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Lemma 2.3 The Cantor-Lebesgue function \( F \) on \( \mathcal{C} \) satisfies a Lipschitz condition with exponent \( \gamma = \log 2/\log 3 \) . | Proof. The function \( F \) was constructed in Section 3.1 of Chapter 3 as the limit of a sequence \( \left\{ {F}_{n}\right\} \) of piecewise linear functions. The function \( {F}_{n} \) increases by at most \( {2}^{-n} \) on each interval of length \( {3}^{-n} \) . So the slope of \( {F}_{n} \) is always bounded by \(... | Yes |
Theorem 2.5 The Sierpinski triangle \( \mathcal{S} \) has strict Hausdorff dimension \( \alpha = \log 3/\log 2 \) . | The inequality \( {m}_{\alpha }\left( \mathcal{S}\right) \leq 1 \) follows immediately from the construction. Given \( \delta > 0 \), choose \( K \) so that \( {2}^{-K} < \delta \) . Since the set \( {S}_{K} \) covers \( \mathcal{S} \) and consists of \( {3}^{K} \) triangles each of diameter \( {2}^{-K} < \delta \), we... | Yes |
Lemma 2.6 Suppose \( B \) is a ball in the covering \( \mathcal{B} \) that satisfies\n\n\[ \n{2}^{-\ell } \leq \operatorname{diam}B < {2}^{-\ell + 1}\;\text{ for some }\ell \leq k.\n\]\n\nThen \( B \) contains at most \( c{3}^{k - \ell } \) vertices of the \( {k}^{\text{th }} \) generation. | Proof of Lemma 2.6. Let \( {B}^{ * } \) denote the ball with same center as \( B \) but three times its diameter, and let \( {\bigtriangleup }_{k} \) be a triangle of the \( {k}^{\text{th }} \) generation whose vertex \( v \) lies in \( B \) . If \( {\bigtriangleup }_{\ell }^{\prime } \) denotes the triangle of the \( ... | Yes |
Theorem 2.7 The function \( \mathcal{K}\left( t\right) \) satisfies a Lipschitz condition of exponent \( \gamma = \log 3/\log 4 \), that is:\n\n\[ \left| {\mathcal{K}\left( t\right) - \mathcal{K}\left( s\right) }\right| \leq M{\left| t - s\right| }^{\gamma }\;\text{ for all }t, s \in \left\lbrack {0,1}\right\rbrack . \... | We have already observed that \( \left| {{K}_{j + 1}\left( t\right) - {K}_{j}\left( t\right) }\right| \leq {3}^{-j} \) . Since \( {K}_{j} \) travels a distance of \( {3}^{-j} \) in \( {4}^{-j} \) units of time, we see that\n\n\[ \left| {{K}_{j}^{\prime }\left( t\right) }\right| \leq {\left( \frac{4}{3}\right) }^{j}\;\t... | Yes |
Lemma 2.8 Suppose \( \\left\\{ {f}_{j}\\right\\} \) is a sequence of continuous functions on the interval \( \\left\\lbrack {0,1}\\right\\rbrack \) that satisfy\n\n\[ \n\\left| {{f}_{j}\\left( t\\right) - {f}_{j}\\left( s\\right) }\\right| \\leq {A}^{j}\\left| {t - s}\\right| \\;\\text{ for some }A > 1, \n\]\n\nand\n\n... | Proof. The continuous limit \( f \) is given by the uniformly convergent series\n\n\[ \nf\\left( t\\right) = {f}_{1}\\left( t\\right) + \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }\\left( {{f}_{k + 1}\\left( t\\right) - {f}_{k}\\left( t\\right) }\\right) \n\]\n\nand therefore\n\n\[ \n\\left| {f\\left( t\\right) - {f}_... | Yes |
