is a finite-valued signed measure that is absolutely continuous with respect to \( \mu \), then there is a measurable function \( h \) such that for all \( A \in \mathcal{A},\nu \left( A\right) = {\int }_{A}{hd\mu } \) .
|
Proof. By the preceding theorem, there exist measures \( {\nu }^{ + } \) and \( {\nu }^{ - } \) such that \( \nu = {\nu }^{ + } - {\nu }^{ - } \) and \( {\nu }^{ + } \bot {\nu }^{ - } \) . Consequently, there exists a measurable set \( P \) for which \( {\nu }^{ + }\left( {X \smallsetminus P}\right) = 0 = {\nu }^{ - }\left( P\right) \) . If \( A \) is a measurable set satisfying \( \mu \left( A\right) = 0 \) , then \( \mu \left( {A \cap P}\right) = 0 \) and \( \nu \left( {A \cap P}\right) = 0 \), by the absolute continuity. Hence\n\n\[ \n{\nu }^{ + }\left( A\right) = {\nu }^{ + }\left( {A \cap P}\right) + {\nu }^{ + }\left( {A \smallsetminus P}\right) = {\nu }^{ + }\left( {A \cap P}\right) \n\]\n\n\[ \n= \left( {\nu + {\nu }^{ - }}\right) \left( {A \cap P}\right) = \nu \left( {A \cap P}\right) = 0 \n\]\n\nThis establishes that \( {\nu }^{ + } \) is absolutely continuous with respect to \( \mu \) . It follows that \( {\nu }^{ - } \) is also absolutely continuous with respect to \( \mu \) . By the earlier Radon-Nikodym Theorem (Theorem 2 in Section 8.8, page 414), there exist nonnegative measurable functions \( {h}_{1} \) and \( {h}_{2} \) such that for \( A \) in \( \mathcal{A} \) ,\n\n\[ \n{\nu }^{ + }\left( A\right) = {\int }_{A}{h}_{1}{d\mu }\;{\nu }^{ - }\left( A\right) = {\int }_{A}{h}_{2}{d\mu } \n\]\n\nIt follows that \( {h}_{1} \) and \( {h}_{2} \) are finite almost everywhere. Thus, there is nothing suspicious in the equation\n\n\[ \n\nu \left( A\right) = {\nu }^{ + }\left( A\right) - {\nu }^{ - }\left( A\right) = {\int }_{A}{h}_{1}{d\mu } - {\int }_{A}{h}_{2}{d\mu } = {\int }_{A}\left( {{h}_{1} - {h}_{2}}\right) {d\mu } \n\]
|
Yes
|
Theorem 3. The Hahn Decomposition. If \( \mu \) is a signed measure on the measurable space \( \left( {X,\mathcal{A}}\right) \), then there is a decomposition of \( X \) into a disjoint pair of measurable sets \( N \) and \( P \) such that \( \mu \left( A\right) \geq 0 \) when \( A \subset P \) and \( \mu \left( A\right) \leq 0 \) when \( A \subset N \) .
|
Proof. Left as a problem.
|
No
|
Lemma 1. Let \( \\left( {X,\\mathcal{A}}\\right) \) and \( \\left( {Y,\\mathcal{B}}\\right) \) be two measurable spaces. If\n\n\( E \\in \\mathcal{A} \\otimes \\mathcal{B} \), then \( {E}_{x} \\in \\mathcal{B} \) for all \( x \\in X \) and \( {E}^{y} \\in \\mathcal{A} \) for all \( y \\in Y \) .
|
Proof. Define\n\n\\[ \n\\mathcal{M} = \\left\\{ {E : E \\subset X \\times Y\\text{ and }{E}^{y} \\in \\mathcal{A}\\text{ for all }y \\in Y}\\right\\} \n\\]\n\nWe shall prove that \( \\mathcal{M} \) is a \( \\sigma \) -algebra containing all rectangles. From this it will follow that \( \\mathcal{M} \\supset \\mathcal{A} \\otimes \\mathcal{B} \), since the latter is the smallest \( \\sigma \) -algebra containing all rectangles. Then, if \( E \\in \\mathcal{A} \\otimes \\mathcal{B} \), we can conclude that \( E \\in \\mathcal{M} \) and that \( {E}^{y} \\in \\mathcal{A} \) for each \( y \) . Now consider any rectangle \( E = A \\times B \) . If \( y \\in B \), then \( {E}^{y} = A \\in \\mathcal{A} \) . If \( y \\notin B \), then \( {E}^{y} = \\varnothing \\in \\mathcal{A} \) . Thus in all cases \( {E}^{y} \\in \\mathcal{A} \) and \( E \\in \\mathcal{M} \) . Next, let \( E \) be any member of \( \\mathcal{M} \) . The equation\n\n(1)\n\n\\[ \n{\\left\\[ \\left( X \\times Y\\right) \\smallsetminus E\\right\\rbrack }^{y} = X \\smallsetminus {E}^{y} \n\\]\n\nshows that \( \\left( {X \\times Y}\\right) \\smallsetminus E \) belongs to \( \\mathcal{M} \) . If \( {E}_{i} \\in \\mathcal{M} \), then by the equation\n\n(2)\n\n\\[ \n{\\left\\[ \\mathop{\\bigcup }\\limits_{{i = 1}}^{\\infty }{E}_{i}\\right\\rbrack }^{y} = \\mathop{\\bigcup }\\limits_{{i = 1}}^{\\infty }{E}_{i}^{y} \n\\]\n\nwe see that \( \\bigcup {E}_{i} \\in \\mathcal{M} \) .
|
Yes
|
Lemma 2. The collection of all unions of finite disjoint families of rectangles constructed from a pair of \( \sigma \) -algebras is an algebra.
|
Proof. Let \( \mathcal{C} \) be the collection referred to, and let \( E \) and \( F \) be members of \( \mathcal{C} \). Then \( E \) and \( F \) have expressions \( E = \mathop{\bigcup }\limits_{{i = 1}}^{n}\left( {{A}_{i} \times {B}_{i}}\right) \) and \( F = \mathop{\bigcup }\limits_{{j = 1}}^{m}\left( {{C}_{j} \times {D}_{j}}\right) \), both being unions of disjoint families. Since\n\n\[ E \cap F = \mathop{\bigcup }\limits_{{i = 1}}^{n}\mathop{\bigcup }\limits_{{j = 1}}^{m}\left\lbrack {\left( {{A}_{i} \cap {C}_{j}}\right) \times \left( {{B}_{i} \cap {D}_{j}}\right) }\right\rbrack \]\n\nwe see that \( E \cap F \in \mathcal{C} \), and that \( \mathcal{C} \) is closed under the taking of intersections. From the equation\n\n\[ \left( {X \times Y}\right) \smallsetminus \left( {A \times B}\right) = \left\lbrack {\left( {X \smallsetminus A}\right) \times B}\right\rbrack \cup \left\lbrack {X \times \left( {Y \smallsetminus B}\right) }\right\rbrack \]\n\nwe get\n\n\[ \left( {X \times Y}\right) \smallsetminus E = \left( {X \times Y}\right) \smallsetminus \mathop{\bigcup }\limits_{{i = 1}}^{n}\left( {{A}_{i} \times {B}_{i}}\right) = \mathop{\bigcap }\limits_{{i = 1}}^{n}\left\lbrack {\left( {X \times Y}\right) \smallsetminus \left( {{A}_{i} \times {B}_{i}}\right) }\right\rbrack \]\n\n\[ = \mathop{\bigcap }\limits_{{i = 1}}^{n}\left\{ {\left\lbrack {\left( {X \smallsetminus {A}_{i}}\right) \times {B}_{i}}\right\rbrack \cup \left\lbrack {X \times \left( {Y \smallsetminus {B}_{i}}\right) }\right\rbrack }\right\} \]\n\nThis shows that the complement of \( E \) belongs to \( \mathcal{C} \), because \( \mathcal{C} \) is closed under finite intersections. By the de Morgan identities, \( \mathcal{C} \) is closed under unions.
|
Yes
|
Lemma 3. In any measure space \( \left( {X,\mathcal{A},\mu }\right) \) the following are true for measurable sets \( {A}_{i} \) :\n\n(1) If \( {A}_{1} \subset {A}_{2} \subset \cdots \), then \( \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mathop{\lim }\limits_{n}\mu \left( {A}_{n}\right) \)\n\n(2) If \( {A}_{1} \supset {A}_{2} \supset \cdots \) and \( \mu \left( {A}_{1}\right) < \infty \), then \( \mu \left( {\mathop{\bigcap }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mathop{\lim }\limits_{n}\mu \left( {A}_{n}\right) \)
|
Proof. Assume the hypothesis in (1), and define \( {B}_{n} = {A}_{n} \smallsetminus {A}_{n - 1} \) . The sequence \( \left\{ {B}_{n}\right\} \) is disjoint, and consequently,\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{B}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {B}_{i}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{i = 1}}^{n}\mu \left( {B}_{i}\right) \]\n\n\[ = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{n}{B}_{i}}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mu \left( {A}_{n}\right) \]\n\nTo establish (2), assume its hypothesis. Then \( \left\{ {{A}_{1} \smallsetminus {A}_{n}}\right\} \) is an increasing sequence, and by part (1) we have\n\n\[ \mu \left( {A}_{1}\right) - \mu \left( {\mathop{\bigcap }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mu \left( {{A}_{1} \smallsetminus \mathop{\bigcap }\limits_{{i = 1}}^{\infty }{A}_{i}}\right) = \mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{A}_{1} \smallsetminus {A}_{i}}\right) }\right) \]\n\n\[ = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mu \left( {{A}_{1} \smallsetminus {A}_{n}}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\mu \left( {A}_{1}\right) - \mu \left( {A}_{n}\right) }\right) \]\n\n\[ = \mu \left( {A}_{1}\right) - \mathop{\lim }\limits_{{n \rightarrow \infty }}\mu \left( {A}_{n}\right) \]
|
Yes
|
Lemma 5. If \( \left( {X,\mathcal{A},\mu }\right) \) and \( \left( {Y,\mathcal{B},\nu }\right) \) are \( \sigma \) -finite measure spaces, then so is \( \left( {X \times Y,\mathcal{A} \otimes \mathcal{B},\mu \otimes \nu }\right) \) .
|
Proof. It is clear that the set function \( \phi \) has the property \( \phi \left( \varnothing \right) = 0 \) and the property \( \phi \left( E\right) \geq 0 \) . If \( \left\{ {E}_{i}\right\} \) is a disjoint sequence of sets in \( \mathcal{A} \otimes \mathcal{B} \), then \( \left\{ {E}_{i}^{y}\right\} \) is a disjoint sequence in \( \mathcal{A} \) . Hence, by the Dominated Convergence Theorem, \[ \phi \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{E}_{i}}\right) = {\int }_{Y}\mu \left( {\left( \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{E}_{i}\right) }^{y}\right) {d\nu }\left( y\right) = {\int }_{Y}\mu \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{E}_{i}^{y}}\right) {d\nu }\left( y\right) \] \[ = {\int }_{Y}\mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {E}_{i}^{y}\right) {d\nu }\left( y\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }{\int }_{Y}\mu \left( {E}_{i}^{y}\right) {d\nu }\left( y\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\phi \left( {E}_{i}\right) \] Thus \( \phi \) is a measure. For the \( \sigma \) -finiteness, observe that if \( X = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{A}_{n} \) and \( Y = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{B}_{n} \), where \( {A}_{1} \subset {A}_{2} \subset \cdots \) and \( {B}_{1} \subset {B}_{2} \subset \cdots \), then \( X \times Y = \) \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }\left( {{A}_{n} \times {B}_{n}}\right) \) . If, further, \( \mu \left( {A}_{n}\right) < \infty \) and \( \nu \left( {B}_{n}\right) < \infty \) for all \( n \), then we have \( \phi \left( {{A}_{n} \times {B}_{n}}\right) = \mu \left( {A}_{n}\right) \nu \left( {B}_{n}\right) < \infty \) .
