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Lemma 7.10.6 Suppose \( M \) is a closed set in \( X \) where \( \left( {X, d}\right) \) is a metric space and suppose \( f : M \rightarrow \left\lbrack {-1,1}\right\rbrack \) is continuous at every point of \( M \) . Then there exists a function, \( g \) which is defined and continuous on all of \( X \) such that \( g... | Proof: Let \( {g}_{1} \) be such that \( {g}_{1}\left( X\right) \subseteq \left\lbrack {-1/3,1/3}\right\rbrack \) and \( {\begin{Vmatrix}f - {g}_{1}\end{Vmatrix}}_{M} \leq \frac{2}{3} \) . Suppose \( {g}_{1},\cdots ,{g}_{m} \) have been chosen such that \( {g}_{j}\left( X\right) \subseteq \left\lbrack {-1/3,1/3}\right\... | Yes |
Theorem 7.10.7 Let \( M \) be a closed nonempty subset of a metric space \( \left( {X, d}\right) \) and let \( f : M \rightarrow \left\lbrack {a, b}\right\rbrack \) is continuous at every point of \( M \) . Then there exists a function, \( g \) continuous on all of \( X \) which coincides with \( f \) on \( M \) such t... | Proof: Let \( {f}_{1}\left( x\right) = 1 + \frac{2}{b - a}\left( {f\left( x\right) - b}\right) \) . Then \( {f}_{1} \) satisfies the conditions of Lemma 7.10.6 and so there exists \( {g}_{1} : X \rightarrow \left\lbrack {-1,1}\right\rbrack \) such that \( g \) is continuous on \( X \) and equals \( {f}_{1} \) on \( M \... | Yes |
Theorem 7.11.2 Let \( f : \left( {X, d}\right) \rightarrow \left( {X, d}\right) \) be a contraction map and let \( \left( {X, d}\right) \) be a complete metric space. Thus Cauchy sequences converge and also \( d\left( {f\left( x\right), f\left( \widehat{x}\right) }\right) \leq \) \( {rd}\left( {x,\widehat{x}}\right) \)... | Proof: Pick \( {x}_{0} \in X \) and consider the sequence of iterates of the map,\n\n\[ {x}_{0}, f\left( {x}_{0}\right) ,{f}^{2}\left( {x}_{0}\right) ,\cdots \text{.} \]\n\nWe argue that this is a Cauchy sequence. For \( m < n \), it follows from the triangle inequality,\n\n\[ d\left( {{f}^{m}\left( {x}_{0}\right) ,{f}... | Yes |
Corollary 7.11.3 Let \( B \) be a closed subset of the complete metric space \( \left( {X, d}\right) \) and let \( f : B \rightarrow X \) be a contraction map\n\n\[ d\left( {f\left( x\right), f\left( \widehat{x}\right) }\right) \leq {rd}\left( {x,\widehat{x}}\right), r < 1.\]\n\nAlso suppose there exists \( {x}_{0} \in... | Proof: By assumption, the sequence of iterates stays in \( B \) . Then, as in the proof of the preceding theorem, for \( m < n \), it follows from the triangle inequality,\n\n\[ d\left( {{f}^{m}\left( {x}_{0}\right) ,{f}^{n}\left( {x}_{0}\right) }\right) \leq \mathop{\sum }\limits_{{k = m}}^{{n - 1}}d\left( {{f}^{k + 1... | Yes |
Corollary 7.11.4 Suppose \( f : X \times \Lambda \rightarrow X \) where \( \Lambda \) is a metric space and \( X \) is a complete metric space. Suppose \( f \) satisfies\n\n1. \( d\left( {f\left( {x,\lambda }\right), f\left( {y,\lambda }\right) }\right) \leq {rd}\left( {x, y}\right) \) for each \( \lambda \in \Lambda \... | Proof: Pick \( {x}_{0} \in X \) and consider the above sequence of iterates, \( \left\{ {{f}^{n}\left( {x,\lambda }\right) }\right\} \) . Let \( \rho \) be the metric on \( \Lambda \) . Then there is a fixed point and if \( x\left( \lambda \right) \) is this unique fixed point,\n\n\[ d\left( {x\left( \lambda \right) ,{... | Yes |
Theorem 7.11.5 Let \( f : \left( {X, d}\right) \rightarrow \left( {X, d}\right) \) have the property that for some \( n \in \mathbb{N} \) , \( {f}^{n} \) is a contraction map and let \( \left( {X, d}\right) \) be a complete metric space. Then there is a unique fixed point for \( f \) . As in the earlier theorem the seq... | Proof: From Theorem 7.11.2 there is a unique fixed point for \( {f}^{n} \) . Thus\n\n\[ \n{f}^{n}\left( x\right) = x \n\] \n\nThen \n\n\[ \n{f}^{n}\left( {f\left( x\right) }\right) = {f}^{n + 1}\left( x\right) = f\left( x\right) \n\] \n\nBy uniqueness, \( f\left( x\right) = x \) .\n\nNow consider the sequence of iterat... | Yes |
Lemma 7.11.7 Suppose \( {x}_{n} \rightarrow x \) and \( {y}_{n} \rightarrow y \) . Then \( d\left( {{x}_{n},{y}_{n}}\right) \rightarrow d\left( {x, y}\right) \) . | Proof: Consider the following.\n\n\[ d\left( {x, y}\right) \leq d\left( {x,{x}_{n}}\right) + d\left( {{x}_{n}, y}\right) \leq d\left( {x,{x}_{n}}\right) + d\left( {{x}_{n},{y}_{n}}\right) + d\left( {{y}_{n}, y}\right) \]\n\nso\n\n\[ d\left( {x, y}\right) - d\left( {{x}_{n},{y}_{n}}\right) \leq d\left( {x,{x}_{n}}\right... | Yes |
Theorem 7.11.8 Let \( f : \left( {X, d}\right) \rightarrow \left( {Y,\rho }\right) \) be a continuous function and let \( K \) be a compact subset of \( X \) . Then the restriction of \( f \) to \( K \) is uniformly continuous. | Proof: First of all, \( K \) is a metric space and \( f \) restricted to \( K \) is continuous. Now suppose it fails to be uniformly continuous. Then there exists \( \varepsilon > 0 \) and pairs of points \( {x}_{n},{\widehat{x}}_{n} \) such that \( d\left( {{x}_{n},{\widehat{x}}_{n}}\right) < 1/n \) but \( \rho \left(... | Yes |
Theorem 7.11.11 Let \( {f}_{n} : X \rightarrow Y \) where \( \left( {X, d}\right) ,\left( {Y,\rho }\right) \) are two metric spaces and suppose each \( {f}_{n} \) is continuous at \( x \in X \) and also that \( {f}_{n} \) converges uniformly to \( f \) on \( X \) . Then \( f \) is also continuous at \( x \) . In additi... | Proof: Let \( \varepsilon > 0 \) be given. Then\n\n\[ \rho \left( {f\left( x\right), f\left( \widehat{x}\right) }\right) \leq \rho \left( {f\left( x\right) ,{f}_{n}\left( x\right) }\right) + \rho \left( {{f}_{n}\left( x\right) ,{f}_{n}\left( \widehat{x}\right) }\right) + \rho \left( {{f}_{n}\left( \widehat{x}\right), f... | Yes |
Proposition 7.12.2 Let \( X \) be a set and let \( \mathcal{B} \) be a basis for a topology as defined above and let \( \tau \) be the set of open sets determined by \( \mathcal{B} \). Then\n\n\[ \varnothing \in \tau, X \in \tau \]\n\n\( \left( {7.12.12}\right) \)\n\n\[ \text{If}\mathcal{C} \subseteq \tau \text{, then}... | Proof: If \( p \in \varnothing \) then there exists \( B \in \mathcal{B} \) such that \( p \in B \subseteq \varnothing \) because there are no points in \( \varnothing \). Therefore, \( \varnothing \in \tau \). Now if \( p \in X \), then by part 2.) of Definition 7.12.1 \( p \in B \subseteq X \) for some \( B \in \math... | Yes |
Theorem 7.12.6 A subset, \( E \), of \( X \) is closed if and only if it contains all its limit points. | Proof: Suppose first that \( E \) is closed and let \( x \) be a limit point of \( E \) . Is \( x \in E \) ? If \( x \notin E \), then \( {E}^{C} \) is an open set containing \( x \) which contains no points of \( E \), a contradiction. Thus \( x \in E \) . Now suppose \( E \) contains all its limit points. Is the comp... | Yes |
Theorem 7.12.7 If \( \left( {X,\tau }\right) \) is a Hausdorff space and if \( p \in X \), then \( \{ p\} \) is a closed set. | Proof: If \( x \neq p \), there exist open sets \( U \) and \( V \) such that \( x \in U, p \in V \) and \( U \cap V = \varnothing \) . Therefore, \( \{ p{\} }^{C} \) is an open set so \( \{ p\} \) is closed. | Yes |
