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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... | 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... | 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 \( \si... | 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... | 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} ... | 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... | 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} ... | 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... | 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} ... | 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... | 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} ... | 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... | 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} ... | 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 \... | 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 \(... | 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}}^... | 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 \( \mu \) . Furthermor... | 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 \subs... | 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... | 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 ev... | 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 coun... | 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( {{... | 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 proper... | 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}}^{ * }\... | 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}... | 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}(\\... | 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. \( {... | 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}}... | 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 se... | 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}\... | 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}... | 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 }\l... | 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 \... | 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... | 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 satisfyi... | 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... | 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... | 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... | 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} =... | 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 ... | 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 a... | 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 ... | 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 ... | 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 \... | 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 \( \... | 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 \( p = 1 \) ... | 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 \( \lef... | 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 ... | 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 w... | 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 sequenc... | 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 \( \... | 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... | 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... | 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 \... | 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 \math... | 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 }^{ - }\... | 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\righ... | 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}... | 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} \time... | 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) \... | 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}}\... | 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 disjoi... | 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 \\m... | 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}\\l... | 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 \( X \times Y \) . ... | 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 in... | 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 \( \sig... | 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 \( \... | 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 projectio... | 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 \d... | 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... | 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\ri... | 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 \( \O... | 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 ... | 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\... | 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}... | 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 bi... | 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 \( {\sigma }_{A}\left( x\right) \subseteq {\sigma }_{B}\left( x\right) \) ... | 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 \... | 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... | 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) ,\;{\lef... | 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( {... | 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}\... | 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 ... | 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 ... | 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 =... | 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... | 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 Hilbe... | 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... | 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}^{ ... | 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 \( {\mathb... | 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 \... | 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... | 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 in... | 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 p... | 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... | 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{Vmatri... | 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 ... | 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}... | 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 \... | 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 equiva... | 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 \... | 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, si... | 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) =... | 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}^... | 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... | 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... | 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 establish... | 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 ... | 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) = \p... | 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... | 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 ... | 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... | 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... | 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... | 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}_{... | 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}... | 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| - ... | 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 }\... | 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 ... | 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( {\;\oper... | 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 necess... | 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 |
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