Theorem 2.9 Suppose \( {S}_{1},{S}_{2},\ldots ,{S}_{m} \) are \( m \) similartities, each with the same ratio \( r \) that satisfies \( 0 < r < 1 \) . Then there exists a unique nonempty compact set \( F \) such that\n\n\[ F = {S}_{1}\left( F\right) \cup \cdots \cup {S}_{m}\left( F\right) \] | The proof of this theorem is in the nature of a fixed point argument. We shall begin with some large ball \( B \) and iteratively apply the mappings \( {S}_{1},\ldots ,{S}_{m} \) . The fact that the similarities have ratio \( r < 1 \) will suffice to imply that this process contracts to a unique set \( F \) with the de... | Yes |
Lemma 2.10 There exists a closed ball \( B \) so that \( {S}_{j}\left( B\right) \subset B \) for all \( j = 1,\ldots, m \) . | Proof. Indeed, we note that if \( S \) is a similarity with ratio \( r \), then\n\n\[ \left| {S\left( x\right) }\right| \leq \left| {S\left( x\right) - S\left( 0\right) }\right| + \left| {S\left( 0\right) }\right| \]\n\n\[ \leq r\left| x\right| + \left| {S\left( 0\right) }\right| \text{.} \]\n\nIf we require that \( \l... | Yes |
Lemma 2.11 The distance function dist defined on compact subsets of \( {\mathbb{R}}^{d} \) satisfies\n\n(i) \( \operatorname{dist}\left( {A, B}\right) = 0 \) if and only if \( A = B \) .\n\n(ii) \( \operatorname{dist}\left( {A, B}\right) = \operatorname{dist}\left( {B, A}\right) \) .\n\n(iii) \( \operatorname{dist}\lef... | The proof of the lemma is simple and may be left to the reader. | No |
Theorem 2.12 Suppose \( {S}_{1},{S}_{2},\ldots ,{S}_{m} \) are \( m \) separated similarities with the common ratio \( r \) that satisfies \( 0 < r < 1 \) . Then the set \( F \) has Hausdorff dimension equal to \( \log m/\log \left( {1/r}\right) \) . | We now turn to the proof of Theorem 2.12, which will follow the same approach used in the case of the Sierpinski triangle. If \( \alpha = \log m/\log \left( {1/r}\right) \) , we claim that \( {m}_{\alpha }\left( F\right) < \infty \), hence \( \dim F \leq \alpha \) . Moreover, this inequality holds even without the sepa... | Yes |
Lemma 2.13 Suppose \( B \) is a ball in the covering \( \mathcal{B} \) that satisfies\n\n\[ \n{r}^{\ell } \leq \operatorname{diam}B < {r}^{\ell - 1}\;\text{ for some }\ell \leq k.\n\]\n\nThen \( B \) contains at most \( {\mathrm{{cm}}}^{k - \ell } \) vertices of the \( {k}^{\text{th }} \) generation. | Proof. If \( v \) is a vertex of the \( {k}^{\text{th }} \) generation with \( v \in B \), and \( \mathcal{O}\left( v\right) \) denotes the corresponding open set of the \( {k}^{\text{th }} \) generation, then, for some fixed dilate \( {B}^{ * } \) of \( B \), properties (a) and (b) above guarantee that \( \mathcal{O}\... | Yes |
Proposition 3.3 Chains of quartic intervals satisfy the following properties:\n\n(i) If \( \\left\\{ {I}^{k}\\right\\} \) is a chain of quartic intervals, then there exists a unique \( t \\in \\left\\lbrack {0,1}\\right\\rbrack \) such that \( t \\in \\mathop{\\bigcap }\\limits_{k}{I}^{k} \) . | Proof. Part (i) follows from the fact that \( \\left\\{ {I}^{k}\\right\\} \) is a decreasing sequence of compact sets whose diameters go to 0 . | Yes |