|
Yes
|
Theorem 2. Second Fubini Theorem. Let \( \\left( {X,\\mathcal{A},\\mu }\\right) \) and \( \\left( {Y,\\mathcal{B},\\nu }\\right) \) be two \( \\sigma \) -finite measure spaces. Let \( f \) be a nonnegative function on \( X \\times Y \) that is measurable with respect to \( \\left( {X \\times Y,\\mathcal{A} \\otimes \\mathcal{B}}\\right) \) . Then\n\n(1) For each \( x, y \\mapsto f\\left( {x, y}\\right) \) is measurable\n\n(2) For each \( y, x \\mapsto f\\left( {x, y}\\right) \) is measurable\n\n(3) \( y \\mapsto {\\int }_{X}f\\left( {x, y}\\right) {d\\mu }\\left( x\\right) \) is measurable\n\n(4) \( x \\mapsto {\\int }_{Y}f\\left( {x, y}\\right) {d\\nu }\\left( y\\right) \) is measurable\n\n\\left( 5\\right) \\;{\\int }_{X \\times Y}f\\left( {x, y}\\right) \\;{d\\phi } = {\\int }_{X}{\\int }_{Y}f\\left( {x, y}\\right) \\;{d\\nu }\\;{d\\mu } = {\\int }_{Y}{\\int }_{X}f\\left( {x, y}\\right) \\;{d\\mu }\\;{d\\nu }
|
Proof. If \( f \) is the characteristic function of a measurable set \( E \), then (1) is true because \( f\\left( {x, y}\\right) = {x}_{{E}_{x}}\\left( y\\right) \) . Part (2) is true by the symmetry in the situation. Since\n\n\\[ \n{\\int }_{X}f\\left( {x, y}\\right) {d\\mu }\\left( x\\right) = {\\int }_{X}{x}_{E}\\left( {x, y}\\right) {d\\mu }\\left( x\\right) = {\\int }_{X}{x}_{E}y\\left( x\\right) {d\\mu }\\left( x\\right) = \\mu \\left( {E}^{y}\\right) \n\\]\n\nthe preceding lemma asserts that (3) is also true in this case. Part (4) is true by symmetry. For part (5), write\n\n\\[ \n{\\int }_{X \\times Y}f\\left( {x, y}\\right) {d\\phi } = \\phi \\left( E\\right) = {\\int }_{Y}\\mu \\left( {E}^{y}\\right) {d\\nu }\\left( y\\right) \n\\]\n\n\\[ \n= {\\int }_{Y}{\\int }_{X}f\\left( {x, y}\\right) {d\\mu }\\left( x\\right) {d\\nu }\\left( y\\right) \n\\]\n\nThe other equality is similar. Thus, Theorem 2 is true when \( f \) is the characteristic function of a measurable set.\n\nIf \( f \) is a simple function, then \( f \) has properties (1) to (5) by the linearity of the integrals.\n\nIf \( f \) is an arbitrary nonnegative measurable function, then there exist simple functions \( {f}_{n} \) such that \( {f}_{n} \\uparrow f \) . Since the limit of a sequence of measurable functions is measurable, \( f \) has properties (1) to (4). By the Monotone Convergence Theorem, property (5) follows for \( f \) .
|
Yes
|
Theorem 3. Fubini's Theorem for Complex Measures. Let \( \left( {X,\mathcal{A}}\right) \) and \( \left( {Y,\mathcal{B}}\right) \) be two measurable spaces, and let \( \mu \) and \( \nu \) be complex measures on \( X \) and \( Y \), respectively. Let \( f \) be a complex-valued measurable function on \( X \times Y \) . If \( {\int }_{Y}{\int }_{X}\left| {f\left( {x, y}\right) }\right| d\left| \mu \right| d\left| \nu \right| < \infty \), then\n\n\[ \n{\int }_{Y}{\int }_{X}f\left( {x, y}\right) {d\mu }\left( x\right) {d\nu }\left( y\right) = {\int }_{X}{\int }_{Y}f\left( {x, y}\right) {d\nu }\left( y\right) {d\mu }\left( x\right) \n\]
|
This theorem is to be found in [DS].
|
No
|
Corollary 1. \( {A}^{-1} \) is an open set in \( A \) and \( x \mapsto {x}^{-1} \) is a continuous map of \( {A}^{-1} \) to itself.
|
Proof. To see that \( {A}^{-1} \) is open, choose an invertible element \( {x}_{0} \) and an arbitrary element \( h \in A \) . We have \( {x}_{0} + h = {x}_{0}\left( {\mathbf{1} + {x}_{0}^{-1}h}\right) \) . So if \( \begin{Vmatrix}{{x}_{0}^{-1}h}\end{Vmatrix} < 1 \) then by the preceding theorem \( {x}_{0} + h \) is invertible. In particular, if \( \parallel h\parallel < \) \( {\begin{Vmatrix}{x}_{0}^{-1}\end{Vmatrix}}^{-1} \), then this condition is satisfied, proving that \( {x}_{0} + h \) is invertible when \( \parallel h\parallel \) is sufficiently small.\n\nSupposing that \( h \) has been so chosen, we can write\n\n\[{\left( {x}_{0} + h\right) }^{-1} - {x}_{0}^{-1} = {\left( {x}_{0}\left( \mathbf{1} + {x}_{0}^{-1}h\right) \right) }^{-1} - {x}_{0}^{-1} = \left\lbrack {{\left( \mathbf{1} + {x}_{0}^{-1}h\right) }^{-1} - \mathbf{1}}\right\rbrack \cdot {x}_{0}^{-1}.\n\]\n\nThus for \( \parallel h\parallel < {\begin{Vmatrix}{x}_{0}^{-1}\end{Vmatrix}}^{-1} \) we have\n\n\[ \begin{Vmatrix}{{\left( {x}_{0} + h\right) }^{-1} - {x}_{0}^{-1}}\end{Vmatrix} \leq \begin{Vmatrix}{{\left( \mathbf{1} + {x}_{0}^{-1}h\right) }^{-1} - \mathbf{1}}\end{Vmatrix} \cdot \begin{Vmatrix}{x}_{0}^{-1}\end{Vmatrix} \leq \frac{\begin{Vmatrix}{{x}_{0}^{-1}h}\end{Vmatrix} \cdot \begin{Vmatrix}{x}_{0}^{-1}\end{Vmatrix}}{1 - \begin{Vmatrix}{{x}_{0}^{-1}h}\end{Vmatrix}},\]\n\nand the last term obviously tends to zero as \( \parallel h\parallel \rightarrow 0 \) .
|
Yes
|
Proposition 1.6.2. For every \( x \in A,\sigma \left( x\right) \) is a closed subset of the disk \( \{ z \in \mathbb{C} : \left| z\right| \leq \parallel x\parallel \} \)
|
Proof. The complement of the spectrum is given by\n\n\[ \mathbb{C} \smallsetminus \sigma \left( x\right) = \left\{ {\lambda \in \mathbb{C} : x - \lambda \in {A}^{-1}}\right\} .\n\]\n\nSince \( {A}^{-1} \) is open and the map \( \lambda \in \mathbb{C} \mapsto x - \lambda \in A \) is continuous, the complement of \( \sigma \left( x\right) \) must be open.\n\nTo prove the second assertion, we will show that no complex number \( \lambda \) with \( \left| \lambda \right| > \parallel x\parallel \) can belong to \( \sigma \left( x\right) \) . Indeed, for such a \( \lambda \) the formula\n\n\[ x - \lambda = \left( {-\lambda }\right) \left( {1 - {\lambda }^{-1}x}\right) \]\n\n\ntogether with the fact that \( \begin{Vmatrix}{{\lambda }^{-1}x}\end{Vmatrix} < 1 \), implies that \( x - \lambda \) is invertible.
|
Yes
|
Corollary 1. An element \( x \) of a unital Banach algebra \( A \) is quasinilpotent iff \( \sigma \left( x\right) = \{ 0\} \) .
|
Proof. \( x \) is quasinilpotent \( \Leftrightarrow r\left( x\right) = 0 \Leftrightarrow \sigma \left( x\right) = \{ 0\} \) .
|
Yes
|
Proposition 1.8.2. Let \( A \) be a Banach algebra with normalized unit \( \mathbf{1} \) and let \( I \) be a proper ideal in \( A \) . Then for every \( z \in I \) we have \( \parallel \mathbf{1} + z\parallel \geq 1 \) . In particular, the closure of a proper ideal is a proper ideal.
|
Proof. If there is an element \( z \in I \) with \( \parallel \mathbf{1} + z\parallel < 1 \), then by Theorem 1.5.2 \( z \) must be invertible in \( A \) ; hence \( \mathbf{1} = {z}^{-1}z \in I \), which implies that \( I \) cannot be a proper ideal. The second assertion follows from the continuity of the norm; if \( \parallel \mathbf{1} + z\parallel \geq 1 \) for all \( z \in I \), then \( \parallel \mathbf{1} + z\parallel \geq 1 \) persists for all \( z \) in the closure of \( I \) .
|
Yes
|
Proposition 1.8.4. Every bounded homomorphism of Banach algebras \( \omega : A \rightarrow B \) has a unique factorization \( \omega = \dot{\omega } \circ \pi \), where \( \dot{\omega } \) is an injective homomorphism of \( A/\ker \omega \) to \( B \) and \( \pi : A \rightarrow A/\ker \omega \) is the natural projection. One has \( \parallel \dot{\omega }\parallel = \parallel \omega \parallel \) .
|
Proof. The assertions in the first sentence are straightforward, and we prove \( \parallel \dot{\omega }\parallel = \parallel \omega \parallel \) . From the factorization \( \omega = \dot{\omega } \circ \pi \) and the fact that \( \parallel \pi \parallel \leq 1 \) we have \( \parallel \omega \parallel \leq \parallel \dot{\omega }\parallel \) ; the opposite inequality follows from\n\n\[ \parallel \dot{\omega }\left( \dot{x}\right) \parallel = \parallel \omega \left( x\right) \parallel = \parallel \omega \left( {x + z}\right) \parallel \leq \parallel \omega \parallel \parallel x + z\parallel ,\;z \in \ker \omega ,\]\n\nafter the infimum is taken over all \( z \in \ker \omega \) .
|
Yes
|
Proposition 1.9.3. In its relative weak*-topology, \( \operatorname{sp}\left( A\right) \) is a compact Hausdorff space.
|
Proof. It suffices to show that \( \operatorname{sp}\left( A\right) \) is a weak*-closed subset of the unit ball of the dual of \( A \) . Notice that a linear functional \( f : A \rightarrow \mathbb{C} \) belongs to \( \operatorname{sp}\left( A\right) \) iff \( \parallel f\parallel \leq 1, f\left( \mathbf{1}\right) = 1 \), and \( f\left( {yz}\right) = f\left( y\right) f\left( z\right) \) for all \( y, z \in A \) . These conditions obviously define a weak*-closed subset of the unit ball of \( {A}^{\prime } \) .
|
Yes
|
Proposition 1.11.1. Let \( B \) be a Banach subalgebra of \( A \) that contains the unit of \( A \) . For every element \( x \in B \) we have \( {\sigma }_{A}\left( x\right) \subseteq {\sigma }_{B}\left( x\right) \) .
|
Proof. This is an immediate consequence of the fact that invertible elements of \( B \) are invertible elements of \( A \) .
|
No
|
Corollary 1. Let \( {\mathbf{1}}_{A} \in B \subseteq A \) be as above, let \( x \in A \), and let \( \Omega \) be a bounded component of \( \mathbf{C} \smallsetminus {\sigma }_{A}\left( x\right) \) . Then either \( \Omega \cap {\sigma }_{B}\left( x\right) = \varnothing \) or \( \Omega \subseteq {\sigma }_{B}\left( x\right) \) .
|
Proof. Let \( \Omega \) be a hole of \( {\sigma }_{A}\left( x\right) \) . Consider \( X = \Omega \cap {\sigma }_{B}\left( x\right) \) as a closed subspace of the topological space \( \Omega \) . Since \( \Omega \) is an open set in \( \mathbb{C} \), the boundary \( {\partial }_{\Omega }X \) of \( X \) relative to \( \Omega \) is contained in\n\n\[ \n\partial {\sigma }_{B}\left( x\right) \subseteq {\sigma }_{A}\left( x\right) \subseteq \mathbb{C} \smallsetminus \Omega .\n\]\n\nHence \( {\partial }_{\Omega }X = \varnothing \) . Lemma 1.11.4 implies that either \( X = \varnothing \) or \( X = \Omega \), as asserted.