Theorem 7.12.12 \( \bar{E} = E \cup \{ \) limit points of \( E\} \) . | Proof: Let \( x \in \bar{E} \) and suppose that \( x \notin E \) . If \( x \) is not a limit point either, then there exists an open set, \( U \), containing \( x \) which does not intersect \( E \) . But then \( {U}^{C} \) is a closed set which contains \( E \) which does not contain \( x \), contrary to the definitio... | Yes |
Theorem 7.12.16 The set \( \mathcal{B} \) of Definition 7.12.15 is a basis for a topology. | Proof: Suppose \( \mathbf{x} \in \mathop{\prod }\limits_{{i = 1}}^{n}{A}_{i} \cap \mathop{\prod }\limits_{{i = 1}}^{n}{B}_{i} \) where \( {A}_{i} \) and \( {B}_{i} \) are open sets. Say\n\n\[ \mathbf{x} = \left( {{x}_{1},\cdots ,{x}_{n}}\right) \]\n\nThen \( {x}_{i} \in {A}_{i} \cap {B}_{i} \) for each \( i \) . Theref... | Yes |
Theorem 7.12.18 If \( \left( {X,\tau }\right) \) is a Hausdorff space, then every compact subset must also be a closed set. | Proof: Suppose \( p \notin K \) . For each \( x \in X \), there exist open sets, \( {U}_{x} \) and \( {V}_{x} \) such that\n\n\[ x \in {U}_{x}, p \in {V}_{x} \]\n\nand\n\n\[ {U}_{x} \cap {V}_{x} = \varnothing \]\n\nIf \( K \) is assumed to be compact, there are finitely many of these sets, \( {U}_{{x}_{1}},\cdots ,{U}_... | Yes |
Lemma 7.12.20 If \( \left( {X,\tau }\right) \) is a locally compact Hausdorff space, then \( \left( {\widetilde{X},\widetilde{\tau }}\right) \) is a compact Hausdorff space. Also if \( U \) is an open set of \( \widetilde{\tau } \), then \( U \smallsetminus \{ \infty \} \) is an open set of \( \tau \) . | Proof: Since \( \left( {X,\tau }\right) \) is a locally compact Hausdorff space, it follows \( \left( {\widetilde{X},\widetilde{\tau }}\right) \) is a Hausdorff topological space. The only case which needs checking is the one of \( p \in X \) and \( \infty \) . Since \( \left( {X,\tau }\right) \) is locally compact, th... | Yes |
Theorem 7.12.22 Let \( \mathcal{K} \) be a set whose elements are compact subsets of a Hausdorff topological space, \( \left( {X,\tau }\right) \) . Suppose \( \mathcal{K} \) has the finite intersection property. Then \( \varnothing \neq \cap \mathcal{K} \) . | Proof: Suppose to the contrary that \( \varnothing = \cap \mathcal{K} \) . Then consider\n\n\[ \mathcal{C} \equiv \left\{ {{K}^{C} : K \in \mathcal{K}}\right\} \]\n\nIt follows \( \mathcal{C} \) is an open cover of \( {K}_{0} \) where \( {K}_{0} \) is any particular element of \( \mathcal{K} \) . But then there are fin... | Yes |
Lemma 7.12.23 Let \( \left( {X,\tau }\right) \) be a topological space and let \( \mathcal{B} \) be a basis for \( \tau \) . Then \( K \) is compact if and only if every open cover of basic open sets admits a finite subcover. | Proof: Suppose first that \( X \) is compact. Then if \( \mathcal{C} \) is an open cover consisting of basic open sets, it follows it admits a finite subcover because these are open sets in \( \mathcal{C} \) .\n\nNext suppose that every basic open cover admits a finite subcover and let \( \mathcal{C} \) be an open cove... | Yes |
Theorem 7.13.2 Suppose \( U \) and \( V \) are connected sets having nonempty intersection. Then \( U \cup V \) is also connected. | Proof: Suppose \( U \cup V = A \cup B \) where \( \bar{A} \cap B = \bar{B} \cap A = \varnothing \) . Consider the sets, \( A \cap U \) and \( B \cap U \) . Since\n\n\[ \overline{\left( A \cap U\right) } \cap \left( {B \cap U}\right) = \left( {A \cap U}\right) \cap \left( \overline{B \cap U}\right) = \varnothing ,\]\n\n... | Yes |
Theorem 7.13.3 Let \( f : X \rightarrow Y \) be continuous where \( X \) and \( Y \) are topological spaces and \( X \) is connected. Then \( f\left( X\right) \) is also connected. | Proof: To do this you show \( f\left( X\right) \) is not separated. Suppose to the contrary that \( f\left( X\right) = A \cup B \) where \( A \) and \( B \) separate \( f\left( X\right) \) . Then consider the sets, \( {f}^{-1}\left( A\right) \) and \( {f}^{-1}\left( B\right) \) . If \( z \in {f}^{-1}\left( B\right) \),... | Yes |
Theorem 7.13.5 Let \( {C}_{p} \) be a connected component of a set \( S \) in a general topological space. Then \( {C}_{p} \) is a connected set and if \( {C}_{p} \cap {C}_{q} \neq \varnothing \), then \( {C}_{p} = {C}_{q} \) . | Proof: Let \( \mathcal{C} \) denote the connected subsets of \( S \) which contain \( p \) . If \( {C}_{p} = A \cup B \) where\n\n\[ \bar{A} \cap B = \bar{B} \cap A = \varnothing ,\]\n\nthen \( p \) is in one of \( A \) or \( B \) . Suppose without loss of generality \( p \in A \) . Then every set of \( \mathcal{C} \) ... | Yes |
Theorem 7.13.6 A set, \( C \) in \( \mathbb{R} \) is connected if and only if \( C \) is an interval. | Proof: Let \( C \) be connected. If \( C \) consists of a single point, \( p \), there is nothing to prove. The interval is just \( \left\lbrack {p, p}\right\rbrack \) . Suppose \( p < q \) and \( p, q \in C \) . You need to show \( \left( {p, q}\right) \subseteq C \) . If\n\n\[ x \in \left( {p, q}\right) \smallsetminu... | Yes |
Theorem 7.13.7 Let \( U \) be an open set in \( \mathbb{R} \) . Then there exist countably many disjoint open sets, \( {\left\{ \left( {a}_{i},{b}_{i}\right) \right\} }_{i = 1}^{\infty } \) such that \( U = { \cup }_{i = 1}^{\infty }\left( {{a}_{i},{b}_{i}}\right) \) . | Proof: Let \( p \in U \) and let \( z \in {C}_{p} \), the connected component determined by \( p \) . Since \( U \) is open, there exists, \( \delta > 0 \) such that \( \left( {z - \delta, z + \delta }\right) \subseteq U \) . It follows from Theorem 7.13.2 that\n\n\[ \left( {z - \delta, z + \delta }\right) \subseteq {C... | Yes |
Proposition 7.13.9 If a topological space is arcwise connected, then it is connected. | Proof: Let \( X \) be an arcwise connected space and suppose it is separated. Then \( X = A \cup B \) where \( A, B \) are two separated sets. Pick \( p \in A \) and \( q \in B \) . Since \( X \) is given to be arcwise connected, there must exist a continuous function \( \gamma : \left\lbrack {a, b}\right\rbrack \right... | Yes |
Theorem 7.13.10 Let \( U \) be an open subset of a locally arcwise connected topological space, \( X \) . Then \( U \) is arcwise connected if and only if \( U \) if connected. Also the connected components of an open set in such a space are open sets, hence arcwise connected. | Proof: By Proposition 7.13.9 it is only necessary to verify that if \( U \) is connected and open in the context of this theorem, then \( U \) is arcwise connected. Pick \( p \in U \) . Say \( x \in U \) satisfies \( \mathcal{P} \) if there exists a continuous function, \( \gamma : \left\lbrack {a, b}\right\rbrack \rig... | Yes |
Corollary 7.13.11 Let \( f : \Omega \rightarrow \mathbb{Z} \) be continuous where \( \Omega \) is a connected open set. Then \( f \) must be a constant. | Proof: Suppose not. Then it achieves two different values, \( k \) and \( l \neq k \) . Then \( \Omega = {f}^{-1}\left( l\right) \cup {f}^{-1}\left( {\{ m \in \mathbb{Z} : m \neq l\} }\right) \) and these are disjoint nonempty open sets which separate \( \Omega \) . To see they are open, note\n\n\[ \n{f}^{-1}\left( {\{... | Yes |