Theorem 3.5 Given a dyadic correspondence \( \Phi \), there exist sets \( {Z}_{1} \subset \left\lbrack {0,1}\right\rbrack \) and \( {Z}_{2} \subset \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \), each of measure zero, so that: (i) \( {\Phi }^{ * } \) is a bijection on \( \left\lbrack {0,1}\ri... | Proof. First, let \( {\mathcal{N}}_{1} \) denote the collection of chains of those quartic intervals arising in (iii) of Proposition 3.3, those for which the points in \( I = \left\lbrack {0,1}\right\rbrack \) are not uniquely representable. Similarly, let \( {\mathcal{N}}_{2} \) denote the collection of chains of thos... | Yes |
Lemma 3.6 Let\n\n\\[ \n{E}_{0} = \\left\\{ {x = \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }{a}_{k}/{4}^{k},\\;}\\right. \\text{where}\\left. {{a}_{k} \\neq {f}_{k}\\text{for all sufficiently large}k}\\right\\} \\text{.} \n\\]\n\nThen \\( m\\left( {E}_{0}\\right) = 0 \\) . | Indeed, if we fix \\( r \\), then \\( m\\left( \\left\\{ {x : {a}_{r} \\neq {f}_{r}}\\right\\} \\right) = 3/4 \\), and\n\n\\[ \nm\\left( \\left\\{ {x : {a}_{r} \\neq {f}_{r}\\text{ and }{a}_{r + 1} \\neq {f}_{r + 1}}\\right\\} \\right) = {\\left( 3/4\\right) }^{2},\\;\\text{ etc. } \n\\]\n\nThus \\( m\\left( \\left\\{ ... | Yes |
Lemma 3.8 If \( \Phi \) is the dyadic correspondence in Lemma 3.7, then \( {\Phi }^{ * }\left( t\right) = \) \( \mathcal{P}\left( t\right) \) for every \( 0 \leq t \leq 1 \) . | Proof. First, we observe that \( {\Phi }^{ * }\left( t\right) \) is unambiguously defined for every \( t \) . Indeed, suppose \( t \in \mathop{\bigcap }\limits_{k}{I}^{k} \) and \( t \in \mathop{\bigcap }\limits_{k}{J}^{k} \) are two chains of quartic intervals; then \( {I}^{k} \) and \( {J}^{k} \) must be adjacent for... | Yes |
Theorem 4.3 There exists a set \( \mathcal{B} \) in \( {\mathbb{R}}^{2} \) that:\n\n(i) is compact,\n\n(ii) has Lebesgue measure zero,\n\n(iii) contains a translate of every unit line segment. | Note that with \( F = \mathcal{B} \) and \( \gamma \in {S}^{1} \) one has \( {m}_{1}\left( {F \cap {\mathcal{P}}_{{t}_{0},\gamma }}\right) \geq 1 \) for some \( {t}_{0} \). If \( {m}_{1}\left( {F \cap {\mathcal{P}}_{t,\gamma }}\right) \) were continuous in \( t \), then this measure would be strictly positive for an in... | No |
Lemma 4.7 If \( f \) is continuous with compact support, then for every \( \gamma \in {S}^{d - 1} \) we have\n\n\[ \widehat{\mathcal{R}}\left( f\right) \left( {\lambda ,\gamma }\right) = \widehat{f}\left( {\lambda \gamma }\right) \] | Proof. For each unit vector \( \gamma \) we use the adapted coordinate system described above: \( x = \left( {{x}_{1},\ldots ,{x}_{d}}\right) \) where \( \gamma \) coincides with the \( {x}_{d} \) direction. We can then write each \( x \in {\mathbb{R}}^{d} \) as \( x = \left( {u, t}\right) \) with \( u \in {\mathbb{R}}... | Yes |
Lemma 4.8 If \( f \) is continuous with compact support, then\n\n\[{\int }_{{S}^{d - 1}}\left( {{\int }_{-\infty }^{\infty }{\left| \widehat{\mathcal{R}}\left( f\right) \left( \lambda ,\gamma \right) \right| }^{2}{\left| \lambda \right| }^{d - 1}{d\lambda }}\right) {d\sigma }\left( \gamma \right) = 2{\int }_{{\mathbb{R... | Proof. The Plancherel formula in Chapter 5 guarantees that\n\n\[2{\int }_{{\mathbb{R}}^{d}}{\left| f\left( x\right) \right| }^{2}{dx} = 2{\int }_{{\mathbb{R}}^{d}}{\left| \widehat{f}\left( \xi \right) \right| }^{2}{d\xi }.\]\n\nChanging to polar coordinates \( \xi = {\lambda \gamma } \) where \( \lambda > 0 \) and \( \... | Yes |
