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Proposition 2.7 Suppose \( f \in {H}_{0}^{1}\left( \Omega \right) \) . There exists a unique function \( u \) from \( \left( {0, + \infty }\right) \) to \( {H}_{0}^{1}\left( \Omega \right) \), differentiable in \( \left( {0, + \infty }\right) \) and satisfying the following conditions:\n\n\[ - {u}^{\prime }\left( t\rig... | Proof. Suppose \( u \) satisfies the conditions of the statement. Then\n\n\[ \frac{d}{dt}{\left( \parallel u\left( t\right) {\parallel }_{{L}^{2}}\right) }^{2} = 2\operatorname{Re}{\left( {u}^{\prime }\left( t\right) \mid u\left( t\right) \right) }_{{L}^{2}} = - 2{\left( \parallel u\left( t\right) {\parallel }_{{H}_{0}... | Yes |
Proposition 2.8 Suppose \( f, g \in {H}_{0}^{1}\left( \Omega \right) \) . There exists at most one function \( u \) from \( \mathbb{R} \) to \( {H}_{0}^{1}\left( \Omega \right) \), twice differentiable on \( \mathbb{R} \) and satisfying these conditions:\n\n\[ \n- {u}^{\prime \prime }\left( t\right) = {\Delta u}\left( ... | Proof. Let \( u \) satisfy the conditions of the statement. By Proposition 2.1,\n\n\[ \n\frac{d}{dt}{\left( {\begin{Vmatrix}{u}^{\prime }\left( t\right) \end{Vmatrix}}_{{L}^{2}}\right) }^{2} = 2\operatorname{Re}{\left( {u}^{\prime \prime }\left( t\right) \mid {u}^{\prime }\left( t\right) \right) }_{{L}^{2}}\n\]\n\n\[ \... | Yes |
Lemma 1.1 If a rectangle is the almost disjoint union of finitely many other rectangles, say \( R = \mathop{\bigcup }\limits_{{k = 1}}^{N}{R}_{k} \), then\n\n\[ \left| R\right| = \mathop{\sum }\limits_{{k = 1}}^{N}\left| {R}_{k}\right| \] | Proof. We consider the grid formed by extending indefinitely the sides of all rectangles \( {R}_{1},\ldots ,{R}_{N} \) . This construction yields finitely many rectangles \( {\widetilde{R}}_{1},\ldots ,{\widetilde{R}}_{M} \), and a partition \( {J}_{1},\ldots ,{J}_{N} \) of the integers between 1 and \( M \), such that... | Yes |
Theorem 1.3 Every open subset \( \mathcal{O} \) of \( \mathbb{R} \) can be writen uniquely as a countable union of disjoint open intervals. | Proof. For each \( x \in \mathcal{O} \), let \( {I}_{x} \) denote the largest open interval containing \( x \) and contained in \( \mathcal{O} \). More precisely, since \( \mathcal{O} \) is open, \( x \) is contained in some small (non-trivial) interval, and therefore if\n\n\[ \n{a}_{x} = \inf \{ a < x : \left( {a, x}\... | Yes |
Theorem 1.4 Every open subset \( \mathcal{O} \) of \( {\mathbb{R}}^{d}, d \geq 1 \), can be written as a countable union of almost disjoint closed cubes. | Proof. We must construct a countable collection \( \mathcal{Q} \) of closed cubes whose interiors are disjoint, and so that \( \mathcal{O} = \mathop{\bigcup }\limits_{{Q \in \mathcal{Q}}}Q \) . As a first step, consider the grid in \( {\mathbb{R}}^{d} \) formed by taking all closed cubes of side length 1 whose vertices... | Yes |
Property 2 If \( {m}_{ * }\left( E\right) = 0 \), then \( E \) is measurable. In particular, if \( F \) is a subset of a set of exterior measure 0, then \( F \) is measurable. | By Observation 3 of the exterior measure, for every \( \epsilon > 0 \) there exists an open set \( \mathcal{O} \) with \( E \subset \mathcal{O} \) and \( {m}_{ * }\left( \mathcal{O}\right) \leq \epsilon \) . Since \( \left( {\mathcal{O} - E}\right) \subset \mathcal{O} \) , monotonicity implies \( {m}_{ * }\left( {\math... | No |
Property 3 A countable union of measurable sets is measurable. | Suppose \( E = \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j} \), where each \( {E}_{j} \) is measurable. Given \( \epsilon > 0 \), we may choose for each \( j \) an open set \( {\mathcal{O}}_{j} \) with \( {E}_{j} \subset {\mathcal{O}}_{j} \) and \( {m}_{ * }\left( {{\mathcal{O}}_{j} - {E}_{j}}\right) \leq \epsil... | Yes |
Lemma 3.1 If \( F \) is closed, \( K \) is compact, and these sets are disjoint, then \( d\left( {F, K}\right) > 0 \) . | Proof. Since \( F \) is closed, for each point \( x \in K \), there exists \( {\delta }_{x} > 0 \) so that \( d\left( {x, F}\right) > 3{\delta }_{x} \) . Since \( \mathop{\bigcup }\limits_{{x \in K}}{B}_{2{\delta }_{x}}\left( x\right) \) covers \( K \), and \( K \) is compact, we may find a subcover, which we denote by... | Yes |
Property 5 The complement of a measurable set is measurable. | If \( E \) is measurable, then for every positive integer \( n \) we may choose an open set \( {\mathcal{O}}_{n} \) with \( E \subset {\mathcal{O}}_{n} \) and \( {m}_{ * }\left( {{\mathcal{O}}_{n} - E}\right) \leq 1/n \) . The complement \( {\mathcal{O}}_{n}^{c} \) is closed, hence measurable, which implies that the un... | Yes |
Property 6 A countable intersection of measurable sets is measurable. | This follows from Properties 3 and 5, since\n\n\[ \mathop{\bigcap }\limits_{{j = 1}}^{\infty }{E}_{j} = {\left( \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j}^{c}\right) }^{c} \]\n\n | Yes |
Theorem 3.2 If \( {E}_{1},{E}_{2},\ldots \), are disjoint measurable sets, and \( E = \) \( \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j} \), then\n\n\[ m\left( E\right) = \mathop{\sum }\limits_{{j = 1}}^{\infty }m\left( {E}_{j}\right) \] | Proof. First, we assume further that each \( {E}_{j} \) is bounded. Then, for each \( j \), by applying the definition of measurability to \( {E}_{j}^{c} \), we can choose a closed subset \( {F}_{j} \) of \( {E}_{j} \) with \( {m}_{ * }\left( {{E}_{j} - {F}_{j}}\right) \leq \epsilon /{2}^{j} \) . For each fixed \( N \)... | Yes |
Corollary 3.3 Suppose \( {E}_{1},{E}_{2},\ldots \) are measurable subsets of \( {\mathbb{R}}^{d} \). (i) If \( {E}_{k} \nearrow E \), then \( m\left( E\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}m\left( {E}_{N}\right) \). | Proof. For the first part, let \( {G}_{1} = {E}_{1},{G}_{2} = {E}_{2} - {E}_{1} \), and in general \( {G}_{k} = {E}_{k} - {E}_{k - 1} \) for \( k \geq 2 \). By their construction, the sets \( {G}_{k} \) are measurable, disjoint, and \( E = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{G}_{k} \). Hence \[ m\left( E\right... | Yes |
Theorem 3.4 Suppose \( E \) is a measurable subset of \( {\mathbb{R}}^{d} \) . Then, for every \( \epsilon > 0 \) :\n\n(i) There exists an open set \( \mathcal{O} \) with \( E \subset \mathcal{O} \) and \( m\left( {\mathcal{O} - E}\right) \leq \epsilon \) .\n\n(ii) There exists a closed set \( F \) with \( F \subset E ... | Proof. Part (i) is just the definition of measurability. For the second part, we know that \( {E}^{c} \) is measurable, so there exists an open set \( \mathcal{O} \) with \( {E}^{c} \subset \mathcal{O} \) and \( m\left( {\mathcal{O} - {E}^{c}}\right) \leq \epsilon \) . If we let \( F = {\mathcal{O}}^{c} \), then \( F \... | Yes |
Corollary 3.5 A subset \( E \) of \( {\mathbb{R}}^{d} \) is measurable\n\n(i) if and only if \( E \) differs from a \( {G}_{\delta } \) by a set of measure zero,\n\n(ii) if and only if \( E \) differs from an \( {F}_{\sigma } \) by a set of measure zero. | Proof. Clearly \( E \) is measurable whenever it satisfies either (i) or (ii), since the \( {F}_{\sigma },{G}_{\delta } \), and sets of measure zero are measurable.\n\nConversely, if \( E \) is measurable, then for each integer \( n \geq 1 \) we may select an open set \( {\mathcal{O}}_{n} \) that contains \( E \), and ... | Yes |