Theorem 8.1.2 For \( \mathbf{v},\mathbf{w} \in {\mathbb{F}}^{n} \) and \( \alpha ,\beta \) scalars,(real numbers), the following hold.\n\n\[ \mathbf{v} + \mathbf{w} = \mathbf{w} + \mathbf{v} \] | You should verify these properties all hold. For example, consider 8.1.7\n\n\[ \alpha \left( {\mathbf{v} + \mathbf{w}}\right) = \alpha \left( {{v}_{1} + {w}_{1},\cdots ,{v}_{n} + {w}_{n}}\right) \]\n\n\[ = \left( {\alpha \left( {{v}_{1} + {w}_{1}}\right) ,\cdots ,\alpha \left( {{v}_{n} + {w}_{n}}\right) }\right) \]\n\n... | No |
Lemma 8.2.2 A set of vectors \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{p}}\right\} \) is linearly independent if and only if none of the vectors can be obtained as a linear combination of the others. | Proof: Suppose first that \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{p}}\right\} \) is linearly independent. If\n\n\[ \n{\mathbf{x}}_{k} = \mathop{\sum }\limits_{{j \neq k}}{c}_{j}{\mathbf{x}}_{j} \n\]\n\nthen\n\n\[ \n\mathbf{0} = 1{\mathbf{x}}_{k} + \mathop{\sum }\limits_{{j \neq k}}\left( {-{c}_{j}}\right) {\... | Yes |
Theorem 8.2.3 If\n\n\\[ \n\\operatorname{span}\\left( {{\\mathbf{u}}_{1},\\cdots ,{\\mathbf{u}}_{r}}\\right) \\subseteq \\operatorname{span}\\left( {{\\mathbf{v}}_{1},\\cdots ,{\\mathbf{v}}_{s}}\\right) \\equiv V \n\\]\n\nand \\( \\left\\{ {{\\mathbf{u}}_{1},\\cdots ,{\\mathbf{u}}_{r}}\\right\\} \\) are linearly indepe... | Proof: Suppose \\( r > s \\) . Let \\( {F}_{p} \\) denote the first \\( p \\) vectors in \\( \\left\\{ {{\\mathbf{u}}_{1},\\cdots ,{\\mathbf{u}}_{r}}\\right\\} \\) . In case \\( p = 0,{F}_{p} \\) will denote the empty set. Let \\( {E}_{p} \\) denote a finite list of vectors of \\( \\left\\{ {{\\mathbf{v}}_{1},\\cdots ,... | Yes |
Corollary 8.2.5 Let \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{r}}\right\} \) and \( \left\{ {{\mathbf{y}}_{1},\cdots ,{\mathbf{y}}_{s}}\right\} \) be two bases \( {}^{1} \) of \( {\mathbb{F}}^{n} \) . Then \( r = s = n \) . More generally, if you have two bases for a vector space \( V \) then they have the sam... | Proof: From the exchange theorem, if \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{r}}\right\} \) and \( \left\{ {{\mathbf{y}}_{1},\cdots ,{\mathbf{y}}_{s}}\right\} \) are two bases for \( V \), then \( r \leq s \) and \( s \leq r \) . Now note the vectors,\n\n\[ \n{\mathbf{e}}_{i} = \overset{1\text{ is in the }{i... | No |
Lemma 8.2.6 Let \( \\left\\{ {{\\mathbf{v}}_{1},\\cdots ,{\\mathbf{v}}_{r}}\\right\\} \) be a set of vectors. Then \( V \\equiv \\operatorname{span}\\left( {{\\mathbf{v}}_{1},\\cdots ,{\\mathbf{v}}_{r}}\\right\\} \) is a subspace. | Proof: Suppose \( \\alpha ,\\beta \) are two scalars and let \( \\mathop{\\sum }\\limits_{{k = 1}}^{r}{c}_{k}{\\mathbf{v}}_{k} \) and \( \\mathop{\\sum }\\limits_{{k = 1}}^{r}{d}_{k}{\\mathbf{v}}_{k} \) are two elements of \( V \) . What about\n\n\[ \n\\alpha \\mathop{\\sum }\\limits_{{k = 1}}^{r}{c}_{k}{\\mathbf{v}}_{... | No |
Lemma 8.2.8 Suppose \( \mathbf{v} \notin \operatorname{span}\left( {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k}}\right) \) and \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k}}\right\} \) is linearly independent. Then \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k},\mathbf{v}}\right\} \) is also linearly indep... | Proof: Suppose \( \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\mathbf{u}}_{i} + d\mathbf{v} = \mathbf{0} \) . It is required to verify that each \( {c}_{i} = 0 \) and that \( d = 0 \) . But if \( d \neq 0 \), then you can solve for \( \mathbf{v} \) as a linear combination of the vectors, \( \left\{ {{\mathbf{u}}_{1},\cd... | Yes |
Theorem 8.2.9 Let \( V \) be a nonzero subspace of \( Y \) a finite dimensional vector space having dimension \( n \) . Then \( V \) has a basis. | Proof: Let \( {\mathbf{v}}_{1} \in V \) where \( {\mathbf{v}}_{1} \neq 0 \) . If \( \operatorname{span}\left\{ {\mathbf{v}}_{1}\right\} = V \), stop. \( \left\{ {\mathbf{v}}_{1}\right\} \) is a basis for \( V \) . Otherwise, there exists \( {\mathbf{v}}_{2} \in V \) which is not in span \( \left\{ {\mathbf{v}}_{1}\righ... | Yes |
Corollary 8.2.10 Let \( V \) be a subspace of \( Y \), a finite dimensional vector space of dimension \( n \) and let \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \) be a linearly independent set of vectors in \( V \) . Then either it is a basis for \( V \) or there exist vectors, \( {\mathbf{v}}_{r +... | Proof: This follows immediately from the proof of Theorem 8.2.9. You do exactly the same argument except you start with \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \) rather than \( \left\{ {\mathbf{v}}_{1}\right\} \) . ∎ | No |
Theorem 8.2.11 Let \( V \) be a subspace of \( Y \), a finite dimensional vector space of dimension \( n \) and suppose \( \operatorname{span}\left( {{\mathbf{u}}_{1}\cdots ,{\mathbf{u}}_{p}}\right) = V \) where the \( {\mathbf{u}}_{i} \) are nonzero vectors. Then there exist vectors, \( \left\{ {{\mathbf{v}}_{1}\cdots... | Proof: Let \( r \) be the smallest positive integer with the property that for some set, \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \subseteq \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{p}}\right\} \), \[ \operatorname{span}\left( {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right) = V \] Then \(... | Yes |
Lemma 8.3.3 If \( z \in \mathbb{F} \) there exists \( \theta \in \mathbb{F} \) such that \( {\theta z} = \left| z\right| \) and \( \left| \theta \right| = 1 \) . | Proof: Let \( \theta = 1 \) if \( z = 0 \) and otherwise, let \( \theta = \frac{\bar{z}}{\left| z\right| } \) . Recall that for \( z = \) \( x + {iy},\bar{z} = x - {iy} \) and \( \bar{z}z = {\left| z\right| }^{2} \) . In case \( z \) is real, there is no change in the above. \( ▱ \) | Yes |
Theorem 8.3.4 (Cauchy Schwarz)Let \( H \) be an inner product space. The following inequality holds for \( \mathbf{x} \) and \( \mathbf{y} \in H \) .\n\n\[ \left| \left( {\mathbf{x},\mathbf{y}}\right) \right| \leq {\left( \mathbf{x},\mathbf{x}\right) }^{1/2}{\left( \mathbf{y},\mathbf{y}\right) }^{1/2} \]\n\nEquality ho... | Proof: Let \( \theta \in \mathbb{F} \) such that \( \left| \theta \right| = 1 \) and\n\n\[ \theta \left( {\mathbf{x},\mathbf{y}}\right) = \left| \left( {\mathbf{x},\mathbf{y}}\right) \right| \]\n\nConsider \( p\left( t\right) \equiv \left( {\mathbf{x} + \bar{\theta }t\mathbf{y},\mathbf{x} + t\bar{\theta }\mathbf{y}}\ri... | Yes |
\[ \left| {\mathbf{z} + \mathbf{w}}\right| \leq \left| \mathbf{z}\right| + \left| \mathbf{w}\right| \] | Proof: The first two claims are left as exercises. To establish the third,\n\n\[ {\left| \mathbf{z} + \mathbf{w}\right| }^{2} \equiv \left( {\mathbf{z} + \mathbf{w},\mathbf{z} + \mathbf{w}}\right) \]\n\n\[ = \left( {\mathbf{z},\mathbf{z}}\right) + \left( {\mathbf{w},\mathbf{w}}\right) + \left( {\mathbf{w},\mathbf{z}}\r... | No |
Proposition 8.3.9 For \( \mathbf{x},\mathbf{y} \in {\mathbb{C}}^{n} \) , \[ \mathop{\sum }\limits_{{i = 1}}^{n}\left| {x}_{i}\right| \left| {y}_{i}\right| \leq {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\left| {x}_{i}\right| }^{p}\right) }^{1/p}{\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\left| {y}_{i}\right| }^{{p}^{... | The proof will depend on the following lemma shown later. | No |