|
Yes
|
For every bounded complex-valued sesquilinear form \( \left\lbrack {\cdot , \cdot }\right\rbrack \) on \( H \) there is a unique bounded operator \( A \) on \( H \) such that\n\n\[ \left\lbrack {\xi ,\eta }\right\rbrack = \langle {A\xi },\eta \rangle ,\;\xi ,\eta \in H. \]
|
Proof. Fix a vector \( \xi \in H \) and consider the linear functional \( f \) defined on \( H \) by \( f\left( \eta \right) = \overline{\left\lbrack \xi ,\eta \right\rbrack } \), the bar denoting complex conjugation. Since \( f \) is a bounded linear functional, the Riesz lemma in its above form implies that there is a unique vector \( {A\xi } \in H \) satisfying \( f\left( \eta \right) = \langle \eta ,{A\xi }\rangle \) ; and after taking the complex conjugate we find that the function \( A : H \rightarrow H \) that we have defined must satisfy\n\n\[ \left\lbrack {\xi ,\eta }\right\rbrack = \langle {A\xi },\eta \rangle ,\;\xi ,\eta \in H. \]\n\nIt is straightforward to verify that this formula implies that \( A \) is a linear transformation. It is bounded because\n\n\[ \mathop{\sup }\limits_{{\parallel \xi \parallel \leq 1}}\parallel {A\xi }\parallel = \mathop{\sup }\limits_{{\parallel \xi \parallel \leq 1,\parallel \eta \parallel \leq 1}}\left| \left\lbrack {\xi ,\eta }\right\rbrack \right| < \infty . \]\n\nThe uniqueness of the operator \( A \) is evident from the uniqueness assertion of the Riesz lemma for linear functionals.
|
Yes
|
Corollary 1. Let \( H, K \) be Hilbert spaces and let \( A \in \mathcal{B}\left( {H, K}\right) \) be a bounded operator from \( H \) to \( K \) . There is a unique operator \( {A}^{ * } \in \mathcal{B}\left( {K, H}\right) \) satisfying\n\n\[ \langle {A\xi },\eta {\rangle }_{K} = {\left\langle \xi ,{A}^{ * }\eta \right\rangle }_{H},\;\xi \in H,\;\eta \in K. \]
|
Proof. One simply applies the above results to the bounded sesquilin-ear form \( \left\lbrack {\cdot , \cdot }\right\rbrack \) defined on \( K \times H \) by \( \left\lbrack {\eta ,\xi }\right\rbrack = \langle \eta ,{A\xi }\rangle \) to deduce the existence of a unique operator \( {A}^{ * } \in \mathcal{B}\left( {K, H}\right) \) satisfying \( {\left\langle {A}^{ * }\eta ,\xi \right\rangle }_{H} = \langle \eta ,{A\xi }{\rangle }_{K} \), and then takes the complex conjugate of both sides.
|
Yes
|
Proposition 2.2.3. Let \( x, y \) be elements of a unital Banach algebra \( A \) satisfying \( {xy} = {yx} \). Then \( {e}^{x + y} = {e}^{x}{e}^{y} \).
|
Proof. Using formula (2.3), we have\n\n\[ \n{e}^{x}{e}^{y} = \mathop{\sum }\limits_{{p, q = 0}}^{\infty }\frac{1}{p!}{x}^{p}\frac{1}{q!}{y}^{q} = \mathop{\sum }\limits_{{n = 0}}^{\infty }\left( {\mathop{\sum }\limits_{{p + q = n}}\frac{1}{p!q!}{x}^{p}{y}^{q}}\right) .\n\]\n\nSince \( {xy} = {yx} \), the proof of the binomial theorem applies here to give\n\n\[ \n{\left( x + y\right) }^{n} = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {x}^{k}{y}^{n - k} = n!\mathop{\sum }\limits_{{p + q = n}}\frac{1}{p!q!}{x}^{p}{y}^{q}\n\]\n\nhence the right side of the preceding formula becomes\n\n\[ \n\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!}{\left( x + y\right) }^{n} = {e}^{x + y}\n\]
|
Yes
|
Corollary 1. Let \( A \) be a (perhaps noncommutative) unital \( {C}^{ * } \) -algebra. Then the spectrum of any self-adjoint element \( x \) of \( A \) is real.
|
Proof. Choose an element \( x = {x}^{ * } \) of \( A \), and let \( B \) be the norm-closure of the set of all polynomials in \( x \) . Then \( B \) is a commutative \( {C}^{ * } \) -subalgebra of \( A \) that contains the unit of \( A \), hence \( {\sigma }_{A}\left( x\right) \subseteq {\sigma }_{B}\left( x\right) \) . On the other hand, Theorem 2.2.4 implies that \( \omega \left( x\right) \) is real for every \( \omega \in \operatorname{sp}\left( B\right) \), and hence \( {\operatorname{sp}}_{A}\left( x\right) \subseteq {\sigma }_{B}\left( x\right) = \{ \omega \left( x\right) : \omega \in \operatorname{sp}\left( B\right) \} \subseteq \mathbb{R}. \
|
Yes
|
Corollary 2. Let \( A \) be a unital \( {C}^{ * } \) -algebra and let \( B \subseteq A \) be a \( {C}^{ * } \) - subalgebra of \( A \) that contains the unit of \( A \) . Then for every \( x \in B \) we have \( {\sigma }_{B}\left( x\right) = {\sigma }_{A}\left( x\right) \) . In particular, for every self-adjoint \( x \in A \) , \[ \parallel x\parallel = r\left( x\right) . \]
|
Proof. We know that \( {\sigma }_{A}\left( x\right) \subseteq {\sigma }_{B}\left( x\right) \) in general, and to prove the opposite inclusion it suffices to show that for any element \( x \in B \) which is invertible in \( A \) one has \( {x}^{-1} \in B \) . Fix such an \( x \) . Then \( {x}^{ * }x \) is a self-adjoint element of \( B \) that is also invertible in \( A \) . By the preceding corollary, \( {\sigma }_{B}\left( {{x}^{ * }x}\right) \) is real. In particular, every point of \( {\sigma }_{B}\left( {{x}^{ * }x}\right) \) is a boundary point. By Theorem 1.11.3, \( {\sigma }_{B}\left( {{x}^{ * }x}\right) = \) \( \partial {\sigma }_{B}\left( {{x}^{ * }x}\right) \subseteq {\sigma }_{A}\left( {{x}^{ * }x}\right) \) . Since \( 0 \notin {\sigma }_{A}\left( {{x}^{ * }x}\right) ,0 \notin {\sigma }_{B}\left( {{x}^{ * }x}\right) \), and hence \( {x}^{ * }x \) is invertible in \( B \), equivalently, \( {\left( {x}^{ * }x\right) }^{-1} \in B \) . Obviously, \( {\left( {x}^{ * }x\right) }^{-1}{x}^{ * } \) is a left inverse of \( x \) ; hence \( {x}^{-1} = {\left( {x}^{ * }x\right) }^{-1}{x}^{ * } \) must belong to \( B \) . The assertion that \( \parallel x\parallel = r\left( x\right) \) follows after an application of Theorem 2.2.4 to the \( {C}^{ * } \) -subalgebra of \( A \) generated by \( x \) and 1 .
|
Yes
|
Proposition 2.5.4. Every nonunital Banach \( * \) -algebra can be embedded as a maximal ideal of codimension 1 in a unital Banach \( * \) -algebra for which \( \parallel \mathbf{1}\parallel = 1 \) .
|
Proof. Let \( A \) be a nonunital Banach \( * \) -algebra. The vector space \( A \oplus \mathbb{C} \) can be made into a \( * \) -algebra \( \widetilde{A} \) by introducing the operations\n\n\[ \left( {a,\lambda }\right) \cdot \left( {b,\mu }\right) = \left( {{ab} + {\lambda b} + {\mu a},{\lambda \mu }}\right) ,\;{\left( a,\lambda \right) }^{ * } = \left( {{a}^{ * },\bar{\lambda }}\right) .\n\]\n\nThe element \( \mathbf{1} = \left( {0,1}\right) \) is a unit for \( \widetilde{A} \), and we have \( \left( {a,\lambda }\right) = a + \lambda \mathbf{1} \) . Obviously, \( A \) is a maximal ideal of codimension 1 in \( \widetilde{A} \) . \( \widetilde{A} \) becomes a Banach \( * \) -algebra by way of the norm \( \parallel \left( {a,\lambda }\right) \parallel = \parallel a\parallel + \left| \lambda \right| \), with respect to which the inclusion map of \( A \) in \( \widetilde{A} \) is an isometric \( * \) -homomorphism.
|
Yes
|
Proposition 2.7.1. A spectral measure \( P \) has the following properties:\n\n(1) \( {E}_{1} \subseteq {E}_{2} \Rightarrow P\left( {E}_{1}\right) \leq P\left( {E}_{2}\right) \) .\n\n(2) \( E \cap F = \varnothing \Rightarrow P\left( E\right) \bot P\left( F\right) \) .\n\n(3) For every \( E, F \in \mathcal{B}, P\left( {E \cap F}\right) = P\left( E\right) P\left( F\right) \) .
|
Proof. The first assertion follows from finite additivity of \( P \), together with the decomposition \( F = E \cup \left( {F \smallsetminus E}\right) \) and the fact that \( P\left( {F \smallsetminus E}\right) \geq 0 \) .\n\nFor (2), we can write\n\n\[ \mathbf{1} = P\left( {E \cup \left( {\mathbb{C} \smallsetminus E}\right) }\right) = P\left( E\right) + P\left( {\mathbb{C} \smallsetminus E}\right) . \]\n\nHence by \( \left( 1\right), P\left( F\right) \leq P\left( {\mathbb{C} \smallsetminus E}\right) = \mathbf{1} - P\left( E\right) \), the latter being the projection onto \( P\left( E\right) {H}^{ \bot } \) .\n\nTo deduce (3) from (2), one can write \( P\left( E\right) = P\left( {E \cap F}\right) + P\left( {E \smallsetminus F}\right) \), \( P\left( F\right) = P\left( {E \cap F}\right) + P\left( {F \smallsetminus E}\right) \), and observe that because of (2), \( P\left( {E \cap F}\right) \), \( P\left( {E \smallsetminus F}\right) \), and \( P\left( {F \smallsetminus E}\right) \) are mutually orthogonal projections.\n\nThese observations imply that the projections \( P\left( {E}_{1}\right), P\left( {E}_{2}\right) ,\ldots \) appearing on the right of (2.17) are mutually orthogonal, so that the infinite sum has a clear meaning.
|
Yes
|
Proposition 2.8.1. Let \( N \) be a normal operator acting on an infinite-dimensional Hilbert space \( H \) . For every accumulation point \( \lambda \in \sigma \left( N\right) \) there is an orthonormal sequence \( {\xi }_{1},{\xi }_{2},\ldots \) in \( H \) such that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\begin{Vmatrix}{N{\xi }_{n} - \lambda {\xi }_{n}}\end{Vmatrix} = 0 \]
|
Proof. By the Spectral Theorem we may assume that \( H = {L}^{2}\left( {X,\mu }\right) \) has been coordinatized by a \( \sigma \) -finite measure space and that \( N = {M}_{f} \) is multiplication by an \( {L}^{\infty } \) function. By Theorem 2.1.4 the spectrum of \( N \) is the essential range \( \Lambda \) of \( f \) .\n\nSince \( \lambda \) is an accumulation point of \( \Lambda \), we can find a sequence of distinct points \( {\lambda }_{n} \in \Lambda \) that converges to \( \lambda \) . For each \( n \) choose \( {\epsilon }_{n} > 0 \) small enough that \( {\epsilon }_{n} \rightarrow 0 \) and the disks \( {D}_{n} = \left\{ {z \in \mathbb{C} : \left| {z - {\lambda }_{n}}\right| < {\epsilon }_{n}}\right\}, n = 1,2,\ldots \) , are mutually disjoint. For each \( n \) the set \( \left\{ {p \in X : f\left( p\right) \in {D}_{n}}\right\} \) has positive measure because \( {\lambda }_{n} \) belongs to the essential range of \( f \) ; and by \( \sigma \) -finiteness there is a subset \( {E}_{n} \subseteq \left\{ {p \in X : f\left( p\right) \in {D}_{n}}\right\} \) of finite positive measure, \( n = 1,2,\ldots \) Considered as elements of \( {L}^{2}\left( {X,\mu }\right) \), the characteristic functions \( {\chi }_{{E}_{1}},{\chi }_{{E}_{2}},\ldots \) are mutually orthogonal because the sets \( {E}_{1},{E}_{2},\ldots \) are mutually disjoint. Moreover,\n\n\[ \left| {f - \lambda }\right| \cdot {\chi }_{{E}_{n}} \leq \left( {\left| {f - {\lambda }_{n}}\right| + \left| {{\lambda }_{n} - \lambda }\right| }\right) {\chi }_{{E}_{n}} \leq \left( {{\epsilon }_{n} + \left| {{\lambda }_{n} - \lambda }\right| }\right) {\chi }_{{E}_{n}}. \]\n\nIt follows that\n\n\[ \begin{Vmatrix}{\left( {N - \lambda }\right) {\chi }_{{E}_{n}}}\end{Vmatrix} \leq \left( {{\epsilon }_{n} + \left| {{\lambda }_{n} - \lambda }\right| }\right) {\begin{Vmatrix}{\chi }_{{E}_{n}}\end{Vmatrix}}_{2} = \left( {{\epsilon }_{n} + \left| {{\lambda }_{n} - \lambda }\right| }\right) \mu {\left( {E}_{n}\right) }^{1/2}, \]\n\nand the orthonormal sequence can be taken as \( {\xi }_{n} = \mu {\left( {E}_{n}\right) }^{-1/2}{\chi }_{{E}_{n}}, n = \) \( 1,2,\ldots \) .