\[ \mathop{\sup }\limits_{{t \in \mathbb{R}}}\left| {F\left( t\right) }\right| \leq c\left( {A + B}\right) \] | Proof. The first inequality is obtained by considering separately the two cases \( \left| \lambda \right| \leq 1 \) and \( \left| \lambda \right| > 1 \) . We write\n\n\[ F\left( t\right) = {\int }_{\left| \lambda \right| \leq 1}\widehat{F}\left( \lambda \right) {e}^{2\pi i\lambda t}{d\lambda } + {\int }_{\left| \lambda... | Yes |
Theorem 4.10 If \( f \) is continuous with compact support, then\n\n\[ \n{\int }_{{S}^{1}}{\mathcal{R}}_{\delta }^{ * }\left( f\right) \left( \gamma \right) {d\sigma }\left( \gamma \right) \leq c{\left( \log 1/\delta \right) }^{1/2}\left( {\parallel f{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{2}\right) } + \parallel f{\... | The same argument as in the proof of Theorem 4.5 applies here, except that we need a modified version of Lemma 4.9. More precisely, let us set\n\n\[ \n{F}_{\delta }\left( t\right) = {\int }_{-\infty }^{\infty }\widehat{F}\left( \lambda \right) \left( \frac{{e}^{{2\pi i}\left( {t + \delta }\right) \lambda } - {e}^{{2\pi... | Yes |
Theorem 4.12 The set \( F \) is compact and of two-dimensional measure zero. It contains a translate of any unit line segment whose slope is a number \( s \) that lies outside the intervals \( \left( {-1,2}\right) \) . | Let us see how these two assertions imply the theorem. First, we note that the set \( F \) is closed (and hence compact), because both \( {E}_{0} \) and \( {E}_{1} \) are closed. Next observe that with \( 0 < y < 1 \), the slice \( {F}^{y} \) of the set \( F \) is exactly \( \left( {1 - y}\right) \mathcal{C} + \frac{y}... | Yes |
Proposition 4.13 Suppose \( {\lambda }_{0} \) and \( \ell \) are given, with \( 1 \leq {\lambda }_{0} \leq 4 \) and \( \ell \) a positive integer. Then, there exist a \( \bar{\lambda } \) and a pair \( i,{i}^{\prime } \) with \( i \neq {i}^{\prime } \) such that\n\n(9)\n\n\[ \n{\mathcal{K}}_{i}^{\ell }\left( \bar{\lamb... | This is proved on the basis of the following observation.\n\nLemma 4.14 For every | No |
Lemma 4.14 For every \( {\lambda }_{0} \) there is a pair \( 1 \leq {i}_{1},{i}_{2} \leq 4 \), with \( {i}_{1} \neq {i}_{2} \) such that \( {\mathcal{K}}_{{i}_{1}}\left( {\lambda }_{0}\right) \) and \( {\mathcal{K}}_{{i}_{2}}\left( {\lambda }_{0}\right) \) intersect. | Proof. Indeed, if the \( {\mathcal{K}}_{i} \) are disjoint for \( 1 \leq i \leq 4 \) then for sufficiently small \( \delta \) the \( {\mathcal{K}}_{i}^{\delta } \) are also disjoint. Here we have used the notation that \( {F}^{\delta } \) denotes the set of points of distance less than \( \delta \) from \( F \) . (See ... | Yes |
One-dim Lebesgue measure \( {\mathcal{L}}^{1} \) on \( \mathbb{R} \) is defined by\n\n\[ \n{\mathcal{L}}^{1}\left( A\right) = \inf \left\{ {\mathop{\sum }\limits_{{i = 1}}^{\infty }\operatorname{diam}\left( {C}_{i}\right) : A \subseteq \left( {{ \cup }_{i = 1}^{\infty }{C}_{i}}\right) ,{C}_{i} = \left( {{a}_{i},{b}_{i}... | It is easy to check that \( {\mathcal{L}}^{1} \) is a measure on \( {\mathbb{R}}^{1} \) by the above definition 1.1. | No |