Lemma 8.3.10 If \( a, b \geq 0 \) and \( {p}^{\prime } \) is defined by \( \frac{1}{p} + \frac{1}{{p}^{\prime }} = 1 \), then\n\n\[ \n{ab} \leq \frac{{a}^{p}}{p} + \frac{{b}^{{p}^{\prime }}}{{p}^{\prime }}\n\] | Proof of the Proposition: If \( \mathbf{x} \) or \( \mathbf{y} \) equals the zero vector there is nothing to prove. Therefore, assume they are both nonzero. Let \( A = {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\left| {x}_{i}\right| }^{p}\right) }^{1/p} \) and \( B = {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\left| ... | Yes |
Theorem 8.3.11 The \( p \) norms do indeed satisfy the axioms of a norm. | Proof: It is obvious that \( \parallel \cdot {\parallel }_{p} \) does indeed satisfy most of the norm axioms. The only one that is not clear is the triangle inequality. To save notation write \( \parallel \cdot \parallel \) in place of \( \parallel \cdot {\parallel }_{p} \) in what follows. Note also that \( \frac{p}{{... | Yes |
Proposition 8.3.13 Suppose \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{k}}\right\} \) is an orthonormal set of vectors. Then it is linearly independent. | Proof: Suppose \( \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\mathbf{v}}_{i} = \mathbf{0} \) . Then taking inner products with \( {\mathbf{v}}_{j} \) ,\n\n\[ 0 = \left( {\mathbf{0},{\mathbf{v}}_{j}}\right) = \mathop{\sum }\limits_{i}{c}_{i}\left( {{\mathbf{v}}_{i},{\mathbf{v}}_{j}}\right) = \mathop{\sum }\limits_{i}{c}... | Yes |
Corollary 8.4.2 The set \( Q \equiv \left\lbrack {a, b}\right\rbrack + i\left\lbrack {c, d}\right\rbrack \subseteq \mathbb{C} \) is compact, meaning\n\n\[ \{ x + {iy} : x \in \left\lbrack {a, b}\right\rbrack, y \in \left\lbrack {c, d}\right\rbrack \} \] | Proof: Let \( \left\{ {{x}_{n} + i{y}_{n}}\right\} \) be a sequence in \( Q \) . Then there is a subsequence such that\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} = x \in \left\lbrack {a, b}\right\rbrack \]\n\nThere is a further subsequence such that \( \mathop{\lim }\limits_{{l \rightarrow \inft... | Yes |
Corollary 8.4.3 In \( \mathbb{C} \), let \( D\left( {z, r}\right) \equiv \{ w \in \mathbb{C} : \left| {z - w}\right| \leq r\} \) . Then \( D\left( {z, r}\right) \) is compact. | Proof: Note that\n\n\[ D\left( {z, r}\right) \subseteq \left\lbrack {\operatorname{Re}z - r,\operatorname{Re}z + r}\right\rbrack + i\left\lbrack {\operatorname{Im}z - r,\operatorname{Im}z + r}\right\rbrack \]\n\nwhich was just shown to be compact. Also, if \( {w}_{k} \rightarrow w \) where \( {w}_{k} \in D\left( {z, r}... | Yes |
Lemma 8.4.4 Let \( {K}_{i} \) be a nonempty compact set in \( \mathbb{F} \) . Then \( P \equiv \mathop{\prod }\limits_{{i = 1}}^{n}{K}_{i} \) is compact in \( {\mathbb{F}}^{n} \) . | Proof: Let \( \left\{ {\mathbf{x}}_{k}\right\} \) be a sequence in \( P \) . Taking a succession of subsequences as in the proof of Corollary 8.4.2, there exists a subsequence, still denoted as \( \left\{ {\mathbf{x}}_{k}\right\} \) such that if \( {x}_{k}^{i} \) is the \( {i}^{\text{th }} \) component of \( {\mathbf{x... | Yes |
Theorem 8.4.5 A set \( K \subseteq {\mathbb{F}}^{n} \) is compact if it is closed and bounded. If \( f : K \rightarrow \mathbb{R} \) , then \( f \) achieves its maximum and its minimum on \( K \) . | Proof: Say \( K \) is closed and bounded, being contained in \( B\left( {\mathbf{0,}r}\right) \) . Then if \( \mathbf{x} \in K \) , \( \left| {x}_{i}\right| < r \) where \( {x}_{i} \) is the \( {i}^{th} \) component. Hence \( K \subseteq \mathop{\prod }\limits_{{i = 1}}^{n}D\left( {0, r}\right) \), a compact set by Lem... | Yes |
Lemma 8.4.7 There exists \( \delta > 0 \) and \( \Delta \geq \delta \) such that\n\n\[ \n\delta = \min \{ f\left( \mathbf{\alpha }\right) : \left| \mathbf{\alpha }\right| = 1\} ,\Delta = \max \{ f\left( \mathbf{\alpha }\right) : \left| \mathbf{\alpha }\right| = 1\} \n\]\n\nAlso,\n\n\[ \n\delta \left| \mathbf{\alpha }\r... | Proof: These numbers exist thanks to Theorem 8.4.5. It cannot be that \( \delta = 0 \) because if it were, you would have \( \left| \mathbf{\alpha }\right| = 1 \) but \( \mathop{\sum }\limits_{{j = 1}}^{n}{\alpha }_{k}{\mathbf{v}}_{j} = \mathbf{0} \) which is impossible since \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\math... | Yes |
Let \( \left( {V,\parallel \cdot \parallel }\right) \) be a finite dimensional normed linear space. Then the compact sets are exactly those which are closed and bounded. Also \( \left( {V,\parallel \cdot \parallel }\right) \) is complete. If \( K \) is a closed and bounded set in \( \left( {V,\parallel \cdot \parallel ... | Proof: First note that the inequalities 8.4.21 and 8.4.22 show that both \( {\theta }^{-1} \) and \( \theta \) are continuous. Thus these take convergent sequences to convergent sequences.\n\nLet \( {\left\{ {\mathbf{w}}_{k}\right\} }_{k = 1}^{\infty } \) be a Cauchy sequence. Then from \( {8.4.22},{\left\{ \theta {\ma... | Yes |
Theorem 8.4.9 Let \( \parallel \cdot \parallel ,\parallel \parallel \cdot \parallel \parallel \) be two norms on \( V \) a finite dimensional vector space. Then they are equivalent, which means there are constants \( 0 < a < b \) such that for all \( \mathbf{v} \) ,\n\n\[ a\parallel \mathbf{v}\parallel \leq \parallel \... | Proof: In Lemma 8.4.7, let \( \delta ,\Delta \) go with \( \parallel \cdot \parallel \) and \( \widehat{\delta },\widehat{\Delta } \) go with \( \parallel \mid \cdot \parallel \mid \) . Then using the inequalities of this lemma,\n\n\[ \left| \right| \mathbf{v}\left| \right| \leq \Delta \left| {\theta \mathbf{v}}\right|... | Yes |
Corollary 8.4.10 Consider the metric spaces \( \left( {V,\parallel \cdot {\parallel }_{1}}\right) ,\left( {V,\parallel \cdot {\parallel }_{2}}\right) \) where \( V \) has dimension \( n \) . Then a set is closed or open in one of these if and only if it is respectively closed or open in the other. In other words, the t... | Proof: This follows from Theorem 7.6.5, the theorem about the equivalent formulations of continuity. Using this theorem, it follows from Theorem 8.4.9 that the identity map \( I\left( \mathbf{x}\right) \equiv \mathbf{x} \) is continuous. The reason for this is that the inequality of this theorem implies that if \( {\be... | Yes |
Lemma 9.1.2 The following estimate holds for \( x \in \left\lbrack {0,1}\right\rbrack \) . | Proof: By the Binomial theorem,\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{m}\left( \begin{matrix} m \\ k \end{matrix}\right) {\left( {e}^{t}x\right) }^{k}{\left( 1 - x\right) }^{m - k} = {\left( 1 - x + {e}^{t}x\right) }^{m}. \]\n\n(9.1.1)\n\nDifferentiating both sides with respect to \( t \) and then evaluating at \( t =... | Yes |
Lemma 9.1.4 Let \( \\mathbf{f} \) be a continuous function defined on \( {\\left\\lbrack -M, M\\right\\rbrack }^{n} \) having values in a normed linear space. Then there exists a sequence of polynomials, \( \\left\\{ {\\mathbf{p}}_{m}\\right\\} \) converging uniformly to \( \\mathbf{f} \) on \( {\\left\\lbrack -M, M\\r... | Proof: Let \( h\\left( t\\right) = - M + {2Mt} \) so \( h : \\left\\lbrack {0,1}\\right\\rbrack \\rightarrow \\left\\lbrack {-M, M}\\right\\rbrack \) and let \( \\mathbf{h}\\left( \\mathbf{t}\\right) \\equiv \) \( \\left( {h\\left( {t}_{1}\\right) ,\\cdots, h\\left( {t}_{n}\\right) }\\right) \) . Therefore, \( \\mathbf... | Yes |