|
Yes
|
Proposition 2.8.3. The trace has the following properties:\n\n(1) \( \operatorname{trace}{A}^{ * }A = \operatorname{trace}A{A}^{ * } \), for any \( A \in \mathcal{B}\left( H\right) \) .
|
Proof. For (1), consider the double sequence of nonnegative terms \( {\left| \left\langle A{e}_{p},{e}_{q}\right\rangle \right| }^{2} = {\left| \left\langle {e}_{p},{A}^{ * }{e}_{q}\right\rangle \right| }^{2}, p, q = 1,2,\ldots \) Summing first on \( q \) and then on \( p \), we obtain\n\n\[ \mathop{\sum }\limits_{{p = 1}}^{\infty }\mathop{\sum }\limits_{{q = 1}}^{\infty }{\left| \left\langle A{e}_{p},{e}_{q}\right\rangle \right| }^{2} = \mathop{\sum }\limits_{{p = 1}}^{\infty }{\begin{Vmatrix}A{e}_{p}\end{Vmatrix}}^{2} \]\n\nwhile summing in the opposite order gives\n\n\[ \mathop{\sum }\limits_{{q = 1}}^{\infty }\mathop{\sum }\limits_{{p = 1}}^{\infty }{\left| \left\langle {e}_{p},{A}^{ * }{e}_{q}\right\rangle \right| }^{2} = \mathop{\sum }\limits_{{q = 1}}^{\infty }{\begin{Vmatrix}{A}^{ * }{e}_{q}\end{Vmatrix}}^{2}. \]\n\nSince the sum of a nonnegative double sequence is independent of the order of summation, this proves (1).
|
Yes
|
Proposition 2.8.4. Every Hilbert-Schmidt operator \( A \) is compact, and satisfies \( \parallel A{\parallel }^{2} \leq \operatorname{trace}{A}^{ * }A \) .
|
Proof. We first prove the inequality \( \parallel A{\parallel }^{2} \leq \operatorname{trace}{A}^{ * }A \) . Indeed, for every unit vector \( e \) we can find an orthonormal basis \( {e}_{1},{e}_{2},\ldots \) starting with \( {e}_{1} = e \) . Hence \( \parallel {Ae}{\parallel }^{2} \leq \mathop{\sum }\limits_{n}{\begin{Vmatrix}A{e}_{n}\end{Vmatrix}}^{2} = \operatorname{trace}{A}^{ * }A \), and since \( e \) is arbitrary we obtain\n\n\[ \parallel A{\parallel }^{2} = \mathop{\sup }\limits_{{\parallel e\parallel = 1}}\parallel {Ae}{\parallel }^{2} \leq \operatorname{trace}{A}^{ * }A. \]\n\nTo see that every Hilbert-Schmidt operator \( A \) is compact, fix an orthonormal basis \( {e}_{1},{e}_{2},\ldots \) and for every \( n \geq 1 \) let \( {Q}_{n} \) be the projection onto the subspace spanned by \( {e}_{n + 1},{e}_{n + 2},\ldots \) Obviously, \( {F}_{n} = A\left( {1 - {Q}_{n}}\right) \) is a finite-rank operator, and by the preceding paragraph we have\n\n\[ {\begin{Vmatrix}A - {F}_{n}\end{Vmatrix}}^{2} = {\begin{Vmatrix}A{Q}_{n}\end{Vmatrix}}^{2} \leq \operatorname{trace}\left( {{Q}_{n}{A}^{ * }A{Q}_{n}}\right) = \mathop{\sum }\limits_{{k = n + 1}}^{\infty }{\begin{Vmatrix}A{e}_{k}\end{Vmatrix}}^{2}. \]\n\nThe right side tends to 0 as \( n \rightarrow \infty \) because \( \mathop{\sum }\limits_{k}{\begin{Vmatrix}A{e}_{k}\end{Vmatrix}}^{2} < \infty \) . Hence \( A = \mathop{\lim }\limits_{n}{F}_{n} \) is the norm limit of a sequence of finite-rank operators, and is therefore compact.
|
Yes
|
Proposition 2.8.6. Let \( \\left( {X,\\mu }\\right) \) be a separable \( \\sigma \) -finite measure space. For every function \( k \\in {L}^{2}\\left( {X \\times X,\\mu \\times \\mu }\\right) \) there is a unique bounded operator \( {A}_{k} \) on \( {L}^{2}\\left( {X,\\mu }\\right) \) satisfying
|
\[ {A}_{k}\\xi \\left( x\\right) = {\\int }_{X}k\\left( {x, y}\\right) \\xi \\left( y\\right) {d\\mu }\\left( y\\right) ,\\;\\xi \\in {L}^{2}\\left( {X,\\mu }\\right) . \] The map \( k \\mapsto {A}_{k} \) is an isometric isomorphism of the Hilbert space \( {L}^{2}(X \\times \) \( X,\\mu \\times \\mu \) ) onto the Hilbert space \( {\\mathcal{L}}^{2} \) of all Hilbert-Schmidt operators on \( {L}^{2}\\left( {X,\\mu }\\right) \) .
|
Yes
|
The involution in \( {A}^{e} \) satisfies \( \begin{Vmatrix}{{X}^{ * }X}\end{Vmatrix} = \parallel X{\parallel }^{2} \) for every \( X \in {A}^{e} \), and \( {A}^{e} \) is closed in the operator norm of \( \mathcal{B}\left( A\right) \) ; hence it is a unital \( {C}^{ * } \) -algebra. Moreover, the regular representation is an isometric *-isomorphism of \( A \) onto a maximal ideal of codimension one in \( {A}^{e} \).
|
Notice first that \( \begin{Vmatrix}{L}_{a}\end{Vmatrix} = \parallel a\parallel \) for every \( a \in A \) . Indeed, \( \leq \) is true for any Banach algebra, and the opposite inequality follows for an element \( a \) of norm 1 because\n\n\[ \begin{Vmatrix}{L}_{a}\end{Vmatrix} \geq \begin{Vmatrix}{{L}_{a}\left( {a}^{ * }\right) }\end{Vmatrix} = \begin{Vmatrix}{a{a}^{ * }}\end{Vmatrix} = {\begin{Vmatrix}{a}^{ * }\end{Vmatrix}}^{2} = \parallel a{\parallel }^{2} = 1. \]\n\nThe set \( \left\{ {{L}_{a} : a \in A}\right\} \) is obviously an ideal in \( {A}^{e} \) of codimension at most one. If the codimension were zero, then the identity operator would have the form \( {L}_{f} \) for some element \( f \in A \) ; that would imply \( f \) was a unit for \( A \) , contrary to hypothesis. Hence \( \left\{ {{L}_{a} : a \in A}\right\} \) has codimension one. Since \( L \) is an isometry, this ideal must be closed in the operator norm of \( \mathcal{B}\left( A\right) \) ; and since \( {A}^{e} \) is obtained from this ideal by adjoining the one-dimensional space spanned by 1, it follows that \( {A}^{e} \) must also be norm closed.\n\nIt remains to show that the involution in \( {A}^{e} \) satisfies \( \begin{Vmatrix}{{X}^{ * }X}\end{Vmatrix} = \parallel X{\parallel }^{2} \) . By Remark 2.9.1, it is enough to verify the inequality \( \parallel X{\parallel }^{2} \leq \begin{Vmatrix}{{X}^{ * }X}\end{Vmatrix} \) for \( X = {L}_{a} + ╏ \) in \( {A}^{e} \) . For such an \( X \), we have\n\n\[ \parallel X{\parallel }^{2} = \mathop{\sup }\limits_{{\parallel b\parallel \leq 1}}{\begin{Vmatrix}\left( {L}_{a} + \lambda \mathbf{1}\right) \left( b\right) \end{Vmatrix}}^{2} = \mathop{\sup }\limits_{{\parallel b\parallel \leq 1}}\parallel {ab} + {\lambda b}{\parallel }^{2} \]\n\n\[ = \mathop{\sup }\limits_{{\parallel b\parallel \leq 1}}\begin{Vmatrix}{{\left( ab + \lambda b\right) }^{ * }\left( {{ab} + {\lambda b}}\right) }\end{Vmatrix} = \mathop{\sup }\limits_{{\parallel b\parallel \leq 1}}\begin{Vmatrix}{{b}^{ * }\left( {{X}^{ * }X\left( b\right) }\right) }\end{Vmatrix} \]\n\n\[ \leq \mathop{\sup }\limits_{{\parallel b\parallel \leq 1}}\begin{Vmatrix}{{X}^{ * }X\left( b\right) }\end{Vmatrix} \leq \begin{Vmatrix}{{X}^{ * }X}\end{Vmatrix}. \]
|
Yes
|
Proposition 2.9.3. Every \( * \) -homomorphism \( \pi : A \rightarrow B \) of \( {C}^{ * } \) -algebras has norm at most \( 1 \) . If \( \pi \) has trivial kernel, then it is an isometry.
|
Proof. Suppose first that \( A \) has a unit \( {\mathbf{1}}_{A} \) . By passing from \( B \) to the closure of the \( * \) -subalgebra \( \pi \left( A\right) \) if necessary, we may assume that \( \pi \left( A\right) \) is dense in \( B \) . In this case, \( \pi \left( {\mathbf{1}}_{A}\right) \) is the unit \( {\mathbf{1}}_{B} \) of \( B \) . Thus we may argue as we did for nondegenerate representations. For example, since \( \pi \) must map invertible elements of \( A \) to invertible elements of \( B \), it follows that \( \sigma \left( {\pi \left( x\right) }\right) \subseteq \sigma \left( x\right) \) for every element \( x \in A \) . Corollary 2 of Theorem 2.2.4 implies that for self-adjoint elements \( x \in A \) we have\n\n\[ \parallel \pi \left( x\right) \parallel = r\left( {\pi \left( x\right) }\right) \leq r\left( x\right) = \parallel x\parallel \]\n\nso that for general elements \( z \in A \) we have\n\n\[ \parallel \pi \left( z\right) {\parallel }^{2} = \begin{Vmatrix}{\pi {\left( z\right) }^{ * }\pi \left( x\right) }\end{Vmatrix} = \begin{Vmatrix}{\pi \left( {{z}^{ * }z}\right) }\end{Vmatrix} \leq \begin{Vmatrix}{{z}^{ * }z}\end{Vmatrix} = \parallel z{\parallel }^{2}. \]\n\nIf, in addition, \( \pi \) has trivial kernel, then we claim that \( \parallel \pi \left( x\right) \parallel = \parallel x\parallel \) for every \( x \in A \) . As above, this reduces to the case where \( x = {x}^{ * } \) is selfadjoint; and by Corollary 2 of Theorem 2.2.4 it is enough to show that \( x \) and \( \pi \left( x\right) \) have the same spectrum when \( x = {x}^{ * } \) . We have already seen that \( \sigma \left( {\pi \left( x\right) }\right) \subseteq \sigma \left( x\right) \) . For the opposite inclusion, suppose that \( \lambda \) is a point of \( \sigma \left( x\right) \) that does not belong to \( \sigma \left( {\pi \left( x\right) }\right) \) . There is a continuous function \( f : \sigma \left( x\right) \rightarrow \mathbb{R} \) such that \( f \) vanishes on \( \sigma \left( {\pi \left( x\right) }\right) \) and \( f\left( \lambda \right) \neq 0 \) . Since \( f = 0 \) on \( \sigma \left( {\pi \left( x\right) }\right) \), we must have \( f\left( {\pi \left( x\right) }\right) = 0 \) . Notice that \( f\left( {\pi \left( x\right) }\right) = \pi \left( {f\left( x\right) }\right) \) (this is obvious if \( f \) is a polynomial, and it follows for general continuous \( f \) by an application of the Weierstrass approximation theorem and the previously established fact that \( \pi \) is a bounded linear map of \( A \) to \( B \) ). But \( \pi \left( {f\left( x\right) }\right) = 0 \) implies that \( f\left( x\right) = 0 \) because \( \pi \) has trivial kernel; in turn, \( f\left( x\right) = 0 \) implies that \( f = 0 \) on \( \sigma \left( x\right) \), contradicting the fact that \( f\left( \lambda \right) \neq 0 \) .
|
Yes
|
Corollary 1. Let \( A \) be a complex algebra with involution. If there is a norm on \( A \) that makes it into a \( {C}^{ * } \) -algebra, then that norm is unique.