Theorem 3.6 The set \( \mathcal{N} \) is not measurable. | The proof is by contradiction, so we assume that \( \mathcal{N} \) is measurable. Let \( {\left\{ {r}_{k}\right\} }_{k = 1}^{\infty } \) be an enumeration of all the rationals in \( \left\lbrack {-1,1}\right\rbrack \), and consider the translates\n\n\[ \n{\mathcal{N}}_{k} = \mathcal{N} + {r}_{k} \n\]\n\nWe claim that t... | Yes |
If \( f \) is measurable and finite-valued, and \( \Phi \) is continuous, then \( \Phi \circ f \) is measurable. | In fact, \( \Phi \) is continuous, so \( {\Phi }^{-1}\left( \left( {-\infty, a}\right) \right) \) is an open set \( \mathcal{O} \), and hence \( {\left( \Phi \circ f\right) }^{-1}\left( \left( {-\infty, a}\right) \right) = {f}^{-1}\left( \mathcal{O}\right) \) is measurable. | Yes |
Property 3 Suppose \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \) is a sequence of measurable functions. Then\n\n\[ \mathop{\sup }\limits_{n}{f}_{n}\left( x\right) ,\;\mathop{\inf }\limits_{n}{f}_{n}\left( x\right) ,\;\mathop{\limsup }\limits_{{n \rightarrow \infty }},{f}_{n}\left( x\right) \;\text{ and }\;\mathop{... | Proving that \( \mathop{\sup }\limits_{n}{f}_{n} \) is measurable requires noting that \( \left\{ {\mathop{\sup }\limits_{n}{f}_{n} > a}\right\} = \) \( \mathop{\bigcup }\limits_{n}\left\{ {{f}_{n} > a}\right\} \) . This also yields the result for \( \mathop{\inf }\limits_{n}{f}_{n}\left( x\right) \), since this quanti... | Yes |
Property 4 If \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \) is a collection of measurable functions, and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \]\n\nthen \( f \) is measurable. | Since \( f\left( x\right) = \mathop{\limsup }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) \), this property is a consequence of property 3. | Yes |
Property 5 If \( f \) and \( g \) are measurable, then\n\n(i) The integer powers \( {f}^{k}, k \geq 1 \) are measurable.\n\n(ii) \( f + g \) and \( {fg} \) are measurable if both \( f \) and \( g \) are finite-valued. | For (i) we simply note that if \( k \) is odd, then \( \left\{ {{f}^{k} > a}\right\} = \left\{ {f > {a}^{1/k}}\right\} \), and if \( k \) is even and \( a \geq 0 \), then \( \left\{ {{f}^{k} > a}\right\} = \left\{ {f > {a}^{1/k}}\right\} \cup \left\{ {f < - {a}^{1/k}}\right\} \).\n\nFor (ii), we first see that \( f + g... | Yes |
Theorem 4.1 Suppose \( f \) is a non-negative measurable function on \( {\mathbb{R}}^{d} \) . Then there exists an increasing sequence of non-negative simple functions \( {\left\{ {\varphi }_{k}\right\} }_{k = 1}^{\infty } \) that converges pointwise to \( f \), namely,\n\n\[ \n{\varphi }_{k}\left( x\right) \leq {\varp... | Proof. We begin first with a truncation. For \( N \geq 1 \), let \( {Q}_{N} \) denote the cube centered at the origin and of side length \( N \) . Then we define\n\n\[ \n{F}_{N}\left( x\right) = \left\{ \begin{matrix} f\left( x\right) & \text{ if }x \in {Q}_{N}\text{ and }f\left( x\right) \leq N \\ N & \text{ if }x \in... | Yes |
Theorem 4.2 Suppose \( f \) is measurable on \( {\mathbb{R}}^{d} \) . Then there exists a sequence of simple functions \( {\left\{ {\varphi }_{k}\right\} }_{k = 1}^{\infty } \) that satisfies\n\n\[ \left| {{\varphi }_{k}\left( x\right) }\right| \leq \left| {{\varphi }_{k + 1}\left( x\right) }\right| \;\text{ and }\;\ma... | Proof. We use the following decomposition of the function \( f : f\left( x\right) = \) \( {f}^{ + }\left( x\right) - {f}^{ - }\left( x\right) \), where\n\n\[ {f}^{ + }\left( x\right) = \max \left( {f\left( x\right) ,0}\right) \;\text{ and }\;{f}^{ - }\left( x\right) = \max \left( {-f\left( x\right) ,0}\right) . \]\n\nS... | Yes |
Theorem 4.3 Suppose \( f \) is measurable on \( {\mathbb{R}}^{d} \) . Then there exists a sequence of step functions \( {\left\{ {\psi }_{k}\right\} }_{k = 1}^{\infty } \) that converges pointwise to \( f\left( x\right) \) for almost every \( x \) . | Proof. By the previous result, it suffices to show that if \( E \) is a measurable set with finite measure, then \( f = {\chi }_{E} \) can be approximated by step functions. To this end, we recall part (iv) of Theorem 3.4, which states that for every \( \epsilon \) there exist cubes \( {Q}_{1},\ldots ,{Q}_{N} \) such t... | Yes |
Theorem 4.4 (Egorov) Suppose \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \) is a sequence of measurable functions defined on a measurable set \( E \) with \( m\left( E\right) < \infty \), and assume that \( {f}_{k} \rightarrow f \) a.e on \( E \) . Given \( \epsilon > 0 \), we can find a closed set \( {A}_{\epsilon... | Proof. We may assume without loss of generality that \( {f}_{k}\left( x\right) \rightarrow f\left( x\right) \) for every \( x \in E \) . For each pair of non-negative integers \( n \) and \( k \), let\n\n\[ \n{E}_{k}^{n} = \left\{ {x \in E : \left| {{f}_{j}\left( x\right) - f\left( x\right) }\right| < 1/n,\text{ for al... | Yes |
Theorem 4.5 (Lusin) Suppose \( f \) is measurable and finite valued on \( E \) with \( E \) of finite measure. Then for every \( \epsilon > 0 \) there exists a closed set \( {F}_{\epsilon } \), with\n\n\[ \n{F}_{\epsilon } \subset E,\;\text{ and }\;m\left( {E - {F}_{\epsilon }}\right) \leq \epsilon \n\]\n\nand such tha... | Proof. Let \( {f}_{n} \) be a sequence of step functions so that \( {f}_{n} \rightarrow f \) a.e. Then we may find sets \( {E}_{n} \) so that \( m\left( {E}_{n}\right) < 1/{2}^{n} \) and \( {f}_{n} \) is continuous outside \( {E}_{n} \) . By Egorov’s theorem, we may find a set \( {A}_{\epsilon /3} \) on which \( {f}_{n... | Yes |
Proposition 1.1 The integral of simple functions defined above satisfies the following properties:\n\n(i) Independence of the representation. If \( \varphi = \mathop{\sum }\limits_{{k = 1}}^{N}{a}_{k}{\chi }_{{E}_{k}} \) is any representation of \( \varphi \), then\n\n\[ \int \varphi = \mathop{\sum }\limits_{{k = 1}}^{... | Proof. The only conclusion that is a little tricky is the first, which asserts that the integral of a simple function can be calculated by using any of its decompositions as a linear combination of characteristic functions.\n\nSuppose that \( \varphi = \mathop{\sum }\limits_{{k = 1}}^{N}{a}_{k}{\chi }_{{E}_{k}} \), whe... | Yes |
Lemma 1.2 Let \( f \) be a bounded function supported on a set \( E \) of finite measure. If \( {\left\{ {\varphi }_{n}\right\} }_{n = 1}^{\infty } \) is any sequence of simple functions bounded by \( M \) , supported on \( E \), and with \( {\varphi }_{n}\left( x\right) \rightarrow f\left( x\right) \) for a.e. \( x \)... | Proof. The assertions of the lemma would be nearly obvious if we had that \( {\varphi }_{n} \) converges to \( f \) uniformly on \( E \) . Instead, we recall one of Littlewood's principles, which states that the convergence of a sequence of measurable functions is \ | No |