Example 1.4 Any open set of \( \mathbb{R} \) is \( {\mathcal{L}}^{1} \) -measurable | and this fact be proved in Lemma 2.7. | No |
Lemma 1.5 For \( \\left( {X,\\mu }\\right) \), (i) . If \( A \\subseteq B \\subseteq X \), then \( \\mu \\left( A\\right) \\leq \\mu \\left( B\\right) \). | Proof: As in the class. | No |
Lemma 1.7 \( {\left\{ {A}_{k}\right\} }_{k = 1}^{\infty } \) is a sequence of \( \mu \) -measurable sets, then \( \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{A}_{k} \) and \( { \cap }_{k = 1}^{\infty }{A}_{k} \) are \( \mu \) -measurable. | Proof: We firstly show that \( {A}_{i} \cap {A}_{j} \) is \( \mu \) -measurable for any \( i, j \) . It is as in the class. Let \( C \subset X \) and \( {B}_{j} = \mathop{\bigcup }\limits_{{k = 1}}^{j}{A}_{k} \), then \( {B}_{1} \subseteq \cdots \subseteq {B}_{j} \subseteq {B}_{j + 1} \subseteq \cdots \) and we have\n\... | Yes |
Lemma 1.8 \( {\left\{ {A}_{k}\right\} }_{k = 1}^{\infty } \) is a sequence of \( \mu \) -measurable sets,\n\n(i) . If \( \left\{ {A}_{k}\right\} \) are disjoint, then \( \mu \left( {\mathop{\bigcup }\limits_{{k = 1}}^{\infty }{A}_{k}}\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }\mu \left( {A}_{k}\right) \) . | Proof: As in the class. | No |
Lemma 1.11 (i) \( f : X \rightarrow \mathbb{R} \) is \( \mu \) -measurable if and only if \( {f}^{-1}\left( {-\infty, a}\right) \) is \( \mu \) -measurable for any \( a \in \mathbb{R} \) . | Proof: (i). If \( f \) is \( \mu \) -measurable, by \( \left( {-\infty, a}\right) \) is open, we know that \( {f}^{-1}\left( {-\infty, a}\right) \) is \( \mu \) -measurable for any \( a \in \mathbb{R} \) .\n\nIf \( {f}^{-1}\left( {-\infty, a}\right) \) is \( \mu \) -measurable for any \( a \in \mathbb{R} \), then \( {f... | Yes |
Lemma 1.12 If \( {f}_{k} : X \rightarrow \mathbb{R} \) are \( \mu \) -measurable, \( k = 1,2,\cdots \), then \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{f}_{k},\mathop{\lim }\limits_{{k \rightarrow \infty }}{f}_{k} \) are also \( \mu \) -measurable. | Proof: For any \( a \in \mathbb{R} \), let \( {g}_{k} = \mathop{\inf }\limits_{{j \geq k}}{f}_{j} \), then\n\n\[ \n{g}_{k}^{-1}\left( {-\infty, a}\right) = \mathop{\bigcup }\limits_{{j = k}}^{\infty }{f}_{j}^{-1}\left( {-\infty, a}\right) \n\]\n\nfrom \( {f}_{j} \) are \( \mu \) -measurable and Lemma 1.7, we have \( {g... | Yes |
Lemma 1.13 For any \( \mu \) -integrable simple functions \( {g}_{1},{g}_{2} \), if \( {g}_{1} \geq {g}_{2}\mu \) -a.e., then \( \int {g}_{1}{d\mu } \geq \int {g}_{2}{d\mu } \) . | Proof: Assume \( {g}_{1} = \mathop{\sum }\limits_{{i = 1}}^{\infty }{a}_{i} \cdot {\chi }_{{E}_{i}},{g}_{2} = \mathop{\sum }\limits_{{j = 1}}^{\infty }{b}_{j} \cdot {\chi }_{{F}_{j}} \), where \( {E}_{i},{F}_{j} \) are \( \mu \) -measurable and \( {a}_{i},{b}_{j} \in \mathbb{R} \) . Let \( {\Omega }_{i, j} = {E}_{i} \c... | Yes |