Corollary 9.1.5 Let \( \mathbf{f} \) be a continuous function defined on \( \mathop{\prod }\limits_{{i = 1}}^{n}\left\lbrack {{a}_{i},{b}_{i}}\right\rbrack \) having values in a normed linear space. Then there exists a sequence of polynomials, \( \left\{ {\mathbf{p}}_{m}\right\} \) converging uniformly to \( \mathbf{f}... | Proof: You just let \( {h}_{i}\left( t\right) \) map \( \left\lbrack {0,1}\right\rbrack \) one to one and onto \( \left\lbrack {{a}_{i},{b}_{i}}\right\rbrack \) such that \( {h}_{i}^{-1}\left( x\right) \) is a polynomial. Then apply the same argument. | No |
Theorem 9.1.7 Let \( K \) be a compact set in \( {\mathbb{R}}^{n} \) and let \( \mathbf{f} \) be a continuous function defined on \( K \) having values in \( {\mathbb{R}}^{p} \) . Then there exists a sequence of polynomials \( \left\{ {\mathbf{p}}_{m}\right\} \) converging uniformly to \( \mathbf{f} \) on \( K \) . | Proof: Choose \( M \) large enough that \( K \subseteq {\left\lbrack -M, M\right\rbrack }^{n} \) and let \( \widetilde{f} \) denote a continuous function defined on all of \( {\left\lbrack -M, M\right\rbrack }^{n} \) such that \( \widetilde{\mathbf{f}} = \mathbf{f} \) on \( K \) . Such an extension exists by the Tietze... | Yes |
Corollary 9.2.3 On the interval \( \\left\\lbrack {-M, M}\\right\\rbrack \\), there exist polynomials \( {p}_{n} \) such that\n\n\[ \n{p}_{n}\\left( 0\\right) = 0 \n\]\n\nand\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{\\begin{Vmatrix}{p}_{n} - \\left| \\cdot \\right| \\end{Vmatrix}}_{\\infty } = 0 \n\... | Proof: By Corollary 9.2.2 there exists a sequence of polynomials, \( \\left\\{ {\\widetilde{p}}_{n}\\right\\} \) such that \( {\\widetilde{p}}_{n} \\rightarrow \\left| \\cdot \\right| \) uniformly. Then let \( {p}_{n}\\left( t\\right) \\equiv {\\widetilde{p}}_{n}\\left( t\\right) - {\\widetilde{p}}_{n}\\left( 0\\right)... | Yes |
Theorem 9.2.5 Let \( A \) be a compact topological space and let \( \mathcal{A} \subseteq C\left( {A;\mathbb{R}}\right) \) be an algebra of functions which separates points and annihilates no point. Then \( \mathcal{A} \) is dense in \( C\left( {A;\mathbb{R}}\right) \). | Proof: First here is a lemma. | No |
Lemma 9.2.6 Let \( {c}_{1} \) and \( {c}_{2} \) be two real numbers and let \( {x}_{1} \neq {x}_{2} \) be two points of A. Then there exists a function \( {f}_{{x}_{1}{x}_{2}} \) such that\n\n\[ \n{f}_{{x}_{1}{x}_{2}}\left( {x}_{1}\right) = {c}_{1},{f}_{{x}_{1}{x}_{2}}\left( {x}_{2}\right) = {c}_{2}.\n\] | Proof of the lemma: Let \( g \in \mathcal{A} \) satisfy\n\n\[ \ng\left( {x}_{1}\right) \neq g\left( {x}_{2}\right)\n\]\n\nSuch a \( g \) exists because the algebra separates points. Since the algebra annihilates no point, there exist functions \( h \) and \( k \) such that\n\n\[ \nh\left( {x}_{1}\right) \neq 0, k\left(... | Yes |
Lemma 9.2.8 For \( \left( {X,\tau }\right) \) a locally compact Hausdorff space with the above norm, \( {C}_{0}\left( X\right) \) is a complete space. | Proof: Let \( \left( {\widetilde{X},\widetilde{\tau }}\right) \) be the one point compactification described in Lemma 7.12.20.\n\n\[ D \equiv \{ f \in C\left( \widetilde{X}\right) : f\left( \infty \right) = 0\} . \]\n\nThen \( D \) is a closed subspace of \( C\left( \widetilde{X}\right) \) . For \( f \in {C}_{0}\left( ... | Yes |
Theorem 9.2.9 Let \( \mathcal{A} \) be an algebra of functions in \( {C}_{0}\left( {X;\mathbb{R}}\right) \) where \( \left( {X,\tau }\right) \) is a locally compact Hausdorff space which separates the points and annihilates no point. Then \( \mathcal{A} \) is dense in \( {C}_{0}\left( {X;\mathbb{R}}\right) \) . | Proof: Let \( \left( {\widetilde{X},\widetilde{\tau }}\right) \) be the one point compactification as described in Lemma 7.12.20. Let \( \widetilde{\mathcal{A}} \) denote all finite linear combinations of the form\n\n\[ \left\{ {\mathop{\sum }\limits_{{i = 1}}^{n}{c}_{i}{\widetilde{f}}_{i} + {c}_{0} : f \in \mathcal{A}... | Yes |
Proposition 10.0.3 Say \( \left\lbrack {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{r}}\right\rbrack ,\left\lbrack {{\widehat{\mathbf{x}}}_{1},\cdots ,{\widehat{\mathbf{x}}}_{r}}\right\rbrack ,\left\lbrack {{\mathbf{z}}_{1},\cdots ,{\mathbf{z}}_{r}}\right\rbrack \) are all \( r - 1 \) simplices and\n\n\[ \left\lbrack {{\mat... | Proof: If you have \( \mathop{\sum }\limits_{{i = 1}}^{s}{t}_{i}{\mathbf{y}}_{i} + {t}_{s + 1}\mathbf{b} \) in the right side, the \( {t}_{i} \) summing to 1 and nonnegative, then it is obviously in both of the two simplices on the left because of 10.0.2. Thus \( \left\lbrack {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{r},... | Yes |
Corollary 10.1.2 Let \( K \) be a closed convex bounded subset of \( {\mathbb{R}}^{n} \). Let \( \mathbf{f} : K \rightarrow K \) be continuous. Then there exists \( \mathbf{x} \in K \) such that \( \mathbf{f}\left( \mathbf{x}\right) = \mathbf{x} \). | Proof: Let \( S \) be a large simplex containing \( K \) and let \( P \) be the projection map onto \( K \). Consider \( \mathbf{g}\left( \mathbf{x}\right) \equiv \mathbf{f}\left( {P\mathbf{x}}\right) \). Then \( \mathbf{g} \) satisfies the necessary conditions for Theorem 10.1.1 and so there exists \( \mathbf{x} \in S... | Yes |
Lemma 10.2.1 Let \( \\mathbf{f} \) be continuous and map \( \\overline{B\\left( \\mathbf{p}, r\\right) } \\subseteq \\mathbb{R}^{n} \) to \( \\mathbb{R}^{n} \) . Suppose that for all \( \\mathbf{x} \\in \\overline{B\\left( \\mathbf{p}, r\\right) } \), \[ \\left| \\mathbf{f}\\left( \\mathbf{x}\\right) - \\mathbf{x}\\rig... | Proof: This is from the Brouwer fixed point theorem, Corollary 10.1.2. Consider for \( \\mathbf{y} \\in B\\left( \\mathbf{p},\\left( 1 - \\varepsilon \\right) r\\right) \) \[ \\mathbf{h}\\left( \\mathbf{x}\\right) \\equiv \\mathbf{x} - \\mathbf{f}\\left( \\mathbf{x}\\right) + \\mathbf{y} \] Then \( \\mathbf{h} \) is co... | Yes |
Lemma 10.2.2 Let \( K \) be a compact set in \( {\mathbb{R}}^{n} \) and let \( \mathbf{h} : K \rightarrow {\mathbb{R}}^{n} \) be continuous, \( \mathbf{z} \in K \) is fixed. Let \( \delta > 0 \). Then there exists a polynomial \( \mathbf{g} \) (each component a polynomial) such that \[ \parallel \mathbf{g} - \mathbf{h}... | Proof: By the Weierstrass approximation theorem, Theorem 9.2.5, (apply this theorem to the algebra of real polynomials) there exists a polynomial \( \widehat{\mathbf{g}} \) such that \[ \parallel \widehat{\mathbf{g}} - \mathbf{h}{\parallel }_{K} < \frac{\delta }{3} \] Then define for \( \mathbf{y} \in K \) \[ \mathbf{g... | Yes |
Lemma 10.2.3 Let \( \mathbf{f} : \overline{B\left( {\mathbf{p}, r}\right) } \rightarrow {\mathbb{R}}^{n} \) where the ball is also in \( {\mathbb{R}}^{n} \) . Let \( \mathbf{f} \) be one to one, \( \mathbf{f} \) continuous. Then there exists \( \delta > 0 \) such that\n\n\[ \mathbf{f}\left( \overline{B\left( {\mathbf{p... | Proof: Since \( \mathbf{f}\left( \overline{B\left( {\mathbf{p}, r}\right) }\right) \) is compact, it follows that \( {\mathbf{f}}^{-1} : \mathbf{f}\left( \overline{B\left( {\mathbf{p}, r}\right) }\right) \rightarrow \overline{B\left( {\mathbf{p}, r}\right) } \) is continuous. By Lemma 10.2.2, there exists a polynomial ... | Yes |