Proposition 1.3 Suppose \( f \) and \( g \) are bounded functions supported on sets of finite measure. Then the following properties hold.\n\n(i) Linearity. If \( a, b \in \mathbb{R} \), then\n\n\[ \int \left( {{af} + {bg}}\right) = a\int f + b\int g. \]\n\n(ii) Additivity. If \( E \) and \( F \) are disjoint subsets o... | All these properties follow by using approximations by simple functions, and the properties of the integral of simple functions given in Proposition 1.1. | No |
Theorem 1.4 (Bounded convergence theorem) Suppose that \( \left\{ {f}_{n}\right\} \) is a sequence of measurable functions that are all bounded by \( M \), are supported on a set \( E \) of finite measure, and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) a.e. \( x \) as \( n \rightarrow \) \( \infty \) . T... | Proof. From the assumptions one sees at once that \( f \) is bounded by \( M \) almost everywhere and vanishes outside \( E \), except possibly on a set of measure zero. Clearly, the triangle inequality for the integral implies that it suffices to prove that \( \int \left| {{f}_{n} - f}\right| \rightarrow 0 \) as \( n ... | Yes |
Proposition 1.6 The integral of non-negative measurable functions enjoys the following properties:\n\n(i) Linearity. If \( f, g \geq 0 \), and \( a, b \) are positive real numbers, then\n\n\[ \n\int \left( {{af} + {bg}}\right) = a\int f + b\int g.\n\] | Proof. Of the first four assertions, only (i) is not an immediate consequence of the definitions, and to prove it we argue as follows. We take \( a = b = 1 \) and note that if \( \varphi \leq f \) and \( \psi \leq g \), where both \( \varphi \) and \( \psi \) are bounded and supported on sets of finite measure, then \(... | Yes |
Lemma 1.7 (Fatou) Suppose \( \\left\\{ {f}_{n}\\right\\} \) is a sequence of measurable functions with \( {f}_{n} \\geq 0 \) . If \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{f}_{n}\\left( x\\right) = f\\left( x\\right) \) for a.e. \( x \), then | Proof. Suppose \( 0 \\leq g \\leq f \), where \( g \) is bounded and supported on a set \( E \) of finite measure. If we set \( {g}_{n}\\left( x\\right) = \\min \\left( {g\\left( x\\right) ,{f}_{n}\\left( x\\right) }\\right) \), then \( {g}_{n} \) is measurable, supported on \( E \), and \( {g}_{n}\\left( x\\right) \\r... | Yes |
Corollary 1.8 Suppose \( f \) is a non-negative measurable function, and \( \left\{ {f}_{n}\right\} \) a sequence of non-negative measurable functions with \( {f}_{n}\left( x\right) \leq f\left( x\right) \) and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) for almost every \( x \) . Then\n\n\[ \mathop{\lim ... | Proof. Since \( {f}_{n}\left( x\right) \leq f\left( x\right) \) a.e \( x \), we necessarily have \( \int {f}_{n} \leq \int f \) for all \( n \) ; hence\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}\int {f}_{n} \leq \int f \]\n\n## This inequality combined with Fatou's lemma proves the desired limit. | No |
Corollary 1.10 Consider a series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \), where \( {a}_{k}\left( x\right) \geq 0 \) is measurable for every \( k \geq 1 \) . Then\n\n\[ \n\int \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) {dx} = \mathop{\sum }\limits_{{k = 1}}^{\infty }\in... | Proof. Let \( {f}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}\left( x\right) \) and \( f\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k}\left( x\right) \) . The functions \( {f}_{n} \) are measurable, \( {f}_{n}\left( x\right) \leq {f}_{n + 1}\left( x\right) \), and \( {f}_{n}\left... | Yes |
Proposition 1.12 Suppose \( f \) is integrable on \( {\mathbb{R}}^{d} \) . Then for every \( \epsilon > 0 \) :\n\n(i) There exists a set of finite measure \( B \) (a ball, for example) such that\n\n\[{\int }_{{B}^{c}}\left| f\right| < \epsilon\]\n\n(ii) There is a \( \delta > 0 \) such that\n\n\[{\int }_{E}\left| f\rig... | Proof. By replacing \( f \) with \( \left| f\right| \) we may assume without loss of generality that \( f \geq 0 \) .\n\nFor the first part, let \( {B}_{N} \) denote the ball of radius \( N \) centered at the origin, and note that if \( {f}_{N}\left( x\right) = f\left( x\right) {\chi }_{{B}_{N}}\left( x\right) \), then... | Yes |
Theorem 1.13 Suppose \( \left\{ {f}_{n}\right\} \) is a sequence of measurable functions such that \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) a.e. \( x \), as \( n \) tends to infinity. If \( \left| {{f}_{n}\left( x\right) }\right| \leq g\left( x\right) \), where \( g \) is integrable, then\n\n\[ \n\int ... | Proof. For each \( N \geq 0 \) let \( {E}_{N} = \{ x : \left| x\right| \leq N, g\left( x\right) \leq N\} \) . Given \( \epsilon > 0 \), we may argue as in the first part of the previous lemma, to see that there exists \( N \) so that \( {\int }_{{E}_{N}^{c}}g < \epsilon \) . Then the functions \( {f}_{n}{\chi }_{{E}_{N... | Yes |
Theorem 2.4 The following families of functions are dense in \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) :\n\n(i) The simple functions.\n\n(ii) The step functions.\n\n(iii) The continuous functions of compact support. | Proof. Let \( f \) be an integrable function on \( {\mathbb{R}}^{d} \) . First, we may assume that \( f \) is real-valued, because we may approximate its real and imaginary parts independently. If this is the case, we may then write \( f = {f}^{ + } - {f}^{ - } \) , where \( {f}^{ + },{f}^{ - } \geq 0 \), and it now su... | Yes |
Proposition 2.5 Suppose \( f \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then\n\n\[ \n{\begin{Vmatrix}{f}_{h} - f\end{Vmatrix}}_{{L}^{1}} \rightarrow 0\;\text{ as }h \rightarrow 0.\n\] | The proof is a simple consequence of the approximation of integrable functions by continuous functions of compact support as given in Theorem 2.4. In fact for any \( \epsilon > 0 \), we can find such a function \( g \) so that \( \parallel f - g\parallel < \epsilon \) . Now\n\n\[ \n{f}_{h} - f = \left( {{g}_{h} - g}\ri... | Yes |
Theorem 3.1 Suppose \( f\left( {x, y}\right) \) is integrable on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) . Then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) :\n\n(i) The slice \( {f}^{y} \) is integrable on \( {\mathbb{R}}^{{d}_{1}} \) .\n\n(ii) The function defined by \( {\int }_{{\mathbb{R}}... | We first note that we may assume that \( f \) is real-valued, since the theorem then applies to the real and imaginary parts of a complex-valued function. The proof of Fubini's theorem which we give next consists of a sequence of six steps. We begin by letting \( \mathcal{F} \) denote the set of integrable functions on... | No |
Theorem 3.2 Suppose \( f\left( {x, y}\right) \) is a non-negative measurable function on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) . Then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) :\n\n(i) The slice \( {f}^{y} \) is measurable on \( {\mathbb{R}}^{{d}_{1}} \) .\n\n(ii) The function defined by ... | Proof of Theorem 3.2. Consider the truncations\n\n\[ \n{f}_{k}\left( {x, y}\right) = \left\{ \begin{matrix} f\left( {x, y}\right) & \text{ if }\left| \left( {x, y}\right) \right| < k\text{ and }f\left( {x, y}\right) < k, \\ 0 & \text{ otherwise. } \end{matrix}\right. \]\n\nEach \( {f}_{k} \) is integrable, and by part ... | Yes |