|
Proof. Let \( \parallel \cdot {\parallel }_{1} \) and \( \parallel \cdot {\parallel }_{2} \) be two (complete) Banach algebra norms on \( A \) satisfying \( {\begin{Vmatrix}{x}^{ * }x\end{Vmatrix}}_{k} = \parallel x{\parallel }_{k}^{2} \) for \( x \in A \), and let \( {A}_{k} \) be the algebra \( A \) considered as a \( {C}^{ * } \) -algebra in each norm respectively, \( k = 1,2 \) . The identity map of \( A \) can be regarded as a \( * \) -isomorphism of \( {A}_{1} \) onto \( {A}_{2} \) . By Proposition 2.9.3 this map must be an isometry; hence \( \parallel x{\parallel }_{1} = \parallel x{\parallel }_{2} \) for all \( x \in A \) .
|
Yes
|
Corollary 1. A bounded operator \( T \) belongs to \( \mathcal{F}\left( E\right) \) iff there is an operator \( S \in \mathcal{B}\left( E\right) \) such that \( \mathbf{1} - {ST} \) and \( \mathbf{1} - {TS} \) are both compact.
|
Proof of Corollary 1. If \( \dot{T} \) is invertible in \( \mathcal{B}\left( E\right) /\mathcal{K}\left( E\right) \), then its inverse is an element of the form \( \dot{S} \) for some \( S \in \mathcal{B}\left( E\right) \), and the operators \( 1 - {ST} \) and \( 1 - {TS} \) must be compact because they map to 0 in the quotient algebra. The converse follows immediately from Atkinson's theorem.
|
No
|
Corollary 2. The set \( \mathcal{F}\left( E\right) \) of Fredholm operators is open in the norm topology of \( \mathcal{B}\left( E\right) \), it is stable under compact perturbations, it contains all invertible operators of \( \mathcal{B}\left( E\right) \), and it is closed under operator multiplication.
|
Proof of Corollary 2. Atkinson’s theorem implies that \( \mathcal{F}\left( E\right) \) is the inverse image of the general linear group of \( \mathcal{B}\left( E\right) /\mathcal{K}\left( E\right) \) under the continuous homomorphism \( T \mapsto T \) ; hence these assertions all follow from the fact that the set of invertible elements of a unital Banach algebra \( A \) forms a group that is open in the norm topology of \( A \) .
|
Yes
|
Corollary 2 (Stability of index). For every Fredholm operator \( A \in \) \( \mathcal{B}\left( E\right) \) and compact operator \( K \) ,\n\n\[ \text{ind}\left( {A + K}\right) = \text{ind}A\text{.} \]
|
Proof. By Atkinson’s theorem there is a Fredholm operator \( B \in \mathcal{B}\left( E\right) \) such that \( {AB} = 1 + L \) with \( L \in \mathcal{K}\left( E\right) \) . We have \( \left( {A + K}\right) B = 1 + {L}^{\prime } \) where \( {L}^{\prime } = L - {KB} \in \mathcal{K}\left( E\right) \) . As we have already pointed out, the Fredholm alternative implies that \( \operatorname{ind}\left( {\mathbf{1} + L}\right) = \operatorname{ind}\left( {\mathbf{1} + {L}^{\prime }}\right) = 0 \) ; hence ind \( {AB} = \) ind \( \left( {A + K}\right) B = 0 \) . Using Corollary 1 one has\n\n\[ \operatorname{ind}\left( {A + K}\right) + \operatorname{ind}B = \operatorname{ind}\left( {A + K}\right) B = \operatorname{ind}{AB} = \operatorname{ind}A + \operatorname{ind}B \]\n\nand the formula follows after one cancels the integer ind \( B \) .
|
Yes
|
Corollary 3 (Continuity of index). Given a Fredholm operator \( A \) , let \( {A}_{1},{A}_{2},\ldots \) be a sequence of bounded operators that converges to \( A \) , \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\begin{Vmatrix}{{A}_{n} - A}\end{Vmatrix} = 0 \) . There is an \( {n}_{0} \) such that for \( n \geq {n}_{0},{A}_{n} \) is a Fredholm operator with ind \( {A}_{n} = \operatorname{ind}A \) .
|
Proof. By Atkinson’s theorem, \( \mathcal{F}\left( E\right) \) is open, so that \( {A}_{n} \in \mathcal{F}\left( E\right) \) for sufficiently large \( n \) . We can also find a Fredholm operator \( B \) such that \( {AB} = \mathbf{1} + K \) with \( K \) compact. Writing \( {A}_{n} = A + {T}_{n} \) with \( \begin{Vmatrix}{T}_{n}\end{Vmatrix} \rightarrow 0 \) as \( n \rightarrow \infty \), we can find \( {n}_{0} \) so that, for \( n \geq {n}_{0},\begin{Vmatrix}{{T}_{n}B}\end{Vmatrix} < 1 \) and hence \( \mathbf{1} + {T}_{n}B \) is invertible. For such \( n \), we have\n\n\[ \operatorname{ind}{A}_{n} + \operatorname{ind}B = \operatorname{ind}\left( {A + {T}_{n}}\right) B = \operatorname{ind}\left( {\mathbf{1} + {T}_{n}B + K}\right) .\n\]\n\nThe right side vanishes because \( 1 + {T}_{n}B + K \) is a compact perturbation of an invertible operator (see Exercise (1) below). On the other hand,\n\n\[ \operatorname{ind}A + \operatorname{ind}B = \operatorname{ind}{AB} = \operatorname{ind}\left( {\mathbf{1} + K}\right) = 0 \]\n\nby the Fredholm alternative; hence ind \( {A}_{n} = - \operatorname{ind}B = \operatorname{ind}A \) for sufficiently large \( n \) .
|
Yes
|
Corollary 2. Let \( \left\{ {{e}_{n} : n \in \mathbb{Z}}\right\} \) be a bilateral orthonormal basis for a Hilbert space \( H \), and let \( U \) be the bilateral shift defined on \( H \) by \( U{e}_{n} = {e}_{n + 1} \) , \( n \in \mathbb{Z} \) . Then the von Neumann algebra \( {W}^{ * }\left( U\right) \) generated by \( U \) is maximal abelian, and consists of all operators in \( \mathcal{B}\left( H\right) \) that commute with \( U \) .
|
Proof. We have seen that \( U \) is unitarily equivalent to the multiplication operator \( {M}_{\zeta } \) acting on \( {L}^{2}\left( \mathbb{T}\right) \) by \( {M}_{\zeta }\xi \left( z\right) = \zeta \left( z\right) \xi \left( z\right) ,\zeta \) being the current variable \( \zeta \left( z\right) = z, z \in \mathbb{T} \) . Since \( {M}_{\zeta } \) is unitary, any operator commuting with it must also commute with its adjoint \( {M}_{\zeta }^{ * } = {M}_{\zeta }^{-1} \) . On the other hand, since \( \zeta \) separates points of \( \mathbb{T} \), it follows from Theorem 4.1.3 that any operator commuting with \( \left\{ {{M}_{\zeta },{M}_{\zeta }^{ * }}\right\} \) must belong to the multiplication algebra of \( {L}^{2}\left( \mathbb{T}\right) \), and that the multiplication algebra coincides with the von Neumann algebra generated by \( U \) .
|
Yes
|
Proposition 4.2.1. \( {H}^{\infty } = \left\{ {\phi \in {L}^{\infty } : {M}_{\phi }{H}^{2} \subseteq {H}^{2}}\right\} \) .
|
Proof. Let \( \phi \in {L}^{\infty } \) . If \( \phi \in {H}^{\infty } \), then for \( n \geq 0 \) ,\n\n\[ \n{M}_{\phi }{\zeta }^{n} = {\zeta }^{n} \cdot \phi \in {\zeta }^{n} \cdot {H}^{2} \subseteq {H}^{2} \n\] \n\nhence \( {M}_{\phi } \) leaves \( {H}^{2} = \left\lbrack {1,\zeta ,{\zeta }^{2},\ldots }\right\rbrack \) invariant.\n\nConversely, if \( {M}_{\phi }{H}^{2} \subseteq {H}^{2} \), then \( \phi = {M}_{\phi }1 \in {H}^{2} \) .
|
Yes
|
Proposition 4.2.3. Let \( A \) be a bounded operator on \( {H}^{2} \) . The matrix of A relative to the natural basis \( \left\{ {{\zeta }^{n} : n = 0,1,2,\ldots }\right\} \) is a Toeplitz matrix iff \( {S}^{ * }{AS} = A \) .
|
Proof. The hypothesis on the matrix entries \( {a}_{ij} = \left\langle {A{\zeta }^{j},{\zeta }^{i}}\right\rangle \) of \( A \) is equivalent to requiring\n\n\[ \n{a}_{i + 1, j + 1} = {a}_{ij},\;i, j = 0,1,2,\ldots \n\] \n\nNoting that \( S{\zeta }^{n} = {\zeta }^{n + 1} \) for \( n \geq 0 \) we find that this is equivalent to the requirement that \n\n\[ \n\langle {S}^{ * }{AS}{\zeta }^{j},{\zeta }^{i}\rangle = \langle {AS}{\zeta }^{j}, S{\zeta }^{i}\rangle = \langle A{\zeta }^{j + 1},{\zeta }^{i + 1}\rangle = \langle A{\zeta }^{j},{\zeta }^{i}\rangle \n\] \n\nfor all \( i, j \geq 0 \) ; hence it is equivalent to the formula \( {S}^{ * }{AS} = A \) .
|
Yes
|
Every Toeplitz operator \( {T}_{\phi },\phi \in {L}^{\infty } \), satisfies\n\n\[ \inf \left\{ {\begin{Vmatrix}{{T}_{\phi } + K}\end{Vmatrix} : K \in \mathcal{K}}\right\} = \begin{Vmatrix}{T}_{\phi }\end{Vmatrix} = \parallel \phi {\parallel }_{\infty }.\n\]\n\nIn particular, the only compact Toeplitz operator is 0.
|
Proof. Let \( S \) be the unilateral shift acting on \( {H}^{2} \) by \( S{\zeta }^{n} = {\zeta }^{n + 1}, n \geq 0 \) . It suffices to show that for any operator \( A \in \mathcal{B}\left( {H}^{2}\right) \) satisfying \( {S}^{ * }{AS} = A \) and for any compact operator \( K \) we have\n\n\[ \parallel A + K\parallel \geq \parallel A\parallel \text{.}\]\n\nThe hypothesis \( {S}^{ * }{AS} = A \) implies that \( {S}^{*n}A{S}^{n} = A \) for every \( n = 1,2,\ldots \) ; noting that \( {P}_{n} = {S}^{n}{S}^{*n} \) is the projection onto \( \left\lbrack {{\zeta }^{n},{\zeta }^{n + 1},\ldots }\right\rbrack \) we have\n\n\[ \parallel A + K\parallel \geq \begin{Vmatrix}{{P}_{n}\left( {A + K}\right) {P}_{n}}\end{Vmatrix} = \begin{Vmatrix}{{S}^{*n}\left( {A + K}\right) {S}^{n}}\end{Vmatrix} = \begin{Vmatrix}{A + {S}^{*n}K{S}^{n}}\end{Vmatrix}.\n\nThe norm of the compression of \( K \) to the subspace \( \left\lbrack {{\zeta }^{n},{\zeta }^{n + 1},\ldots }\right\rbrack \) is given by \( \begin{Vmatrix}{{P}_{n}K{P}_{n}}\end{Vmatrix} = \begin{Vmatrix}{{S}^{*n}K{S}^{n}}\end{Vmatrix} \), which tends to 0 as \( n \rightarrow \infty \) because \( K \) is compact and \( {P}_{n} \downarrow 0 \) . Thus\n\n\[ \parallel A + K\parallel \geq \mathop{\lim }\limits_{{n \rightarrow \infty }}\begin{Vmatrix}{A + {S}^{*n}K{S}^{n}}\end{Vmatrix} = \parallel A\parallel \]\n\nas asserted.
|
Yes
|
Proposition 4.3.1. Let \( f, g \in {L}^{\infty } \) . If one of the functions \( f, g \) is continuous, then \( {T}_{fg} - {T}_{f}{T}_{g} \in \mathcal{K} \) .