Theorem 10.2.4 Let \( U \) be an open set in \( {\mathbb{R}}^{n} \) and let \( \mathbf{f} : U \rightarrow {\mathbb{R}}^{n} \) be one to one and continuous. Then \( \mathbf{f}\left( U\right) \) is also an open subset in \( {\mathbb{R}}^{n} . | Proof: It suffices to show that if \( \mathbf{p} \in U \) then \( \mathbf{f}\left( \mathbf{p}\right) \) is an interior point of \( \mathbf{f}\left( U\right) \) . Let \( \overline{B\left( {\mathbf{p}, r}\right) } \subseteq U \) . By Lemma 10.2.3, \( \mathbf{f}\left( U\right) \supseteq \mathbf{f}\left( \overline{B\left( ... | Yes |
Theorem 11.1.5 Let \( {\left\{ {E}_{m}\right\} }_{m = 1}^{\infty } \) be a sequence of measurable sets in a measure space \( \left( {\Omega ,\mathcal{F},\mu }\right) \). Then if \( \cdots {E}_{n} \subseteq {E}_{n + 1} \subseteq {E}_{n + 2} \subseteq \cdots \), \[ \mu \left( {{ \cup }_{i = 1}^{\infty }{E}_{i}}\right) = ... | Proof: First note that \( { \cap }_{i = 1}^{\infty }{E}_{i} = {\left( { \cup }_{i = 1}^{\infty }{E}_{i}^{C}\right) }^{C} \in \mathcal{F} \) so \( { \cap }_{i = 1}^{\infty }{E}_{i} \) is measurable. Also note that for \( A \) and \( B \) sets of \( \mathcal{F}, A \smallsetminus B \equiv {\left( {A}^{C} \cup B\right) }^{... | Yes |
Lemma 11.1.6 Let \( f : \Omega \rightarrow ( - \infty ,\infty \rbrack \) where \( \mathcal{F} \) is a \( \sigma \) algebra of subsets of \( \Omega \) . Then the following are equivalent.\n\n\[ \n{f}^{-1}(\left( {d,\infty \rbrack }\right) \in \mathcal{F}\text{for all finite}d\text{,}\n\]\n\n\[ \n{f}^{-1}\left( \left( {-... | Proof: First note that the first and the third are equivalent. To see this, observe\n\n\[ \n{f}^{-1}\left( \left\lbrack {d,\infty }\right\rbrack \right) = { \cap }_{n = 1}^{\infty }{f}^{-1}(\left( {d - 1/n,\infty \rbrack }\right) ,\n\]\n\nand so if the first condition holds, then so does the third.\n\n\[ \n{f}^{-1}(\le... | Yes |
Theorem 11.1.8 Let \( {f}_{n} \) and \( f \) be functions mapping \( \Omega \) to \( ( - \infty ,\infty \rbrack \) where \( \mathcal{F} \) is a \( \sigma \) algebra of measurable sets of \( \Omega \) . Then if \( {f}_{n} \) is measurable, and \( f\left( \omega \right) = \) \( \mathop{\lim }\limits_{{n \rightarrow \inft... | Proof: First it is shown \( {f}^{-1}\left( \left( {a, b}\right) \right) \in \mathcal{F} \) . Let \( {V}_{m} \equiv \left( {a + \frac{1}{m}, b - \frac{1}{m}}\right) \) and \( {\bar{V}}_{m} = \left\lbrack {a + \frac{1}{m}, b - \frac{1}{m}}\right\rbrack \) . Then for all \( m,{V}_{m} \subseteq \left( {a, b}\right) \) and\... | Yes |
Theorem 11.1.9 Let \( \\left\\{ {f}_{n}\\right\\} \) be a sequence of measurable functions mapping \( \\Omega \) to \( \\left( {X, d}\\right) \) where \( \\left( {X, d}\\right) \) is a metric space and \( \\left( {\\Omega ,\\mathcal{F}}\\right) \) is a measure space. Suppose also that \( f\\left( \\omega \\right) = \\m... | Proof: It is required to show \( {f}^{-1}\\left( U\\right) \) is measurable for all \( U \) open. Let\n\n\[ \n{V}_{m} \\equiv \\left\\{ {x \\in U : \\operatorname{dist}\\left( {x,{U}^{C}}\\right) > \\frac{1}{m}}\\right\\} .\n\]\n\nThus\n\[ \n{V}_{m} \\subseteq \\left\\{ {x \\in U : \\operatorname{dist}\\left( {x,{U}^{C... | Yes |
Theorem 11.1.12 Let \( \mathcal{B} \) consist of open cubes of the form\n\n\[ \n{Q}_{\mathbf{x}} \equiv \mathop{\prod }\limits_{{i = 1}}^{n}\left( {{x}_{i} - \delta ,{x}_{i} + \delta }\right)\n\]\n\nwhere \( \delta \) is a positive rational number and \( \mathbf{x} \in {\mathbb{Q}}^{n} \) . Then every open set in \( {\... | Proof: Let \( U \) be an open set and let \( \mathbf{y} \in U \) . Since \( U \) is open, \( B\left( {\mathbf{y}, r}\right) \subseteq U \) for some \( r > 0 \) and it can be assumed \( r/\sqrt{n} \in \mathbb{Q} \) . Let\n\n\[ \n\mathbf{x} \in B\left( {\mathbf{y},\frac{r}{{10}\sqrt{n}}}\right) \cap {\mathbb{Q}}^{n}\n\]\... | Yes |
Theorem 11.1.13 Let \( {f}_{i} : \Omega \rightarrow \mathbb{R} \) for \( i = 1,\cdots, n \) be measurable functions and let \( g : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be continuous where \( \mathbf{f} \equiv {\left( {f}_{1}\cdots {f}_{n}\right) }^{T} \) . Then \( g \circ \mathbf{f} \) is a measurable function fr... | Proof: First it is shown\n\n\[{\left( g \circ \mathbf{f}\right) }^{-1}\left( \left( {a, b}\right) \right) \in \mathcal{F}.\]\n\nNow \( {\left( g \circ \mathbf{f}\right) }^{-1}\left( \left( {a, b}\right) \right) = {\mathbf{f}}^{-1}\left( {{g}^{-1}\left( \left( {a, b}\right) \right) }\right) \) and since \( g \) is conti... | Yes |
Corollary 11.1.14 Sums, products, and linear combinations of measurable functions are measurable. | Proof: To see the product of two measurable functions is measurable, let \( g\left( {x, y}\right) = {xy} \), a continuous function defined on \( {\mathbb{R}}^{2} \) . Thus if you have two measurable functions, \( {f}_{1} \) and \( {f}_{2} \) defined on \( \Omega \) ,\n\n\[ g \circ \left( {{f}_{1},{f}_{2}}\right) \left(... | Yes |
Lemma 11.3.1 Let \( f\left( {a, b}\right) \in \left\lbrack {-\infty ,\infty }\right\rbrack \) for \( a \in A \) and \( b \in B \) where \( A, B \) are sets. Then\n\n\[ \mathop{\sup }\limits_{{a \in A}}\mathop{\sup }\limits_{{b \in B}}f\left( {a, b}\right) = \mathop{\sup }\limits_{{b \in B}}\mathop{\sup }\limits_{{a \in... | Proof: Note that for all \( a, b, f\left( {a, b}\right) \leq \mathop{\sup }\limits_{{b \in B}}\mathop{\sup }\limits_{{a \in A}}f\left( {a, b}\right) \) and therefore, for all \( a \) ,\n\n\[ \mathop{\sup }\limits_{{b \in B}}f\left( {a, b}\right) \leq \mathop{\sup }\limits_{{b \in B}}\mathop{\sup }\limits_{{a \in A}}f\l... | Yes |
Lemma 11.3.3 Let \( {a}_{ij} \geq 0 \) . Then \( \mathop{\sum }\limits_{{i = 1}}^{\infty }\mathop{\sum }\limits_{{j = 1}}^{\infty }{a}_{ij} = \mathop{\sum }\limits_{{j = 1}}^{\infty }\mathop{\sum }\limits_{{i = 1}}^{\infty }{a}_{ij} \) . Also if \( {\left\{ {b}_{j}\right\} }_{j = 1}^{\infty } \) is any enumeration of t... | Proof: First note there is no trouble in defining these sums because the \( {a}_{ij} \) are all nonnegative. If a sum diverges, it only diverges to \( \infty \) and so \( \infty \) is written as the\n\nanswer.\n\[ \mathop{\sum }\limits_{{j = 1}}^{\infty }\mathop{\sum }\limits_{{i = 1}}^{\infty }{a}_{ij} \geq \mathop{\s... | Yes |
Lemma 11.3.4 The following inequality holds. | Let \( N \in \mathbb{N} \) . \[ \mathop{\sum }\limits_{{i = 1}}^{{2N}}\frac{h}{2}\mu \left( \left\lbrack {i\frac{h}{2} < f}\right\rbrack \right) = \mathop{\sum }\limits_{{i = 1}}^{{2N}}\frac{h}{2}\mu \left( \left\lbrack {{ih} < {2f}}\right\rbrack \right) \] \[ = \mathop{\sum }\limits_{{i = 1}}^{N}\frac{h}{2}\mu \left( ... | Yes |