Corollary 3.3 If \( E \) is a measurable set in \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \), then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) the slice\n\n\[ \n{E}^{y} = \left\{ {x \in {\mathbb{R}}^{{d}_{1}} : \left( {x, y}\right) \in E}\right\}\n\]\n\nis a measurable subset of \( {\mathbb{R}}^{{... | This is an immediate consequence of the first part of Theorem 3.2 applied to the function \( {\chi }_{E} \) . Clearly a symmetric result holds for the \( x \) -slices in \( {\mathbb{R}}^{{d}_{2}} \) . | Yes |
Proposition 3.4 If \( E = {E}_{1} \times {E}_{2} \) is a measurable subset of \( {\mathbb{R}}^{d} \), and \( {m}_{ * }\left( {E}_{2}\right) > 0 \), then \( {E}_{1} \) is measurable. | Proof. By Corollary 3.3, we know that for a.e. \( y \in {\mathbb{R}}^{{d}_{2}} \), the slice function\n\n\[ \n{\left( {\chi }_{{E}_{1} \times {E}_{2}}\right) }^{y}\left( x\right) = {\chi }_{{E}_{1}}\left( x\right) {\chi }_{{E}_{2}}\left( y\right) \n\]\n\nis measurable as a function of \( x \) . In fact, we claim that t... | Yes |
Lemma 3.5 If \( {E}_{1} \subset {\mathbb{R}}^{{d}_{1}} \) and \( {E}_{2} \subset {\mathbb{R}}^{{d}_{2}} \), then\n\n\[ \n{m}_{ * }\left( {{E}_{1} \times {E}_{2}}\right) \leq {m}_{ * }\left( {E}_{1}\right) {m}_{ * }\left( {E}_{2}\right) \n\]\n\nwith the understanding that if one of the sets \( {E}_{j} \) has exterior me... | Proof. Let \( \epsilon > 0 \) . By definition, we can find cubes \( {\left\{ {Q}_{k}\right\} }_{k = 1}^{\infty } \) in \( {\mathbb{R}}^{{d}_{1}} \) and \( {\left\{ {Q}_{\ell }^{\prime }\right\} }_{\ell = 1}^{\infty } \) in \( {\mathbb{R}}^{{d}_{2}} \) such that\n\n\[ \n{E}_{1} \subset \mathop{\bigcup }\limits_{{k = 1}}... | Yes |
Proposition 3.6 Suppose \( {E}_{1} \) and \( {E}_{2} \) are measurable subsets of \( {\mathbb{R}}^{{d}_{1}} \) and \( {\mathbb{R}}^{{d}_{2}} \), respectively. Then \( E = {E}_{1} \times {E}_{2} \) is a measurable subset of \( {\mathbb{R}}^{d} \) . Moreover,\n\n\[ m\left( E\right) = m\left( {E}_{1}\right) m\left( {E}_{2... | Proof. It suffices to prove that \( E \) is measurable, because then the assertion about \( m\left( E\right) \) follows from Corollary 3.3. Since each set \( {E}_{j} \) is measurable, there exist sets \( {G}_{j} \subset {\mathbb{R}}^{{d}_{j}} \) of type \( {G}_{\delta } \), with \( {G}_{j} \supset {E}_{j} \) and \( {m}... | Yes |
Corollary 3.7 Suppose \( f \) is a measurable function on \( {\mathbb{R}}^{{d}_{1}} \) . Then the function \( \widetilde{f} \) defined by \( \widetilde{f}\left( {x, y}\right) = f\left( x\right) \) is measurable on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) . | Proof. To see this, we may assume that \( f \) is real-valued, and recall first that if \( a \in \mathbb{R} \) and \( {E}_{1} = \left\{ {x \in {\mathbb{R}}^{{d}_{1}} : f\left( x\right) < a}\right\} \), then \( {E}_{1} \) is measurable by definition. Since\n\n\[ \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{{d}_{1}} \... | Yes |
Corollary 3.8 Suppose \( f\left( x\right) \) is a non-negative function on \( {\mathbb{R}}^{d} \), and let\n\n\[ \mathcal{A} = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{d} \times \mathbb{R} : 0 \leq y \leq f\left( x\right) }\right\} \]\n\nThen:\n\n(i) \( f \) is measurable on \( {\mathbb{R}}^{d} \) if and only if... | Proof. If \( f \) is measurable on \( {\mathbb{R}}^{d} \), then the previous proposition guarantees that the function\n\n\[ F\left( {x, y}\right) = y - f\left( x\right) \]\nis measurable on \( {\mathbb{R}}^{d + 1} \), so we immediately see that \( \mathcal{A} = \{ y \geq 0\} \cap \{ F \leq \) \( 0\} \) is measurable.\n... | Yes |
Proposition 3.9 If \( f \) is a measurable function on \( {\mathbb{R}}^{d} \), then the function \( \widetilde{f}\left( {x, y}\right) = f\left( {x - y}\right) \) is measurable on \( {\mathbb{R}}^{d} \times {\mathbb{R}}^{d} \) . | By picking \( E = \left\{ {z \in {\mathbb{R}}^{d} : f\left( z\right) < a}\right\} \), we see that it suffices to prove that whenever \( E \) is a measurable subset of \( {\mathbb{R}}^{d} \), then \( \widetilde{E} = \{ \left( {x, y}\right) : x - y \in \) \( E\} \) is a measurable subset of \( {\mathbb{R}}^{d} \times {\m... | Yes |
Proposition 4.1 Suppose \( f \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then \( \widehat{f} \) defined by (14) is continuous and bounded on \( {\mathbb{R}}^{d} \) . | In fact, since \( \left| {f\left( x\right) {e}^{-{2\pi ix} \cdot \xi }}\right| = \left| {f\left( x\right) }\right| \), the integral representing \( \widehat{f} \) converges for each \( \xi \) and \( \mathop{\sup }\limits_{{\xi \in {\mathbb{R}}^{d}}}\left| {\widehat{f}\left( \xi \right) }\right| \leq {\int }_{{\mathbb{R... | Yes |
Theorem 4.2 Suppose \( f \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) and assume also that \( \widehat{f} \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then the inversion formula (15) holds for almost every \( x \) . | The proof of the theorem requires only that we adapt the earlier arguments carried out for Schwartz functions in Chapter 5 of Book I to the present context. We begin with the \ | No |
Lemma 4.4 Suppose \( f \) and \( g \) belong to \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then\n\n\[ \n{\int }_{{\mathbb{R}}^{d}}\widehat{f}\left( \xi \right) g\left( \xi \right) {d\xi } = {\int }_{{\mathbb{R}}^{d}}f\left( y\right) \widehat{g}\left( y\right) {dy}.\n\] | Note that both integrals converge in view of the proposition above. Consider the function \( F\left( {\xi, y}\right) = g\left( \xi \right) f\left( y\right) {e}^{-{2\pi i\xi } \cdot y} \) defined for \( \left( {\xi, y}\right) \in {\mathbb{R}}^{d} \times \) \( {\mathbb{R}}^{d} = {\mathbb{R}}^{2d} \) . It is measurable as... | Yes |
Theorem 1.1 Suppose \( f \) is integrable on \( {\mathbb{R}}^{d} \) . Then:\n\n(i) \( {f}^{ * } \) is measurable.\n\n(ii) \( {f}^{ * }\left( x\right) < \infty \) for a.e. \( x \) .\n\n(iii) \( {f}^{ * } \) satisfies\n\n\[ m\left( \left\{ {x \in {\mathbb{R}}^{d} : {f}^{ * }\left( x\right) > \alpha }\right\} \right) \leq... | Before we come to the proof we want to clarify the nature of the main conclusion (iii). As we shall observe, one has that \( {f}^{ * }\left( x\right) \geq \left| {f\left( x\right) }\right| \) for a.e. \( x \) ; the effect of (iii) is that, broadly speaking, \( {f}^{ * } \) is not much larger than \( \left| f\right| \) ... | No |
Lemma 1.2 Suppose \( \mathcal{B} = \left\{ {{B}_{1},{B}_{2},\ldots ,{B}_{N}}\right\} \) is a finite collection of open balls in \( {\mathbb{R}}^{d} \) . Then there exists a disjoint sub-collection \( {B}_{{i}_{1}},{B}_{{i}_{2}},\ldots ,{B}_{{i}_{k}} \) of \( \mathcal{B} \) that satisfies\n\n\[ m\left( {\mathop{\bigcup ... | Proof. The argument we give is constructive and relies on the following simple observation: Suppose \( B \) and \( {B}^{\prime } \) are a pair of balls that intersect, with the radius of \( {B}^{\prime } \) being not greater than that of \( B \) . Then \( {B}^{\prime } \) is contained in the ball \( \widetilde{B} \) th... | Yes |