|
Proof. Since \( {T}_{fg}^{ * } = {T}_{\bar{f}\bar{g}} \) and \( {\left( {T}_{f}{T}_{g}\right) }^{ * } = {T}_{\bar{g}}{T}_{\bar{f}} \), it suffices to prove the following assertion: If \( f \in C\left( \mathbb{T}\right) \) and \( g \in {L}^{\infty } \), then \( {T}_{fg} - {T}_{f}{T}_{g} \in \mathcal{K} \) . Moreover, since \( C\left( \mathbb{T}\right) \) is the norm-closed linear span of the monomials \( {\zeta }^{n}, n \in \mathbb{Z} \) , and \( \mathcal{K} \) is a norm-closed linear space, we may reduce to the case \( f = {\zeta }^{n} \) and \( g \in {L}^{\infty }, n \in \mathbb{Z} \) .\n\nIf \( n \geq 0 \), then \( {\zeta }^{n} \in {H}^{\infty } \), so that by (4.3.1) we have \( {T}_{f{\zeta }^{n}} = {T}_{f}{T}_{{\zeta }^{n}} \) . Thus \( {T}_{fg} - {T}_{f}{T}_{g} = 0 \) in this case.\n\nIf \( n < 0 \), say \( n = - m \) with \( m \geq 1 \), then \( {\zeta }^{n} \) is the complex conjugate of the \( {H}^{\infty } \) function \( {\zeta }^{m} \), and another application of (4.3.1) gives \( {T}_{f{\zeta }^{n}} = \) \( {T}_{{\zeta }^{n}}{T}_{f} = {S}^{*m}{T}_{f} \) . Noting that \( {S}^{*m}{T}_{f}{S}^{m} = {T}_{f} \) (by iterating the basic formula \( {S}^{ * }{T}_{f}S = {T}_{f} \) valid for any Toeplitz operator) we can write\n\n\[ \n{T}_{f}{T}_{{\zeta }^{n}} = {T}_{f}{S}^{*m} = {S}^{*m}{T}_{f}{S}^{m}{S}^{*m} = {S}^{*m}{T}_{f} - {S}^{*m}{T}_{f}\left( {\mathbf{1} - {S}^{m}{S}^{*m}}\right) .\n\]\n\nHence\n\n\[ \n{T}_{f{\zeta }^{n}} - {T}_{f}{T}_{{\zeta }^{n}} = {S}^{*m}{T}_{f} - {T}_{f}{S}^{*m} = - {S}^{*m}{T}_{f}\left( {\mathbf{1} - {S}^{m}{S}^{*m}}\right) ,\n\]\n\nwhich is a finite-rank operator, since \( 1 - {S}^{m}{S}^{*m} \) is the projection onto \( \left\lbrack {1,\zeta ,{\zeta }^{2},\ldots ,{\zeta }^{m - 1}}\right\rbrack . \)
|
Yes
|
Corollary 1. The Fredholm operators in \( \mathcal{T} \) are precisely the operators of the form \( {T}_{f} + K \) where \( f \) is an invertible symbol in \( C{\left( \mathbb{T}\right) }^{-1} \) and \( K \in \mathcal{K} \) .
|
Consider a Fredholm operator in \( \mathcal{T} \), say \( {T}_{f} + K \) where \( f \in C\left( \mathbb{T}\right) \) has no zeros on the circle and \( K \) is a compact operator. By the stability results of Chapter 3 we see that \( {T}_{f} \) is also a Fredholm operator and\n\n\[ \text{ind}\left( {{T}_{f} + K}\right) = \operatorname{ind}\left( {T}_{f}\right) \text{.} \]\n\nWe know that for \( f = \zeta ,{T}_{f} \) is the shift; hence ind \( \left( {T}_{f}\right) = - 1 \) . However, we still lack tools for computing the index of more general Toeplitz operators with symbols in \( C{\left( \mathbb{T}\right) }^{-1} \) . This issue will be taken up in the following section.
|
No
|
Proposition 4.4.1. For every function \( F \in C\left\lbrack {0,1}\right\rbrack \) such that \( F\left( t\right) \neq 0 \) for every \( t \in \left\lbrack {0,1}\right\rbrack \), there is a function \( G \in C\left\lbrack {0,1}\right\rbrack \) such that\n\n\[ F\left( t\right) = {e}^{G\left( t\right) },\;0 \leq t \leq 1. \]
|
Proof. On the domain \( \{ z \in \mathbb{C} : \left| {z - 1}\right| < 1\} \), let \( \log z \) be the principal branch of the logarithm,\n\n\[ \log z = - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( 1 - z\right) }^{n}}{n} \]\n\nThe log function is holomorphic, satisfies \( \log 1 = 0 \), and of course \( {e}^{\log z} = z \) for \( \left| {z - 1}\right| < 1 \) . Let\n\n\[ M = \mathop{\sup }\limits_{{0 \leq t \leq 1}}{\left| F\left( t\right) \right| }^{-1} \]\n\nBy uniform continuity of \( F \), we can find a finite partition of the interval \( \left\lbrack {0,1}\right\rbrack ,0 = {t}_{0} < {t}_{1} < \cdots < {t}_{n} = 1 \), such that\n\n\[ \mathop{\sup }\limits_{{{t}_{k - 1} \leq t \leq {t}_{k}}}\left| {F\left( t\right) - F\left( {t}_{k - 1}\right) }\right| < \frac{1}{2M}. \]\n\nIt follows that for \( k = 1,\ldots, n \) and \( t \in \left\lbrack {{t}_{k - 1},{t}_{k}}\right\rbrack \) ,\n\n(4.10)\n\n\[ \left| {1 - \frac{F\left( t\right) }{F\left( {t}_{k - 1}\right) }}\right| = \frac{\left| F\left( t\right) - F\left( {t}_{k - 1}\right) \right| }{\left| F\left( {t}_{k - 1}\right) \right| } \leq \frac{1}{{2M}\left| {F\left( {t}_{k - 1}\right) }\right| } \leq \frac{1}{2} < 1. \]\n\nSetting\n\n\[ {G}_{k}\left( t\right) = \log \left( {F\left( t\right) /F\left( {t}_{k - 1}\right) }\right) ,\;{t}_{k - 1} \leq t \leq {t}_{k}, \]\n\nwe find that \( {G}_{k} \) is continuous, \( {G}_{k}\left( {t}_{k - 1}\right) = 0,{G}_{k}\left( {t}_{k}\right) = \log \left( {F\left( {t}_{k}\right) /F\left( {t}_{k - 1}\right) }\right) \) , and it satisfies\n\n\[ F\left( t\right) = F\left( {t}_{k - 1}\right) {e}^{{G}_{k}\left( t\right) } \]\n\nthroughout the interval \( \left\lbrack {{t}_{k - 1},{t}_{k}}\right\rbrack \) . There is an obvious way to piece the \( {G}_{k} \) together so as to obtain a continuous function \( G : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{C} \), namely \( G\left( t\right) = {G}_{1}\left( t\right) \) for \( t \in \left\lbrack {0,{t}_{1}}\right\rbrack \) and, for \( k = 2,\ldots, n \),\n\n\[ G\left( t\right) = {G}_{1}\left( {t}_{1}\right) + \cdots + {G}_{k - 1}\left( {t}_{k - 1}\right) + {G}_{k}\left( t\right) ,\;t \in \left\lbrack {{t}_{k - 1},{t}_{k}}\right\rbrack \]\n\nIt follows that\n\n\[ F\left( t\right) = F\left( 0\right) {e}^{G\left( t\right) },\;0 \leq t \leq 1. \]\n\nWriting \( F\left( 0\right) \in {\mathbb{C}}^{ \times } \) as an exponential \( F\left( 0\right) = {e}^{{z}_{0}} \), we obtain a continuous function \( \widetilde{G} \) satisfying \( F = {e}^{\widetilde{G}} \) by way of \( \widetilde{G}\left( t\right) = G\left( t\right) + {z}_{0} \) .
|
Yes
|
Proposition 4.4.2. For \( f, g \in G = C{\left( \mathbb{T}\right) }^{-1} \) ,\n\n(1) \( \# \left( {fg}\right) = \# \left( f\right) + \# \left( g\right) \) .\n\n(2) \( \# \left( f\right) = n \in \mathbb{Z} \) iff there is a function \( h \in C\left( \mathbb{T}\right) \) such that \( f = {\zeta }^{n}{e}^{h} \) .
|
Proof. For (1), pick continuous functions \( F, G : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{C} \) such that\n\n\[ f\left( {e}^{2\pi it}\right) = {e}^{{2\pi iF}\left( t\right) },\;g\left( {e}^{2\pi it}\right) = {e}^{{2\pi iG}\left( t\right) },\;t \in \left\lbrack {0,1}\right\rbrack .\n\]\n\nThen\n\n\[ f\left( {e}^{2\pi it}\right) g\left( {e}^{2\pi it}\right) = {e}^{{2\pi i}\left( {F\left( t\right) + G\left( t\right) }\right) },\;t \in \left\lbrack {0,1}\right\rbrack ,\n\]\n\nand the winding number of \( {fg} \) is given by\n\n\[ F\left( 1\right) + G\left( 1\right) - \left( {F\left( 0\right) + G\left( 0\right) }\right) = \# \left( f\right) + \# \left( g\right) .\n\]\n\nFor (2), consider first the case \( n = 0 \) . If \( f = {e}^{h} \) is the exponential of a function \( h \in C\left( \mathbb{T}\right) \), then we have\n\n(4.13)\n\n\[ f\left( {e}^{2\pi it}\right) = {e}^{{2\pi iF}\left( t\right) },\;0 \leq t \leq 1\n\]\n\nwhere \( F\left( t\right) = {\left( 2\pi \right) }^{-1}h\left( {e}^{2\pi it}\right) \) . Clearly, \( F\left( 1\right) = F\left( 0\right) \), so that \( \# \left( f\right) = F\left( 1\right) - \) \( F\left( 0\right) = 0 \) . Conversely, if \( \# \left( f\right) = 0 \), then there is a function \( F \in C\left\lbrack {0,1}\right\rbrack \) such that (4.13) is satisfied and \( F\left( 1\right) - F\left( 0\right) = \# \left( f\right) = 0 \) . Since \( F \) is periodic, we have \( f = {e}^{h} \), where \( h \in C\left( \mathbb{T}\right) \) is the function \( h\left( {e}^{2\pi it}\right) = {2\pi iF}\left( t\right) ,0 \leq t \leq 1 \) .\n\nTo deal with the case of arbitrary \( n \in \mathbb{Z} \) note first that \( \# \left( \zeta \right) = 1 \) . Indeed, this is immediate from the fact that\n\n\[ \zeta \left( {e}^{2\pi it}\right) = {e}^{2\pi it},\;0 \leq t \leq 1.\n\]\n\nFrom the property (1) it follows that \( \# \left( {\zeta }^{n}\right) = n \) for every \( n \in \mathbb{Z} \) ; hence \( \# \left( {{\zeta }^{n}{e}^{h}}\right) = \# \left( {\zeta }^{n}\right) + \# \left( {e}^{h}\right) = n \), as asserted. Conversely, if \( f \in C\left( \mathbb{T}\right) \) satisfies \( \# \left( f\right) = n \), consider \( g = {\zeta }^{-n}f \in C{\left( \mathbb{T}\right) }^{-1} \) . Using (1) again we have \( \# \left( g\right) = 0 \) , and by the preceding paragraph there is an \( h \in C\left( \mathbb{T}\right) \) such that \( g = {e}^{h} \) . Thus \( f = {\zeta }^{n}g = {\zeta }^{n}{e}^{h} \) has the asserted form.
|
Yes
|
Proposition 4.7.1. Every positive linear functional \( \rho \) on \( A \) satisfies the Schwarz inequality\n\n(4.19)\n\n\[{\left| \rho \left( {y}^{ * }x\right) \right| }^{2} \leq \rho \left( {{x}^{ * }x}\right) \rho \left( {{y}^{ * }y}\right)\]\n\nand moreover, \( \parallel \rho \parallel = \rho \left( \mathbf{1}\right) \) . In particular, every state of \( A \) has norm 1 .