Lemma 11.3.7 Let the nonnegative simple function, \( s \) be defined as\n\n\[ s\left( \omega \right) = \mathop{\sum }\limits_{{i = 1}}^{n}{c}_{i}{\mathcal{X}}_{{E}_{i}}\left( \omega \right) \]\n\nwhere the \( {c}_{i} \) are not necessarily distinct but the \( {E}_{i} \) are disjoint. It follows that\n\n\[ \int s = \mat... | Proof: Let the values of \( s \) be \( \left\{ {{a}_{1},\cdots ,{a}_{m}}\right\} \) . Therefore, since the \( {E}_{i} \) are disjoint, each \( {a}_{i} \) equal to one of the \( {c}_{j} \) . Let \( {A}_{i} \equiv \cup \left\{ {{E}_{j} : {c}_{j} = {a}_{i}}\right\} \) . Then from Lemma 11.3.6 it follows that\n\n\[ \int s ... | Yes |
Lemma 11.3.8 If \( a, b \geq 0 \) and if \( s \) and \( t \) are nonnegative simple functions, then\n\n\[ \n\int {as} + {bt} = a\int s + b\int t \n\] | Proof: Let\n\n\[ \ns\left( \omega \right) = \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{\mathcal{X}}_{{A}_{i}}\left( \omega \right), t\left( \omega \right) = \mathop{\sum }\limits_{{i = 1}}^{m}{\beta }_{j}{\mathcal{X}}_{{B}_{j}}\left( \omega \right) \n\]\n\nwhere \( {\alpha }_{i} \) are the distinct values of \( s... | Yes |
Theorem 11.3.9 Let \( f \geq 0 \) be measurable. Then there exists a sequence of nonnegative simple functions \( \left\{ {s}_{n}\right\} \) satisfying\n\n\[ 0 \leq {s}_{n}\left( \omega \right) \]\n\n(11.3.14)\n\n\[ \cdots {s}_{n}\left( \omega \right) \leq {s}_{n + 1}\left( \omega \right) \cdots \]\n\n\[ f\left( \omega ... | Proof: Letting \( I \equiv \{ \omega : f\left( \omega \right) = \infty \} \), define\n\n\[ {t}_{n}\left( \omega \right) = \mathop{\sum }\limits_{{k = 0}}^{{2}^{n}}\frac{k}{n}{\mathcal{X}}_{\left\lbrack k/n \leq f < \left( k + 1\right) /n\right\rbrack }\left( \omega \right) + n{\mathcal{X}}_{I}\left( \omega \right) .\n\... | Yes |
Theorem 11.3.10 Let \( \left( {\Omega ,\mathcal{F}}\right) \) be a measure space and let \( f : \Omega \rightarrow X \) where \( \left( {X, d}\right) \) is a separable metric space. Then \( f \) is a measurable function if and only if there exists a sequence of simple functions, \( \left\{ {f}_{n}\right\} \) such that ... | Proof: Let \( D = {\left\{ {x}_{k}\right\} }_{k = 1}^{\infty } \) be a countable dense subset of \( X \) . First suppose \( f \) is measurable. Then since in a metric space every open set is the countable intersection of closed sets, it follows \( {f}^{-1} \) (closed set) \( \in \mathcal{F} \) . Now let \( {D}_{n} = {\... | Yes |
Theorem 11.3.12 (Monotone Convergence theorem) Let \( f \) have values in \( \\left\\lbrack {0,\\infty }\\right\\rbrack \) and suppose \( \\left\\{ {f}_{n}\\right\\} \) is a sequence of nonnegative measurable functions having values in \( \\left\\lbrack {0,\\infty }\\right\\rbrack \) and satisfying\n\n\[ \n\\mathop{\\l... | Proof: From Lemmas 11.3.1 and 11.3.2,\n\n\[ \n\\int {fd\\mu } \\equiv \\mathop{\\sup }\\limits_{{h > 0}}\\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }{h\\mu }\\left( \\left\\lbrack {{ih} < f}\\right\\rbrack \\right)\n\]\n\n\[ \n= \\mathop{\\sup }\\limits_{{h > 0}}\\mathop{\\sup }\\limits_{k}\\mathop{\\sum }\\limits_{{i ... | Yes |
Lemma 11.3.15 Let \( \left\{ {a}_{n}\right\} \) be an increasing (decreasing) sequence in \( \left\lbrack {-\infty ,\infty }\right\rbrack \) . Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} \) exists. | Proof: Suppose first \( \left\{ {a}_{n}\right\} \) is increasing. Recall this means \( {a}_{n} \leq {a}_{n + 1} \) for all \( n \) . If the sequence is bounded above, then it has a least upper bound and so \( {a}_{n} \rightarrow a \) where \( a \) is its least upper bound. If the sequence is not bounded above, then for... | Yes |
Theorem 11.3.18 (Fatou’s lemma) Let \( {f}_{n} \) be a nonnegative measurable function with values in \( \left\lbrack {0,\infty }\right\rbrack \) . Let \( g\left( \omega \right) = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( \omega \right) \) . Then \( g \) is measurable and\n\n\[ \int {gd\mu } \leq ... | Proof: Let \( {g}_{n}\left( \omega \right) = \inf \left\{ {{f}_{k}\left( \omega \right) : k \geq n}\right\} \) . Then\n\n\[ {g}_{n}^{-1}\left( \left\lbrack {a,\infty }\right\rbrack \right) = { \cap }_{k = n}^{\infty }{f}_{k}^{-1}\left( \left\lbrack {a,\infty }\right\rbrack \right) \in \mathcal{F}. \]\n\nThus \( {g}_{n}... | Yes |
Theorem 11.3.19 Let \( f, g \) be nonnegative measurable functions and let \( a, b \) be nonnegative numbers. Then\n\n\[ \int \left( {{af} + {bg}}\right) {d\mu } = a\int {fd\mu } + b\int {gd\mu }.\] | Proof: By Theorem 11.3.9 on Page 257 there exist sequences of nonnegative simple functions, \( {s}_{n} \rightarrow f \) and \( {t}_{n} \rightarrow g \) . Then by the monotone convergence theorem and Lemma 11.3.8,\n\n\[ \int \left( {{af} + {bg}}\right) {d\mu } = \mathop{\lim }\limits_{{n \rightarrow \infty }}\int a{s}_{... | Yes |
Lemma 11.4.3 The definition, 11.4.2 is well defined. Furthermore, I is linear on the vector space of complex simple functions. Also the triangle inequality holds, | Proof: Suppose \( \mathop{\sum }\limits_{{k = 1}}^{n}{c}_{k}{\mathcal{X}}_{{E}_{k}}\left( \omega \right) = 0 \) . Does it follow that \( \mathop{\sum }\limits_{k}{c}_{k}\mu \left( {E}_{k}\right) = 0 \) ? The supposition implies\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{n}\operatorname{Re}{c}_{k}{\mathcal{X}}_{{E}_{k}}\lef... | Yes |
Lemma 11.4.5 Definition 11.4.4 is well defined. | Proof: There are several things which need to be verified. First suppose 11.4.23. Then by Lemma 11.4.3\n\n\[ \left| {I\left( {s}_{n}\right) - I\left( {s}_{m}\right) }\right| = \left| {I\left( {{s}_{n} - {s}_{m}}\right) }\right| \leq I\left( \left| {{s}_{n} - {s}_{m}}\right| \right) \]\n\nand for \( m, n \) large enough... | Yes |
Lemma 11.4.6 Suppose \( f \) has values in \( \lbrack 0,\infty ) \) and \( f \in {L}^{1}\left( \Omega \right) \) . Then \( f \) is measurable and \[ I\left( f\right) = \int {fd\mu } \] | Proof: Since \( f \) is the pointwise limit of a sequence of complex simple functions, \( \left\{ {s}_{n}\right\} \) having the properties described in Definition 11.4.4, it follows \( f\left( \omega \right) = \) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\operatorname{Re}{s}_{n}\left( \omega \right) \) and so \... | Yes |
Theorem 11.4.7 \( \int {d\mu } \) is linear on \( {L}^{1}\left( \Omega \right) \) and \( {L}^{1}\left( \Omega \right) \) is a complex vector space. If \( f \in {L}^{1}\left( \Omega \right) \), then \( \operatorname{Re}f,\operatorname{Im}f \), and \( \left| f\right| \) are all in \( {L}^{1}\left( \Omega \right) \) . Fur... | Proof: First it is necessary to verify that \( {L}^{1}\left( \Omega \right) \) is really a vector space because it makes no sense to speak of linear maps without having these maps defined on a vector space. Let \( f, g \) be in \( {L}^{1}\left( \Omega \right) \) and let \( a, b \in \mathbb{C} \) . Then let \( \left\{ {... | Yes |
Theorem 11.4.9 (Dominated Convergence theorem) Let \( {f}_{n} \in {L}^{1}\left( \Omega \right) \) and suppose\n\n\[ f\left( \omega \right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( \omega \right) \]\n\nand there exists a measurable function \( g \), with values in \( \left\lbrack {0,\infty }\right\... | Proof: \( f \) is measurable by Theorem 11.1.8. Since \( \left| f\right| \leq g \), it follows that\n\n\[ f \in {L}^{1}\left( \Omega \right) \text{and}\left| {f - {f}_{n}}\right| \leq {2g}\text{.} \]\n\nBy Fatou's lemma (Theorem 11.3.18),\n\n\[ \int {2gd\mu } \leq \lim \mathop{\inf }\limits_{{n \rightarrow \infty }}\in... | Yes |