Theorem 1.3 If \( f \) is integrable on \( {\mathbb{R}}^{d} \), then\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} = f\left( x\right) \;\text{ for a.e. }x. \] | Proof. It suffices to show that for each \( \alpha > 0 \) the set\n\n\[ {E}_{\alpha } = \left\{ {x : \mathop{\limsup }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\left| {\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} - f\left( x\right) }\right| > {2\alpha }}\right\} \]\n\nhas measure... | Yes |
Theorem 1.4 If \( f \in {L}_{\text{loc }}^{1}\left( {\mathbb{R}}^{d}\right) \), then\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} = f\left( x\right) ,\;\text{ for a.e. }x. \] | Proof. An application of Theorem 1.4 to the function \( \left| {f\left( y\right) - r}\right| \) shows that for each rational \( r \), there exists a set \( {E}_{r} \) of measure zero, such that\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\... | Yes |
Corollary 1.6 If \( f \) is locally integrable on \( {\mathbb{R}}^{d} \), then almost every point belongs to the Lebesgue set of \( f \) . | Proof. An application of Theorem 1.4 to the function \( \left| {f\left( y\right) - r}\right| \) shows that for each rational \( r \), there exists a set \( {E}_{r} \) of measure zero, such that\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\... | Yes |
Corollary 1.6 If \( f \) is locally integrable on \( {\mathbb{R}}^{d} \), then almost every point belongs to the Lebesgue set of \( f \) . | Proof. An application of Theorem 1.4 to the function \( \left| {f\left( y\right) - r}\right| \) shows that for each rational \( r \), there exists a set \( {E}_{r} \) of measure zero, such that\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\... | Yes |
Corollary 1.7 Suppose \( f \) is locally integrable on \( {\mathbb{R}}^{d} \) . If \( \left\{ {U}_{\alpha }\right\} \) shrinks regularly to \( \bar{x} \), then\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( {U}_{\alpha }\right) \rightarrow 0} \\ {x \in {U}_{\alpha }} }}\frac{1}{m\left( {U}_{\alpha }\right) }{\int }_{{... | The proof is immediate once we observe that if \( \bar{x} \in B \) with \( {U}_{\alpha } \subset B \) and \( m\left( {U}_{\alpha }\right) \geq {cm}\left( B\right) \), then\n\n\[ \frac{1}{m\left( {U}_{\alpha }\right) }{\int }_{{U}_{\alpha }}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq \frac{1}{{c... | Yes |
Theorem 2.1 If \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is an approximation to the identity and \( f \) is integrable on \( {\mathbb{R}}^{d} \), then\n\n\[ \n\left( {f * {K}_{\delta }}\right) \left( x\right) \rightarrow f\left( x\right) \;\text{ as }\delta \rightarrow 0 \n\]\n\nfor every \( x \) in the Lebes... | Since the integral of each kernel \( {K}_{\delta } \) is equal to 1, we may write\n\n\[ \n\left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) = \int \left\lbrack {f\left( {x - y}\right) - f\left( x\right) }\right\rbrack {K}_{\delta }\left( y\right) {dy}. \n\]\n\nConsequently,\n\n\[ \n\left| {\left( {f ... | Yes |
Lemma 2.2 Suppose that \( f \) is integrable on \( {\mathbb{R}}^{d} \), and that \( x \) is a point of the Lebesgue set of \( f \) . Let\n\n\[ \mathcal{A}\left( r\right) = \frac{1}{{r}^{d}}{\int }_{\left| y\right| \leq r}\left| {f\left( {x - y}\right) - f\left( x\right) }\right| {dy},\;\text{ whenever }r > 0. \]\n\nThe... | Proof. The continuity of \( \mathcal{A}\left( r\right) \) follows by invoking the absolute continuity in Proposition 1.12 of Chapter 2.\n\nThe fact that \( \mathcal{A}\left( r\right) \) tends to 0 as \( r \) tends to 0 follows since \( x \) belongs to the Lebesgue set of \( f \), and the measure of a ball of radius \( ... | Yes |
Theorem 2.3 Suppose that \( f \) is integrable on \( {\mathbb{R}}^{d} \) and that \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is an approximation to the identity. Then, for each \( \delta > 0 \), the convolution\n\n\[ \n\left( {f * {K}_{\delta }}\right) \left( x\right) = {\int }_{{\mathbb{R}}^{d}}f\left( {x - y... | The proof is merely a repetition in a more general context of the argument in the special case where \( {K}_{\delta }\left( x\right) = {\delta }^{-d/2}{e}^{-\pi {\left| x\right| }^{2}/\delta } \) given in Section \( {4}^{ * } \) , Chapter 2, and so will not be repeated. | No |
Theorem 3.1 A curve parametrized by \( \left( {x\left( t\right), y\left( t\right) }\right), a \leq t \leq b \), is rectifiable if and only if both \( x\left( t\right) \) and \( y\left( t\right) \) are of bounded variation. | The proof is immediate once we observe that if \( F\left( t\right) = x\left( t\right) + {iy}\left( t\right) \), then\n\n\[ F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) = \left( {x\left( {t}_{j}\right) - x\left( {t}_{j - 1}\right) }\right) + i\left( {y\left( {t}_{j}\right) - y\left( {t}_{j - 1}\right) }\right) ,\... | Yes |
Lemma 3.2 Suppose \( F \) is real-valued and of bounded variation on \( \left\lbrack {a, b}\right\rbrack \) . Then for all \( a \leq x \leq b \) one has \[ F\left( x\right) - F\left( a\right) = {P}_{F}\left( {a, x}\right) - {N}_{F}\left( {a, x}\right) ,\] and \[ {T}_{F}\left( {a, x}\right) = {P}_{F}\left( {a, x}\right)... | Proof. Given \( \epsilon > 0 \) there exists a partition \( a = {t}_{0} < \cdots < {t}_{N} = x \) of \( \left\lbrack {a, x}\right\rbrack \), such that \[ \left| {{P}_{F} - \mathop{\sum }\limits_{\left( +\right) }F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right| < \epsilon \text{ and }\left| {{N}_{F} - \matho... | Yes |
Theorem 3.3 A real-valued function \( F \) on \( \left\lbrack {a, b}\right\rbrack \) is of bounded variation if and only if \( F \) is the difference of two increasing bounded functions. | Proof. Clearly, if \( F = {F}_{1} - {F}_{2} \), where each \( {F}_{j} \) is bounded and increasing, then \( F \) is of bounded variation.\n\nConversely, suppose \( F \) is of bounded variation. Then, we let \( {F}_{1}\left( x\right) = \) \( {P}_{F}\left( {a, x}\right) + F\left( a\right) \) and \( {F}_{2}\left( x\right)... | Yes |
Lemma 3.5 Suppose \( G \) is real-valued and continuous on \( \mathbb{R} \) . Let \( E \) be the set of points \( x \) such that\n\n\[ G\left( {x + h}\right) > G\left( x\right) \;\text{ for some }h = {h}_{x} > 0.\]\n\nIf \( E \) is non-empty, then it must be open, and hence can be written as a countable disjoint union ... | Proof. Since \( G \) is continuous, it is clear that \( E \) is open whenever it is non-empty and can therefore be written as a disjoint union of countably many open intervals (Theorem 1.3 in Chapter 1). If \( \left( {{a}_{k},{b}_{k}}\right) \) denotes a finite interval in this decomposition, then \( {a}_{k} \notin E \... | Yes |
If \( F \) is increasing and continuous, then \( {F}^{\prime } \) exists almost everywhere. Moreover \( {F}^{\prime } \) is measurable, non-negative, and\n\n\[{\int }_{a}^{b}{F}^{\prime }\left( x\right) {dx} \leq F\left( b\right) - F\left( a\right)\]\n\nIn particular, if \( F \) is bounded on \( \mathbb{R} \), then \( ... | Proof. For \( n \geq 1 \), we consider the quotient\n\n\[{G}_{n}\left( x\right) = \frac{F\left( {x + 1/n}\right) - F\left( x\right) }{1/n}.\n\]\n\nBy the previous theorem, we have that \( {G}_{n}\left( x\right) \rightarrow {F}^{\prime }\left( x\right) \) for a.e. \( x \), which shows in particular that \( {F}^{\prime }... | Yes |