|
Proof. Considering \( A \) as a complex vector space,\n\n\[x, y \in A \mapsto \left\lbrack {x, y}\right\rbrack = \rho \left( {{y}^{ * }x}\right)\]\n\ndefines a sesquilinear form which is positive semidefinite in the sense that \( \left\lbrack {x, x}\right\rbrack \geq 0 \) for every \( x \) . The argument that establishes the Schwarz inequality for complex inner product spaces applies verbatim in this context, and we deduce (4.19) from \( {\left| \left\lbrack x, y\right\rbrack \right| }^{2} \leq \left\lbrack {x, x}\right\rbrack \left\lbrack {y, y}\right\rbrack \) .\n\nClearly, \( \rho \left( \mathbf{1}\right) = \rho \left( {{\mathbf{1}}^{ * }\mathbf{1}}\right) \geq 0 \), and we claim that \( \parallel \rho \parallel \leq \rho \left( \mathbf{1}\right) \) . Indeed, for every \( x \in A \) the Schwarz inequality (4.19) implies\n\n\[{\left| \rho \left( x\right) \right| }^{2} = \left| {\rho \left( {{\mathbf{1}}^{ * }x}\right) }\right| \leq \rho \left( {{x}^{ * }x}\right) \rho \left( \mathbf{1}\right) .\n\nIf, in addition, \( \parallel x\parallel \leq 1 \), then \( {x}^{ * }x \) is a self-adjoint element in \( A \) of norm at most 1; consequently, \( 1 - {x}^{ * }x \) must have a self-adjoint square root \( y \in A \) (see Exercise (2b) below). It follows that \( \rho \left( {\mathbf{1} - {x}^{ * }x}\right) = \rho \left( {y}^{2}\right) \geq 0 \), i.e., \( 0 \leq \rho \left( {{x}^{ * }x}\right) \leq \rho \left( \mathbf{1}\right) \) . Substitution into the previous inequality gives \( {\left| \rho \left( x\right) \right| }^{2} \leq \) \( \rho \left( {{x}^{ * }x}\right) \rho \left( \mathbf{1}\right) \leq \rho {\left( \mathbf{1}\right) }^{2} \), and \( \parallel \rho \parallel \leq \rho \left( \mathbf{1}\right) \) follows. Since the inequality \( \parallel \rho \parallel \geq \rho \left( \mathbf{1}\right) \) is obvious, we conclude that \( \parallel \rho \parallel = \rho \left( \mathbf{1}\right) \) .
|
No
|
Corollary 1. Let \( \rho \) be a linear functional on a unital \( {C}^{ * } \) -algebra \( A \) satisfying \( \parallel \rho \parallel = \rho \left( \mathbf{1}\right) = 1 \) . Then \( \rho \) is a state.
|
Proof. We have to show that \( \rho \left( {{a}^{ * }a}\right) \geq 0 \) for every \( a \in A \) . By Theorem 4.8.3 it is enough to show that for every self-adjoint element \( x \in A \) having nonnegative spectrum, we have \( \rho \left( x\right) \geq 0 \) . More generally, we claim that for every normal element \( z \in A \) ,\n\n\[ \rho \left( z\right) \in \overline{\operatorname{conv}}\sigma \left( z\right) \]\n\nTo see this, let \( B \) be the commutative \( {C}^{ * } \) -subalgebra generated by \( z \) and 1 . The restriction \( {\rho }_{0} \) of \( \rho \) to \( B \) satisfies the same hypotheses \( \begin{Vmatrix}{\rho }_{0}\end{Vmatrix} = {\rho }_{0}\left( \mathbf{1}\right) = 1 \) . By Theorem 2.2.4, \( B \) is isometrically \( * \) -isomorphic to \( C\left( X\right) \), and for \( C\left( X\right) \) this is the result of Lemma 1.10.3.
|
Yes
|
For every element \( x \) in a unital \( {C}^{ * } \) -algebra \( A \) there is a state \( \rho \) such that \( \rho \left( {{x}^{ * }x}\right) = \parallel x{\parallel }^{2} \) .
|
Proof. Consider the self-adjoint element \( y = {x}^{ * }x \), and let \( B \) be the sub \( {C}^{ * } \) -algebra generated by \( y \) and the identity. Again, since \( B \cong C\left( X\right) \) there is a complex homomorphism \( \omega \in \operatorname{sp}\left( B\right) \) such that \( \omega \left( y\right) = \parallel y\parallel \) . Let \( \rho \) be any extension of \( \omega \) to a linear functional on \( A \) with \( \parallel \rho \parallel = \parallel \omega \parallel = 1 \) . We also have \( \rho \left( \mathbf{1}\right) = \omega \left( \mathbf{1}\right) = 1 \) . Thus \( \parallel \rho \parallel = \rho \left( \mathbf{1}\right) = 1 \), and the preceding corollary implies that \( \rho \) is a state.
|
Yes
|
Theorem. A connected algebraic group \( G \) is semisimple if and only if \( \mathfrak{g} \) is semisimple.
|
Proof. Let \( \mathfrak{g} \) be semisimple. If \( N \) is a closed connected commutative normal subgroup of \( G \), then \( n \) is a commutative ideal of \( \mathfrak{g} \) (use the easy half of Theorem 13.3 and the remarks above). So \( \mathfrak{n} = 0 \), forcing \( N = e \). Conversely, let \( G \) be semisimple and let \( \mathfrak{n} \) be a commutative ideal of \( \mathfrak{g} \). Define \( H = {C}_{G}{\left( \mathfrak{n}\right) }^{ \circ } \). Then \( \mathfrak{h} = {\mathfrak{c}}_{\mathfrak{g}}\left( \mathfrak{n}\right) \) (this follows from Theorem 13.2, cf. Exercise 2). Since \( \mathfrak{n} \) is an ideal, the Jacobi identity shows at once that \( \mathfrak{h} \) is also an ideal (including \( n \), since \( n \) is commutative, as part of its center). Thanks to Theorem 13.3, \( H \) must be normal in \( G \); as a result \( Z = Z{\left( H\right) }^{ \circ } \) is also normal in \( G \). By Theorem 13.4(b), \( \mathfrak{z} \) is the center of \( \mathfrak{h} \), and therefore includes \( \mathfrak{n} \). But \( G \) is semisimple, so \( Z = e,\mathfrak{z} = 0 \). This forces \( \mathfrak{n} = 0 \). \( ▱ \)
|
Yes
|
The extension of \( {\varphi }_{T} : T \rightarrow H \) to an isomorphism \( \varphi : G \rightarrow H \) was carried out under several assumptions involving the canonical rank 1 or 2 subsystems \( {\Phi }_{\alpha \beta } \) of \( \Phi \left( {\alpha ,\beta \in \Delta }\right) \) . These assumptions were formulated as follows in (32.4):\n\n(A) \( {\varphi }_{T}\left( {t}_{\alpha \beta }\right) = {\left( {h}_{\alpha }{h}_{\beta }\right) }^{m\left( {\alpha ,\beta }\right) } \) for all \( \alpha ,\beta \in \Delta \), where \( {t}_{\alpha \beta } = {\left( {n}_{\alpha }{n}_{\beta }\right) }^{m\left( {\alpha ,\beta }\right) }, m\left( {\alpha ,\beta }\right) = \) order of \( {\sigma }_{\alpha }{\sigma }_{\beta } \) in \( W \) .\n\n(B) For each pair \( \alpha ,\beta \in \Delta \), there is an isomorphism \( {\varphi }_{\alpha \beta } \) (of algebraic groups) from \( {U}_{\alpha \beta } \) onto the corresponding subgroup of \( H \), such that \( {\varphi }_{\alpha \beta } \) extends both \( {\varphi }_{\alpha } \) and \( {\varphi }_{\beta } \) .\n\n(C) Let \( \alpha ,\beta \in \Delta \) . If \( \gamma \in {\Phi }_{\alpha \beta }^{ + } \) with \( \gamma \neq \alpha \), then Int \( {h}_{\alpha } \circ {\varphi }_{\alpha \beta } \) agrees on \( {U}_{\gamma } \) with \( {\varphi }_{\alpha \beta } \circ \operatorname{Int}{n}_{\alpha } \) .\n\nNotice that when \( \alpha = \beta \) ,(B) and (C) are vacuous, while (A) is satisfied by virtue of the classification of rank 1 groups in (32.3).
|
Ostensibly these three statements involve the pair of groups \( G, H \) . But to verify them it is actually sufficient to work inside \( G \) alone. All we have to show is that the choices made \( \left( {{\varepsilon }_{\alpha },{\varepsilon }_{-\alpha }}\right. \), hence \( \left. {{n}_{\alpha }\text{, for}\alpha \in \Delta }\right) \) completely determine the elements \( {t}_{\alpha \beta } \), the commutation in \( {U}_{\alpha \beta } \), and the action of Int \( \alpha \) on certain root groups. Then the corresponding choices in \( H \) must lead to corresponding results. (This approach amounts to providing generators and relations for \( G \) .) With this in mind, let us reformulate (A),(B),(C).\n\nTo prove (A) it will be enough to express \( {t}_{\alpha \beta } \) in terms of \( {t}_{\alpha } \) and \( {t}_{\beta } \) in a way depending only on which root system of rank 2 is involved, since we already know that \( {t}_{\alpha } \) and \( {t}_{\beta } \) are well determined by the given data.\n\nFor (B) we shall prescribe \( {\varepsilon }_{\gamma } : {\mathbf{G}}_{a} \rightarrow {U}_{\gamma } \) for each \( \gamma \in {\Phi }_{\alpha \beta }^{ + } \) in a way depending only on the root system, then show that the constants occurring in the commutator formulas (Lemma 32.5) for all pairs of roots in \( {\Phi }_{\alpha \beta }^{ + } \) are completely determined. So the group-theoretic structure of \( {U}_{\alpha \beta } \) will just depend on the root system. On the other hand, it is easy to construct an isomorphism of varieties from \( {U}_{\alpha \beta } \) onto the corresponding subgroup of \( H \) (cf. (32.5)); the use of the \( {\varepsilon }_{\gamma } \) and corresponding \( {\varepsilon }_{\gamma }^{\prime } \), will insure that this is a group homomorphism. As an example of how \( {\varepsilon }_{\gamma } \) is to be prescribed, consider type \( {A}_{2} \), where the positive roots are \( \alpha ,\beta ,\alpha + \beta \) . We are already given \( {\varepsilon }_{\alpha } \) and \( {n}_{\beta } \), so we just set \( {\varepsilon }_{\alpha + \beta } = {\varepsilon }_{\alpha } \circ {n}_{\beta } \) .
|
Yes
|
Proposition 1.1. Let \( f, g \in A \) with \( g \neq 0 \) . Then there exist elements \( q, r \in A \) such that \( f = {qg} + r \) and \( r \) is either 0 or \( \deg \left( r\right) < \deg \left( g\right) \) . Moreover, \( q \) and \( r \) are uniquely determined by these conditions.
|
Proof. Let \( n = \deg \left( f\right), m = \deg \left( g\right) ,\alpha = \operatorname{sgn}\left( f\right) ,\beta = \operatorname{sgn}\left( g\right) \) . We give the proof by induction on \( n = \deg \left( f\right) \) . If \( n < m \), set \( q = 0 \) and \( r = f \) . If \( n \geq m \), we note that \( {f}_{1} = f - \alpha {\beta }^{-1}{T}^{n - m}g \) has smaller degree than \( f \) . By induction, there exist \( {q}_{1},{r}_{1} \in A \) such that \( {f}_{1} = {q}_{1}g + {r}_{1} \) with \( {r}_{1} \) being either 0 or with degree less than \( \deg \left( g\right) \) . In this case, set \( q = \alpha {\beta }^{-1}{T}^{n - m} + {q}_{1} \) and \( r = {r}_{1} \) and we are done.\n\nIf \( f = {qg} + r = {q}^{\prime }g + {r}^{\prime } \), then \( g \) divides \( r - {r}^{\prime } \) and by degree considerations we see \( r = {r}^{\prime } \) . In this case, \( {qg} = {q}^{\prime }g \) so \( q = {q}^{\prime } \) and the uniqueness is established.\n\nThis proposition shows that \( A \) is a Euclidean domain and thus a principal ideal domain and a unique factorization domain. It also allows a quick proof of the finiteness of the residue class rings.
|
Yes
|
Proposition 1.2. Suppose \( g \in A \) and \( g \neq 0 \) . Then \( A/{gA} \) is a finite ring with \( {q}^{\deg \left( g\right) } \) elements.
|
Proof. Let \( m = \deg \left( g\right) \) . By Proposition 1.1 one easily verifies that \( \{ r \in \) \( A \mid \deg \left( r\right) < m\} \) is a complete set of representatives for \( A/{gA} \) . Such elements look like\n\n\[ r = {\alpha }_{0}{T}^{m - 1} + {\alpha }_{1}{T}^{m - 2} + \cdots + {\alpha }_{m - 1}\;\text{ with }{\alpha }_{i} \in \mathbb{F}. \]\n\nSince the \( {\alpha }_{i} \) vary independently through \( \mathbb{F} \) there are \( {q}^{m} \) such polynomials and the result follows.
|
Yes
|
Proposition 1.3. The group of units in \( A \) is \( {\mathbb{F}}^{ * } \) . In particular, it is a finite cyclic group with \( q - 1 \) elements.
|
Proof. The only thing left to prove is the cyclicity of \( {\mathbb{F}}^{ * } \) . This follows from the very general fact that a finite subgroup of the multiplicative group of a field is cyclic.