Corollary 11.4.10 Suppose \( {f}_{n} \in {L}^{1}\left( \Omega \right) \) and \( f\left( \omega \right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( \omega \right) \) . Suppose also there exist measurable functions, \( {g}_{n}, g \) with values in \( \left\lbrack {0,\infty }\right\rbrack \) such that\n... | Proof: It is just like the above. This time \( g + {g}_{n} - \left| {f - {f}_{n}}\right| \geq 0 \) and so by Fatou's lemma,\n\n\[ \int {2gd\mu } - \lim \mathop{\sup }\limits_{{n \rightarrow \infty }}\int \left| {f - {f}_{n}}\right| {d\mu } = \]\n\n\[ \lim \mathop{\inf }\limits_{{n \rightarrow \infty }}\int \left( {{g}_... | Yes |
Lemma 11.5.2 If \( \mathfrak{S} \) is uniformly integrable, then \( \left| \mathfrak{S}\right| \equiv \{ \left| f\right| : f \in \mathfrak{S}\} \) is uniformly integrable. Also \( \mathfrak{S} \) is uniformly integrable if \( \mathfrak{S} \) is finite. | Proof: Let \( \varepsilon > 0 \) be given and suppose \( \mathfrak{S} \) is uniformly integrable. First suppose the functions are real valued. Let \( \delta \) be such that if \( \mu \left( E\right) < \delta \), then\n\n\[ \left| {{\int }_{E}{fd\mu }}\right| < \frac{\varepsilon }{2} \]\n\nfor all \( f \in \mathfrak{S} ... | Yes |
Theorem 11.5.3 Let \( \left\{ {f}_{n}\right\} \) be a uniformly integrable set of complex valued functions, \( \mu \left( \Omega \right) < \infty \), and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) a.e. where \( f \) is a measurable complex valued function. Then \( f \in {L}^{1}\left( \Omega \right) \) an... | Proof: First it will be shown that \( f \in {L}^{1}\left( \Omega \right) \) . By uniform integrability, there exists \( \delta > 0 \) such that if \( \mu \left( E\right) < \delta \), then\n\n\[ {\int }_{E}\left| {f}_{n}\right| {d\mu } < 1 \]\n\nfor all \( n \) . By Egoroff’s theorem, there exists a set, \( E \) of meas... | Yes |
Lemma 11.6.4 Let \( \mu \) be a finite measure on a \( \sigma \) algebra containing \( \mathcal{B}\left( X\right) \), the Borel sets of \( X \), a separable complete metric space. Then if \( C \) is a closed set,\n\n\[ \mu \left( C\right) = \sup \{ \mu \left( K\right) : K \subseteq C\text{ and }K\text{ is compact. }\} ... | Proof: Let \( \left\{ {a}_{k}\right\} \) be a countable dense subset of \( C \) . Thus \( { \cup }_{k = 1}^{\infty }B\left( {{a}_{k},\frac{1}{n}}\right) \supseteq C \) . Therefore, there exists \( {m}_{n} \) such that\n\n\[ \mu \left( {C \smallsetminus { \cup }_{k = 1}^{{m}_{n}}\overline{B\left( {{a}_{k},\frac{1}{n}}\r... | Yes |
Corollary 11.6.8 Let \( \Omega \) be a complete metric space which is the countable union of compact sets \( {K}_{n} \) and suppose, for \( \mu \) a Borel measure, \( \mu \left( {K}_{n}\right) \) is finite. Then \( \mu \) must be regular. In particular, if \( \Omega \) is a metric space and the closure of each ball is ... | Proof: Let the compact sets be increasing without loss of generality, and let \( {\mu }_{n}\left( E\right) \equiv \mu \left( {{K}_{n} \cap E}\right) \) . Thus \( {\mu }_{n} \) is a finite measure defined on the Borel sets of a Polish space so it is regular. Letting \( l < \mu \left( E\right) \), there exists \( n \) su... | Yes |
Lemma 11.7.2 Let \( \left( {\Omega, d}\right) \) be a metric space in which closed balls are compact. Then if \( K \) is a compact subset of an open set \( V \), then there exists \( \phi \) such that \( K \prec \phi \prec V \) . | Proof: Since \( K \) is compact, the distance between \( K \) and \( {V}^{C} \) is positive, \( \delta > 0 \) . Otherwise there would be \( {x}_{n} \in K \) and \( {y}_{n} \in {V}^{C} \) with \( d\left( {{x}_{n},{y}_{n}}\right) < 1/n \) . Taking a subsequence, still denoted with \( n \), we can assume \( {x}_{n} \right... | Yes |
For each \( E \in \mathcal{F} \), there is an \( {F}_{\sigma } \) set \( F \) and a \( {G}_{\delta } \) set \( G \) such that \( F \subseteq E \subseteq G \) and \( \mu \left( {G \smallsetminus F}\right) = 0 \) . | Let \( {R}_{n} \equiv B\left( {{x}_{0}, n}\right) ,{R}_{0} = \varnothing \) . If \( E \) is Lebesgue measurable, let \( {E}_{n} \equiv E \cap \left( {{R}_{n} \smallsetminus {R}_{n - 1}}\right) \) . Thus these \( {E}_{n} \) are disjoint and their union is \( E \) . By outer regularity, there exists open \( {U}_{n} \sups... | Yes |
Lemma 12.1.3 If \( A \) is \( \mu \) measurable, then \( A \) is \( \mu \lfloor S \) measurable. | Proof: Suppose \( A \) is \( \mu \) measurable. It is desired to to show that for all \( T \subseteq \Omega \) ,\n\n\[ \left( {\mu \lfloor S}\right) \left( T\right) = \left( {\mu \lfloor S}\right) \left( {T \cap A}\right) + \left( {\mu \lfloor S}\right) \left( {T \smallsetminus A}\right) . \]\n\nThus it is desired to s... | Yes |
Theorem 12.1.4 The collection of \( \mu \) measurable sets, \( \mathcal{S} \), forms a \( \sigma \) algebra and\n\n\[ \text{If}{F}_{i} \in \mathcal{S},{F}_{i} \cap {F}_{j} = \varnothing \text{, then}\mu \left( {{ \cup }_{i = 1}^{\infty }{F}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {F}_{i}\right... | Proof: First note that \( \varnothing \) and \( \Omega \) are obviously in \( \mathcal{S} \). Now suppose \( A, B \in \mathcal{S} \). I will show \( A \smallsetminus B \equiv A \cap {B}^{C} \) is in \( \mathcal{S} \). To do so, consider the following picture.\n\n \) be a measure space. Let \( \bar{\mu } \) be the outer measure determined by \( \mu \) . Also denote as \( \overline{\mathcal{F}} \), the \( \sigma \) algebra of \( \bar{\mu } \) measurable sets. Thus \( \left( {\Omega ,\overline{\mathcal{F}},\bar{\mu... | Proof: All that remains to show is the last claim. But this is obvious because if \( S \) is a set,\n\n\[ \bar{\mu }\left( S\right) \leq \bar{\mu }\left( {S \cap E}\right) + \bar{\mu }\left( {S \smallsetminus E}\right) \]\n\n\[ \leq \bar{\mu }\left( E\right) + \bar{\mu }\left( {S \smallsetminus E}\right) \]\n\n\[ = \ba... | Yes |
Lemma 12.1.6 Let \( \Omega \) be a Hausdorff space and suppose \( \mu \) is an outer measure satisfying \( \mu \) is finite on compact sets and the following conditions,\n\n1. \( \mu \left( E\right) = \inf \{ \mu \left( V\right), V \supseteq E, V \) open \( \} \) for all \( E \) . (Outer regularity.)\n\n2. For every op... | Proof: First we establish 1 and 2 and use them to establish the last assertion. Consider 2. Suppose it is not true. Then there exists an open set \( V \) having \( \mu \left( V\right) < \) \( \infty \) but for all \( K \subseteq V,\mu \left( {V \smallsetminus K}\right) \geq \varepsilon \) for some \( \varepsilon > 0 \)... | Yes |
Lemma 12.1.11 Let \( \mu \) be a finite measure defined on \( \mathcal{B}\left( E\right) \) where \( E \) is a closed subset of \( {\mathbb{R}}^{n} \) . Then for every \( F \in \mathcal{B}\left( E\right) \) , \[ \mu \left( F\right) = \sup \{ \mu \left( K\right) : K \subseteq F, K\text{ is closed }\} \] \[ \mu \left( F\... | Proof: For convenience, I will call a measure which satisfies the above two conditions \ | No |
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