Theorem 3.8 If \( F \) is absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \), then \( {F}^{\prime }\left( x\right) \) exists almost everywhere. Moreover, if \( {F}^{\prime }\left( x\right) = 0 \) for a.e. \( x \), then \( F \) is constant. | Since an absolutely continuous function is the difference of two continuous monotonic functions, as we have seen above, the existence of \( {F}^{\prime }\left( x\right) \) for a.e. \( x \) follows from what we have already proved. To prove that \( {F}^{\prime }\left( x\right) = 0 \) a.e. implies \( F \) is constant req... | Yes |
Corollary 3.10 We can arrange the choice of the balls so that\n\n\[ m\left( {E - \mathop{\bigcup }\limits_{{i = 1}}^{N}{B}_{i}}\right) < {2\delta } \] | In fact, let \( \mathcal{O} \) be an open set, with \( \mathcal{O} \supset E \) and \( m\left( {\mathcal{O} - E}\right) < \delta \) . Since we are dealing with a Vitali covering of \( E \), we can restrict all of our choices above to balls contained in \( \mathcal{O} \) . If we do this, then \( \left( {E - \mathop{\big... | Yes |
Theorem 3.11 Suppose \( F \) is absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then \( {F}^{\prime } \) exists almost everywhere and is integrable. Moreover,\n\n\[ F\left( x\right) - F\left( a\right) = {\int }_{a}^{x}{F}^{\prime }\left( y\right) {dy},\;\text{ for all }a \leq x \leq b. \] | Proof. Since we know that a real-valued absolutely continuous function is the difference of two continuous increasing functions, Corollary 3.7 shows that \( {F}^{\prime } \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) . Now let \( G\left( x\right) = {\int }_{a}^{x}{F}^{\prime }\left( y\right) {dy} \) . Then... | Yes |
Lemma 3.12 A bounded increasing function \( F \) on \( \left\lbrack {a, b}\right\rbrack \) has at most countably many discontinuities. | Proof. If \( F \) is discontinuous at \( x \), we may choose a rational number \( {r}_{x} \) so that \( F\left( {x}^{ - }\right) < {r}_{x} < F\left( {x}^{ + }\right) \) . If \( f \) is discontinuous at \( x \) and \( z \) with \( x < z \), we must have \( F\left( {x}^{ + }\right) \leq F\left( {z}^{ - }\right) \), hence... | Yes |
Lemma 3.13 If \( F \) is increasing and bounded on \( \left\lbrack {a, b}\right\rbrack \), then:\n\n(i) \( J\left( x\right) \) is discontinuous precisely at the points \( \left\{ {x}_{n}\right\} \) and has a jump at \( {x}_{n} \) equal to that of \( F \) .\n\n(ii) The difference \( F\left( x\right) - J\left( x\right) \... | Proof. If \( x \neq {x}_{n} \) for all \( n \), each \( {j}_{n} \) is continuous at \( x \), and since the series converges uniformly, \( J \) must be continuous at \( x \) . If \( x = {x}_{N} \) for some \( N \), then we write\n\n\[ J\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{\alpha }_{n}{j}_{n}\left( x\rig... | Yes |
Theorem 3.14 If \( J \) is the jump function considered above, then \( {J}^{\prime }\left( x\right) \) exists and vanishes almost everywhere. | Proof. Given any \( \epsilon > 0 \), we note that the set \( E \) of those \( x \) where\n\n(10)\n\n\[ \mathop{\limsup }\limits_{{h \rightarrow 0}}\frac{J\left( {x + h}\right) - J\left( x\right) }{h} > \epsilon \]\n\nis a measurable set. (The proof of this little fact is outlined in Exercise 14 below.) Suppose \( \delt... | No |
Theorem 4.1 Suppose \( \left( {x\left( t\right), y\left( t\right) }\right) \) is a curve defined for \( a \leq t \leq b \) . If both \( x\left( t\right) \) and \( y\left( t\right) \) are absolutely continuous, then the curve is rectifiable, and if \( L \) denotes its length, we have\n\n\[ L = {\int }_{a}^{b}{\left( {x}... | Note that if \( F\left( t\right) = x\left( t\right) + {iy}\left( t\right) \) is absolutely continuous then it is automatically of bounded variation, and hence the curve is rectifiable. The identity (12) is an immediate consequence of the proposition below, which can be viewed as a more precise version of Corollary 3.7 ... | Yes |
Proposition 4.2 Suppose \( F \) is complex-valued and absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \n{T}_{F}\left( {a, b}\right) = {\int }_{a}^{b}\left| {{F}^{\prime }\left( t\right) }\right| {dt} \n\] | In fact, because of Theorem 3.11, for any partition \( a = {t}_{0} < {t}_{1} < \cdots < \) \( {t}_{N} = b \) of \( \left\lbrack {a, b}\right\rbrack \), we have\n\n\[ \n\mathop{\sum }\limits_{{j = 1}}^{N}\left| {F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right| = \mathop{\sum }\limits_{{j = 1}}^{N}\left| {{\i... | Yes |
Theorem 4.3 Suppose \( \left( {x\left( t\right), y\left( t\right) }\right), a \leq t \leq b \), is a rectifiable curve that has length \( L \) . Consider the arc-length parametrization \( \widetilde{z}\left( s\right) = \left( {\widetilde{x}\left( s\right) ,\widetilde{y}\left( s\right) }\right) \) described above. Then ... | Proof. We noted that \( \left| {\widetilde{z}\left( {s}_{1}\right) - \widetilde{z}\left( {s}_{2}\right) }\right| \leq \left| {{s}_{1} - {s}_{2}}\right| \), so it follows immediately that \( \widetilde{z}\left( s\right) \) is absolutely continuous, hence differentiable almost everywhere. Moreover, this inequality also p... | Yes |
Theorem 4.4 Suppose \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is a quasi-simple curve. The Minkowski content of \( \Gamma \) exists if and only if \( \Gamma \) is rectifiable. When this is the case and \( L \) is the length of the curve, then \( \mathcal{M}\left( \Gamma \right) = L \) . | To prove the theorem, we also consider for any compact set \( K \)\n\n\[ \n{\mathcal{M}}^{ * }\left( K\right) = \mathop{\limsup }\limits_{{\delta \rightarrow 0}}\frac{m\left( {K}^{\delta }\right) }{2\delta }\;\text{ and }\;{\mathcal{M}}_{ * }\left( K\right) = \mathop{\liminf }\limits_{{\delta \rightarrow 0}}\frac{m\lef... | No |
Lemma 4.6 If \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is any curve, and \( \Delta = \left| {z\left( b\right) - z\left( a\right) }\right| \) is the distance between its end-points, then \( m\left( {\Gamma }^{\delta }\right) \geq {2\delta \Delta } \) . | Proof. Since the distance function and the Lebesgue measure are invariant under translations and rotations (see Section 3 in Chapter 1 and Problem 4 in Chapter 2) we may transform the situation by an appropriate composition of these motions. Therefore we may assume that the end-points of the curve have been placed on t... | Yes |
Proposition 4.7 Suppose \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is a rectifiable curve with length \( L \) . Then\n\n\[{\mathcal{M}}^{ * }\left( \Gamma \right) \leq L\] | The quantities \( {\mathcal{M}}^{ * }\left( \Gamma \right) \) and \( L \) are of course independent of the parametrization used; since the curve is rectifiable, it will be convenient to use the arclength parametrization. Thus we write the curve as \( z\left( s\right) = \left( {x\left( s\right), y\left( s\right) }\right... | Yes |