|
No
|
Proposition 1.4. Let \( {m}_{1},{m}_{2},\ldots ,{m}_{t} \) be elements of \( A \) which are pairwise relatively prime. Let \( m = {m}_{1}{m}_{2}\ldots {m}_{t} \) and \( {\phi }_{i} \) be the natural homomorphism from \( A/{mA} \) to \( A/{m}_{i}A \) . Then, the map \( \phi : A/{mA} \rightarrow A/{m}_{1}A \oplus A/{m}_{2}A \oplus \cdots \oplus A/{m}_{t}A \) given by \[ \phi \left( a\right) = \left( {{\phi }_{1}\left( a\right) ,{\phi }_{2}\left( a\right) ,\ldots ,{\phi }_{t}\left( a\right) }\right) \] is a ring isomorphism.
|
Proof. This is a standard result which holds in any principal ideal domain (properly formulated it holds in much greater generality).
|
No
|
Proposition 1.5. Let \( P \in A \) be an irreducible polynomial. Then, \( {\left( A/PA\right) }^{ * } \) is a cyclic group with \( \left| P\right| - 1 \) elements.
|
Proof. Since \( A \) is a principal ideal domain, \( {PA} \) is a maximal ideal and so \( A/{PA} \) is a field. A finite subgroup of the multiplicative group of a field is cyclic. Thus \( {\left( A/PA\right) }^{ * } \) is cyclic. That the order of this group is \( \left| P\right| - 1 \) is immediate.
|
Yes
|
Let \( P \in A \) be an irreducible polynomial and \( e \) a positive integer. The order of \( {\left( A/{P}^{e}A\right) }^{ * } \) is \( {\left| P\right| }^{e - 1}\left( {\left| P\right| - 1}\right) \) . Let \( {\left( A/{P}^{e}A\right) }^{\left( 1\right) } \) be the kernel of the natural map from \( {\left( A/{P}^{e}A\right) }^{ * } \) to \( {\left( A/PA\right) }^{ * } \) . It is a p-group of order \( {\left| P\right| }^{e - 1} \) . As e tends to infinity, the minimal number of generators of \( {\left( A/{P}^{e}A\right) }^{\left( 1\right) } \) tends to infinity.
|
The ring \( A/{P}^{e}A \) has only one maximal ideal \( {PA}/{P}^{e}A \) which has \( {\left| P\right| }^{e - 1} \) elements. Thus, \( {\left( A/{P}^{e}A\right) }^{ * } = A/{P}^{e}A - {PA}/{P}^{e}A \) has \( {\left| P\right| }^{e} - {\left| P\right| }^{e - 1} = \) \( {\left| P\right| }^{e - 1}\left( {\left| P\right| - 1}\right) \) number of elements. Since \( {\left( A/{P}^{e}A\right) }^{ * } \rightarrow {\left( A/PA\right) }^{ * } \) is onto, and the latter group has \( \left| P\right| - 1 \) elements the assertion about the size of \( {\left( A/{P}^{e}A\right) }^{\left( 1\right) } \) follows. It remains to prove the assertion about the minimal number of generators.\n\nIt is instructive to first consider the case \( e = 2 \) . Every element in \( {\left( A/{P}^{2}A\right) }^{\left( 1\right) } \) can be represented by a polynomial of the form \( a = 1 + {bP} \) . Since we are working in characteristic \( p \) we have \( {a}^{p} = 1 + {b}^{p}{P}^{p} \equiv 1 \) \( \left( {\;\operatorname{mod}\;{P}^{2}}\right) \) . So, we have a group of order \( \left| P\right| \) with exponent \( p \) . If \( q = {p}^{f} \) it follows that \( {\left( A/{P}^{2}A\right) }^{\left( 1\right) } \) is a direct sum of \( f\deg \left( P\right) \) number of copies of \( \mathbb{Z}/p\mathbb{Z} \) . This is a cyclic group only under the very restrictive conditions that \( q = p \) and \( \deg \left( P\right) = 1 \) .\n\nTo deal with the general case, suppose that \( s \) is the smallest integer such that \( {p}^{s} \geq e \) . Since \( {\left( 1 + bP\right) }^{{p}^{s}} = 1 + {\left( bP\right) }^{{p}^{s}} \equiv 1\left( {\;\operatorname{mod}\;{P}^{e}}\right) \) we have that raising to the \( {p}^{s} \) -power annihilates \( G = {\left( A/{P}^{e}A\right) }^{\left( 1\right) } \) . Let \( d \) be the minimal number of generators of this group. It follows that there is an onto map from \( {\left( \mathbb{Z}/{p}^{s}\mathbb{Z}\right) }^{d} \) onto \( G \) . Thus, \( {p}^{ds} \geq {p}^{f\deg \left( P\right) \left( {e - 1}\right) } \) and so\n\n\[ d \geq \frac{f\deg \left( P\right) \left( {e - 1}\right) }{s}. \]\n\nSince \( s \) is the smallest integer bigger than or equal to \( {\log }_{p}\left( e\right) \) it is clear that \( d \rightarrow \infty \) as \( e \rightarrow \infty \).
|
Yes
|
Proposition 1.7.\n\n\[ \Phi \left( f\right) = \left| f\right| \mathop{\prod }\limits_{{P \mid f}}\left( {1 - \frac{1}{\left| P\right| }}\right) \]
|
Proof. Let \( f = \alpha {P}_{1}^{{e}_{1}}{P}_{2}^{{e}_{2}}\ldots {P}_{t}^{{e}_{t}} \) be the prime decomposition of \( f \) . By the corollary to Propositions 1.4 and by Proposition 1.6, we see that\n\n\[ \Phi \left( f\right) = \mathop{\prod }\limits_{{i = 1}}^{t}\Phi \left( {P}_{i}^{{e}_{i}}\right) = \mathop{\prod }\limits_{{i = 1}}^{t}\left( {{\left| {P}_{i}\right| }^{{e}_{i}} - {\left| {P}_{i}\right| }^{{e}_{i} - 1}}\right) ,\]\n\nfrom which the result follows immediately.
|
Yes
|
Proposition 1.8. If \( f \in A, f \neq 0 \), and \( a \in A \) is relatively prime to \( f \), i.e., \( \left( {a, f}\right) = 1 \), then\n\n\[ \n{a}^{\Phi \left( f\right) } \equiv 1\;\left( {\;\operatorname{mod}\;f}\right) \n\]
|
Proof. The group \( {\left( A/fA\right) }^{ * } \) has \( \Phi \left( f\right) \) elements. The coset of \( a \) modulo \( f,\bar{a} \) , lies in this group. Thus, \( {\bar{a}}^{\Phi \left( f\right) } = \overline{1} \) and this is equivalent to the congruence in the proposition.
|
Yes
|
Proposition 1.9. Let \( P \in A \) be irreducible of degree \( d \) . Suppose \( X \) is an indeterminate. Then,\n\n\[ \n{X}^{\left| P\right| - 1} - 1 \equiv \mathop{\prod }\limits_{{0 \leq \deg \left( f\right) < d}}\left( {X - f}\right) \;\left( {\;\operatorname{mod}\;P}\right) .\n\]
|
Proof. Recall that \( \{ f \in A \mid \deg \left( f\right) < d\} \) is a set of representatives for the cosets of \( A/{PA} \) . If we throw out \( f = 0 \) we get a set of representatives for \( {\left( A/PA\right) }^{ * } \) . We find\n\n\[ \n{X}^{\left| P\right| - 1} - \overline{1} = \mathop{\prod }\limits_{{0 \leq \deg \left( f\right) < d}}\left( {X - \bar{f}}\right)\n\]\n\nwhere the bars denote cosets modulo \( P \) . This follows from the corollary to Proposition 1.8 since both sides of the equation are monic polynomials in \( X \) with the same set of roots in the field \( A/{PA} \) . Since there are \( \left| P\right| - 1 \) roots and the difference of the two sides has degree less than \( \left| P\right| - 1 \), the difference of the two sides must be 0 . The congruence in the Proposition is equivalent to this assertion.
|
Yes
|
Corollary 1. Let \( d \) divide \( \left| P\right| - 1 \) . The congruence \( {X}^{d} \equiv 1\left( {\;\operatorname{mod}\;P}\right) \) has exactly \( d \) solutions. Equivalently, the equation \( {X}^{d} = \overline{1} \) has exactly \( d \) solutions in \( {\left( A/PA\right) }^{ * } \) .
|
Proof. We prove the second assertion. Since \( d\left| \right| P \mid - 1 \) it follows that \( {X}^{d} - 1 \) divides \( {X}^{\left| P\right| - 1} - 1 \) . By the proposition, the latter polynomial splits as a product of distinct linear factors. Thus so does the former polynomial. This establishes the result.
|
No
|
Corollary 2. With the same notation,\n\n\[ \mathop{\prod }\limits_{{0 \leq \deg \left( f\right) < \deg P}}f \equiv - 1\;\left( {\;\operatorname{mod}\;P}\right) \]
|
Proof. Just set \( X = 0 \) in the proposition. If the characteristic of \( \mathbb{F} \) is odd \( \left| P\right| - 1 \) is even and the result follows. If the characteristic is 2 then the result also follows since in characteristic 2 we have \( - 1 = 1 \) .
|
Yes
|
Proposition 1.10. Let \( P \) be irreducible and \( a \in A \) not divisible by \( P \) . Assume \( d \) divides \( \left| P\right| - 1 \) . The congruence \( {X}^{d} \equiv a\left( {\;\operatorname{mod}\;{P}^{e}}\right) \) is solvable if and only if\n\n\[ \n{a}^{\frac{\left| P\right| - 1}{d}} \equiv 1\;\left( {\;\operatorname{mod}\;P}\right) \n\]\n\nThere are \( \frac{\Phi \left( {P}^{e}\right) }{d}d \) -th power residues modulo \( {P}^{e} \) .
|
Proof. Assume to begin with that \( e \doteq 1 \) .\n\nIf \( {b}^{d} \equiv a\left( {\;\operatorname{mod}\;P}\right) \), then \( {a}^{\frac{\left| P\right| - 1}{d}} \equiv {b}^{\left| P\right| - 1} \equiv 1\left( {\;\operatorname{mod}\;P}\right) \) by the corollary to Proposition 1.8. This shows the condition is necessary. To show it is sufficient recall that by Corollary 1 to Proposition 1.9 all the \( d \) -th roots of unity are in the field \( A/{PA} \) . Consider the homomorphism from \( {\left( A/PA\right) }^{ * } \) to itself given by raising to the \( d \) -th power. It’s kernel has order \( d \) and its image is the \( d \) -th powers. Thus, there are precisely \( \frac{\left| P\right| - 1}{d}d \) -th powers in \( {\left( A/PA\right) }^{ * } \) . We have seen that they all satisfy \( {X}^{\frac{\left| P\right| - 1}{d}} - 1 = 0 \) . Thus, they are precisely the roots of this equation. This proves all assertions in the case \( e = 1 \) .\n\nTo deal with the remaining cases, we employ a little group theory. The natural map (i.e., reduction modulo \( P \) ) is a homomorphism from \( {\left( A/{P}^{e}A\right) }^{ * } \) onto \( {\left( A/PA\right) }^{ * } \) and the kernel is a \( p \) -group as follows from Proposition 1.6. Since the order of \( {\left( A/PA\right) }^{ * } \) is \( \left| P\right| - 1 \) which is prime to \( p \) it follows that \( {\left( A/{P}^{e}A\right) }^{ * } \) is the direct product of a \( p \) -group and a copy of \( {\left( A/PA\right) }^{ * } \) . Since \( \left( {d, p}\right) = 1 \), raising to the \( d \) -th power in an abelian \( p \) -group is an automorphism. Thus, \( a \in A \) is a \( d \) -th power modulo \( {P}^{e} \) if and only if it is a \( d \) -th power modulo \( P \) . The latter has been shown to hold if and only if \( {a}^{\frac{\left| P\right| - 1}{d}} \equiv 1\left( {\;\operatorname{mod}\;P}\right) \) . Now consider the homomorphism from \( {\left( A/{P}^{e}A\right) }^{ * } \) to itself given by raising to the \( d \) -th power. It easily follows from what has been said that the kernel has \( d \) elements and the image is the subgroup of \( d \) -th powers. It follows that the latter group has order \( \frac{\widetilde{\Phi }\left( {P}^{e}\right) }{d} \) . This concludes the proof.
|
Yes
|
Proposition 2.1.\n\n\[ \mathop{\sum }\limits_{{d \mid n}}d{a}_{d} = {q}^{n} \]
|
This formula is often attributed to Richard Dedekind. It is interesting to note that it appears, with essentially the above proof, in a manuscript of C.F. Gauss (unpublished in his lifetime), \
|
No
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.