Proposition 1.1 The space \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) has the following properties:\n\n(i) \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) is a vector space.\n\n(ii) \( f\left( x\right) \overline{g\left( x\right) } \) is integrable whenever \( f, g \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), and the Cauch... | Proof. If \( f, g \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then since \( \left| {f\left( x\right) + g\left( x\right) }\right| \leq 2\max \left( {\left| {f\left( x\right) }\right| ,\left| {g\left( x\right) }\right| }\right) \), we have\n\n\[{\left| f\left( x\right) + g\left( x\right) \right| }^{2} \leq 4\left( {{\l... | Yes |
Theorem 1.3 The space \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) is separable, in the sense that there exists a countable collection \( \left\{ {f}_{k}\right\} \) of elements in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) such that their linear combinations are dense in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . | Proof. Consider the family of functions of the form \( r{\chi }_{R}\left( x\right) \), where \( r \) is a complex number with rational real and imaginary parts, and \( R \) is a rectangle in \( {\mathbb{R}}^{d} \) with rational coordinates. We claim that finite linear combinations of these type of functions are dense i... | Yes |
Proposition 2.1 If \( f \bot g \), then \( \parallel f + g{\parallel }^{2} = \parallel f{\parallel }^{2} + \parallel g{\parallel }^{2} \) . | Proof. It suffices to note that \( \left( {f, g}\right) = 0 \) implies \( \left( {g, f}\right) = 0 \), and therefore\n\n\[ \n\parallel f + g{\parallel }^{2} = \left( {f + g, f + g}\right) = \parallel f{\parallel }^{2} + \left( {f, g}\right) + \left( {g, f}\right) + \parallel g{\parallel }^{2}\n\]\n\n\[ \n= \parallel f{... | Yes |
Proposition 2.2 If \( {\left\{ {e}_{k}\right\} }_{k = 1}^{\infty } \) is orthonormal, and \( f = \sum {a}_{k}{e}_{k} \in \mathcal{H} \) where the sum is finite, then\n\n\[ \parallel f{\parallel }^{2} = \sum {\left| {a}_{k}\right| }^{2} \] | The proof is a simple application of the Pythagorean theorem. | No |
Corollary 2.5 Any two infinite-dimensional Hilbert spaces are unitarily equivalent. | Proof. If \( \mathcal{H} \) and \( {\mathcal{H}}^{\prime } \) are two infinite-dimensional Hilbert spaces, we may select for each an orthonormal basis, say\n\n\[ \left\{ {{e}_{1},{e}_{2},\ldots }\right\} \subset \mathcal{H}\;\text{ and }\;\left\{ {{e}_{1}^{\prime },{e}_{2}^{\prime },\ldots }\right\} \subset {\mathcal{H... | Yes |
Corollary 2.6 Any two finite-dimensional Hilbert spaces are unitarily equivalent if and only if they have the same dimension. | Thus every finite-dimensional Hilbert space over \( \mathbb{C} \) (or over \( \mathbb{R} \) ) is equivalent with \( {\mathbb{C}}^{d} \) (or \( {\mathbb{R}}^{d} \) ), for some \( d \) . | No |
Proposition 2.7 Suppose we are given a pre-Hilbert space \( {\mathcal{H}}_{0} \) with inner product \( {\left( \cdot , \cdot \right) }_{0} \) . Then we can find a Hilbert space \( \mathcal{H} \) with inner product \( \left( {\cdot , \cdot }\right) \) such that\n\n(i) \( {\mathcal{H}}_{0} \subset \mathcal{H} \) .\n\n(ii... | A Hilbert space satisfying properties like \( \mathcal{H} \) in the above proposition is called a completion of \( {\mathcal{H}}_{0} \) . We shall only sketch the construction of \( \mathcal{H} \), since it follows closely Cantor’s familiar method of obtaining the real numbers as the completion of the rationals in term... | Yes |
Theorem 3.1 Suppose \( f \) is integrable on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) .\n\n(i) If \( {a}_{n} = 0 \) for all \( n \), then \( f\left( x\right) = 0 \) for a.e. \( x \) .\n\n(ii) \( \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{r}^{\left| n\right| }{e}^{inx} \) tends to \( f\left( x\right) \... | Proof. The first conclusion is an immediate consequence of the second. To prove the latter we recall the identity\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{r}^{\left| n\right| }{e}^{iny} = {P}_{r}\left( y\right) = \frac{1 - {r}^{2}}{1 - {2r}\cos y + {r}^{2}} \]\n\nfor the Poisson kernel; see Book I, Chapt... | Yes |
Theorem 3.2 Suppose \( f \in {L}^{2}\left( \left\lbrack {-\pi ,\pi }\right\rbrack \right) \) . Then:\n\n(i) We have Parseval's relation\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\left| {a}_{n}\right| }^{2} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{\left| f\left( x\right) \right| }^{2}{dx} \] | To apply the previous results, we let \( \mathcal{H} = {L}^{2}\left( \left\lbrack {-\pi ,\pi }\right\rbrack \right) \) with inner product \( \left( {f, g}\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( x\right) \overline{g\left( x\right) }{dx} \), and take the orthonormal set \( {\left\{ {e}_{k}\right\} }_{k = 1... | Yes |
Theorem 3.3 A bounded holomorphic function \( F\left( {r{e}^{i\theta }}\right) \) on the unit disc has radial limits at almost every \( \theta \) . | Proof. We know that \( F\left( z\right) \) has a power series expansion \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \) in \( \mathbb{D} \) that converges absolutely and uniformly whenever \( z = r{e}^{i\theta } \) and \( r < 1 \) . In fact, for \( r < 1 \) the series \( \mathop{\sum }\limits_{{n = 0}}^{\... | Yes |
Proposition 4.2 If \( \mathcal{S} \) is a closed subspace of a Hilbert space \( \mathcal{H} \), then\n\n\[ \mathcal{H} = \mathcal{S} \oplus {\mathcal{S}}^{ \bot } \]\n\nThe notation in the proposition means that every \( f \in \mathcal{H} \) can be written uniquely as \( f = g + h \), where \( g \in \mathcal{S} \) and ... | The proof of the proposition relies on the previous lemma giving the closest element of \( f \) in \( \mathcal{S} \) . In fact, for any \( f \in \mathcal{H} \), we choose \( {g}_{0} \) as in the lemma and write\n\n\[ f = {g}_{0} + \left( {f - {g}_{0}}\right) \]\n\nBy construction \( {g}_{0} \in \mathcal{S} \), and the ... | Yes |
Lemma 5.1 \( \parallel T\parallel = \sup \{ \left| \left( {{Tf}, g}\right) \right| : \parallel f\parallel \leq 1,\parallel g\parallel \leq 1\} \) | Proof. If \( \parallel T\parallel \leq M \), the Cauchy-Schwarz inequality gives\n\n\[ \left| \left( {{Tf}, g}\right) \right| \leq M\;\text{ whenever }\parallel f\parallel \leq 1\text{ and }\parallel g\parallel \leq 1 \]\n\nthus \( \sup \{ \left| \left( {{Tf}, g}\right) \right| : \parallel f\parallel \leq 1,\parallel g... | Yes |
Proposition 5.2 A linear operator \( T : {\mathcal{H}}_{1} \rightarrow {\mathcal{H}}_{2} \) is bounded if and only if it is continuous. | Proof. If \( T \) is bounded, then \( {\begin{Vmatrix}T\left( f\right) - T\left( {f}_{n}\right) \end{Vmatrix}}_{{\mathcal{H}}_{2}} \leq M{\begin{Vmatrix}f - {f}_{n}\end{Vmatrix}}_{{\mathcal{H}}_{1}} \) , hence \( T \) is continuous. Conversely, suppose that \( T \) is continuous but not bounded. Then for each \( n \) t... | Yes |
Proposition 5.4 Let \( T : \mathcal{H} \rightarrow \mathcal{H} \) be a bounded linear transformation. There exists a unique bounded linear transformation \( {T}^{ * } \) on \( \mathcal{H} \) so that:\n\n(i) \( \left( {{Tf}, g}\right) = \left( {f,{T}^{ * }g}\right) \) ,\n\n(ii) \( \parallel T\parallel = \begin{Vmatrix}{... | The linear operator \( {T}^{ * } : \mathcal{H} \rightarrow \mathcal{H} \) satisfying the above conditions is called the adjoint of \( T \).\n\nTo prove the existence of an operator satisfying (i) above, we observe that for each fixed \( g \in \mathcal{H} \), the linear functional \( \ell = {\ell }_{g} \), defined by\n\... | Yes |
Proposition 5.5 Let \( T \) be a Hilbert-Schmidt operator on \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) with kernel \( K \).\n\n(i) If \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then for almost every \( x \) the function \( y \mapsto K\left( {x, y}\right) f\left( y\right) \) is integrable.\n\n(ii) The operator... | Proof. By Fubini’s theorem we know that for almost every \( x \), the function \( y \mapsto {\left| K\left( x, y\right) \right| }^{2} \) is integrable. Then, part (i) follows directly from an application of the Cauchy-Schwarz inequality.\n\nFor (ii), we make use again of the Cauchy-Schwarz inequality as follows\n\n\[ \... | Yes |
Proposition 6.1 Suppose \( T \) is a bounded linear operator on \( \mathcal{H} \). (i) If \( S \) is compact on \( \mathcal{H} \), then \( {ST} \) and \( {TS} \) are also compact. | Proof. Part (i) is immediate. | No |
Lemma 6.3 Suppose \( T \) is a bounded symmetric linear operator on a Hilbert space \( \mathcal{H} \). (i) If \( \lambda \) is an eigenvalue of \( T \), then \( \lambda \) is real. (ii) If \( {f}_{1} \) and \( {f}_{2} \) are eigenvectors corresponding to two distinct eigenvalues, then \( {f}_{1} \) and \( {f}_{2} \) ar... | Proof. To prove (i), we first choose a non-zero eigenvector \( f \) such that \( T\left( f\right) = {\lambda f} \). Since \( T \) is symmetric (that is, \( T = {T}^{ * } \)), we find that \[ \lambda \left( {f, f}\right) = \left( {{Tf}, f}\right) = \left( {f,{Tf}}\right) = \left( {f,{\lambda f}}\right) = \bar{\lambda }\... | Yes |
Lemma 6.5 Suppose \( T \neq 0 \) is compact and symmetric. Then either \( \parallel T\parallel \) or \( - \parallel T\parallel \) is an eigenvalue of \( T \) . | Proof. By the observation (7) made earlier, either\n\n\[ \parallel T\parallel = \sup \{ \left( {{Tf}, f}\right) : \parallel f\parallel = 1\} \;\text{ or }\; - \parallel T\parallel = \inf \{ \left( {{Tf}, f}\right) : \parallel f\parallel = 1\} .\n\]\n\nWe assume the first case, that is,\n\n\[ \lambda = \parallel T\paral... | Yes |
Theorem 1.1 The Fourier transform \( {\mathcal{F}}_{0} \), initially defined on \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \), has a (unique) extension \( \mathcal{F} \) to a unitary mapping of \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) to itself. In particular,\n\n\[ \parallel \mathcal{F}\left( f\right) {\parallel ... | The extension \( \mathcal{F} \) will be given by a limiting process: if \( \left\{ {f}_{n}\right\} \) is a sequence in the Schwartz space that converges to \( f \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then \( \left\{ {{\mathcal{F}}_{0}\left( {f}_{n}\right) }\right\} \) will converge to an element in \( {L}^{... | Yes |
Lemma 1.2 The space \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) is dense in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . In other words, given any \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), there exists a sequence \( \left\{ {f}_{n}\right\} \subset \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) such that\n\n... | For the proof of the lemma, we fix \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) and \( \epsilon > 0 \) . Then, for each \( M > 0 \), we define\n\n\[ \n{g}_{M}\left( x\right) = \left\{ \begin{matrix} f\left( x\right) & \text{ if }\left| x\right| \leq M\text{ and }\left| {f\left( x\right) }\right| \leq M, \\ 0 & \te... | No |
Lemma 1.3 Let \( {\mathcal{H}}_{1} \) and \( {\mathcal{H}}_{2} \) denote Hilbert spaces with norms \( \parallel \cdot {\parallel }_{1} \) and \( \parallel \cdot {\parallel }_{2} \), respectively. Suppose \( \mathcal{S} \) is a dense subspace of \( {\mathcal{H}}_{1} \) and \( {T}_{0} : \mathcal{S} \rightarrow {\mathcal{... | Proof. Given \( f \in {\mathcal{H}}_{1} \), let \( \left\{ {f}_{n}\right\} \) be a sequence in \( \mathcal{S} \) that converges to \( f \) , and define \[ T\left( f\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{T}_{0}\left( {f}_{n}\right) \] where the limit is taken in \( {\mathcal{H}}_{2} \) . To see that \... | Yes |
Theorem 2.1 The elements \( F \) in \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \) are exactly the functions given by (6), with \( {\widehat{F}}_{0} \in {L}^{2}\left( {0,\infty }\right) \) . Moreover \[ \parallel F{\parallel }_{{H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) } = {\begin{Vmatrix}{\widehat{F}}_{0}\end{Vma... | This shows incidentally that \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \) is a Hilbert space that is isomorphic to \( {L}^{2}\left( {0,\infty }\right) \) via the correspondence (6). The crucial point in the proof of the theorem is the following fact. For any fixed strictly positive \( y \), we let \( {\widehat{F}}... | No |
Lemma 2.2 If \( F \) belongs to \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \), then \( F \) is bounded in any proper half-plane \( \{ z = x + {iy}, y \geq \delta \} \), where \( \delta > 0 \) . | To prove this we exploit the mean-value property of holomorphic functions. This property may be stated in two alternative ways. First, in terms of averages over circles,\n\n(8)\n\n\[ F\left( \zeta \right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }F\left( {\zeta + r{e}^{i\theta }}\right) {d\theta }\;\text{ if }0 < r \leq \del... | Yes |
Theorem 2.3 Suppose \( F \) belongs to \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \) . Then \( \mathop{\lim }\limits_{{y \rightarrow 0}}F\left( {x + {iy}}\right) = \) \( {F}_{0}\left( x\right) \) exists in the following two senses:\n\n(i) As a limit in the \( {L}^{2}\left( \mathbb{R}\right) \) -norm.\n\n(ii) As a l... | Thus \( F \) has boundary values (denoted by \( {F}_{0} \) ) in either of the two senses above. The function \( {F}_{0} \) is sometimes referred to as the boundary-value function of \( f \) . The proof of (i) is immediate from what we already know. Indeed, if \( {F}_{0} \) is the \( {L}^{2} \) function whose Fourier tr... | Yes |
Lemma 3.1 The space \( {C}_{0}^{\infty }\left( \Omega \right) \) is dense in \( {L}^{2}\left( \Omega \right) \) in the norm \( \parallel \cdot {\parallel }_{{L}^{2}\left( \Omega \right) } \) . | The proof is essentially a repetition of that of Lemma 1.2. We take the precaution of modifying the definition of \( {g}_{M} \) given there to be: \( {g}_{M}\left( x\right) = \) \( f\left( x\right) \) if \( \left| x\right| \leq M, d\left( {x,{\Omega }^{c}}\right) \geq 1/M \) and \( \left| {f\left( x\right) }\right| \le... | Yes |
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