Lemma 1.16 If \( f : \left( {X,\mu }\right) \rightarrow \lbrack 0,\infty ) \) satisfies \( {\int }_{X}^{ * }{fd\mu } = 0 \), then \( f = {0\mu } \) -a.e. and \( f \) is \( \mu \) -measurable. | Proof: It is easy to see that if \( f = {0\mu } \) -a.e., then \( f \) is \( \mu \) -measurable function. In the rest, we only need to show that \( f = {0\mu } \) -a.e.\n\nLet \( A = \{ x \in X : f\left( x\right) > 0\} \). By contradiction, if \( \mu \left( A\right) > 0 \), note \( A = \mathop{\bigcup }\limits_{{i \in ... | Yes |
Lemma 1.18 If \( f, g \) are \( \mu \) -integrable functions on \( \left( {X,\mu }\right) \), and none of \( {\int }_{X}f \) and \( {\int }_{X}g \) is infinity, then for any \( a, b \in \mathbb{R} \), we have \( {\int }_{X}{af} + {bgd\mu } = a{\int }_{X}{fd\mu } + b{\int }_{X}{gd\mu } \) . | Proof: Left to the reader. | No |
Lemma 1.19 Any \( \mu \) -measurable function \( f \geq 0 \) is \( \mu \) -integrable. | Proof: Let \( {E}_{i} = {f}^{-1}\left( {{2}^{i},{2}^{i + 1}}\right\rbrack, i \in \mathbb{Z} \) . If \( \mu \left( {E}_{i}\right) = \infty \) for some \( i \in \mathbb{Z} \), we can choose \( g = {2}^{i} \cdot {\chi }_{{E}_{i}} \), then \( g \leq f \) \( \mu \) -a.e. we have\n\n\[{\int }_{ * }f \geq \int g = \int {2}^{i... | Yes |
Lemma 1.21 (Fatou’s Lemma) \( {f}_{k} : X \rightarrow \lbrack 0,\infty ) \) is \( \mu \) -measurable, \( k = 1,2,\cdots \), then\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{X}{f}_{k}{d\mu } \geq {\int }_{X}\mathop{\lim }\limits_{{k \rightarrow \infty }}{f}_{k}{d\mu } \] | Proof: From Lemma 1.12, we know that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{f}_{k} \) is non-negative \( \mu \) -measurable function. Then from Lemma 1.19, we know that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{f}_{k} \) is \( \mu \) -integrable.\n\nTake \( g = \mathop{\sum }\limits_{{j = 1}}^{\inf... | Yes |
Lemma 1.22 (Monotone Convergence Theorem) \( {f}_{k} : X \rightarrow \lbrack 0,\infty ) \) is \( \mu \) -measurable, \( k = 1,2,\cdots \), with \( {f}_{1} \leq \cdots \leq {f}_{k} \leq {f}_{k + 1} \leq \cdots \), then\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{X}{f}_{k}{d\mu } = {\int }_{X}\mathop{\l... | Proof: From \( {f}_{1} \leq \cdots \leq {f}_{k} \leq {f}_{k + 1} \leq \cdots \), we know that \( \int {f}_{k}{d\mu } \) is a monotonic sequence, hence \( \mathop{\lim }\limits_{{k \rightarrow \infty }}\int {f}_{k}{d\mu } \) exists. Similarly, \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{f}_{k} \) exists.\n\nBy Le... | Yes |
Lemma 1.24 If \( f \in {\mathcal{L}}^{\infty }\left( {X,\mu }\right) \), then \( \left| f\right| \leq \parallel f{\parallel }_{\infty }\mu \) -a.e. | Proof: By definition of \( \parallel f{\parallel }_{\infty } \), there exists \( {C}_{i} \in \left( {\parallel f{\parallel }_{\infty },\parallel f{\parallel }_{\infty } + \frac{1}{i}}\right) \) such that \( \left| f\right| \leq {C}_{i}\mu \) -a.e.\n\nLet \( {E}_{i} = \left\{ {x \in X : \left| {f\left( x\right) }\right|... | Yes |
Lemma 1.25 (Dominated Convergence Theorem) Assume \( g,{\left\{ {g}_{k}\right\} }_{k = 1}^{\infty } \in {\mathcal{L}}^{1}\left( {X,\mu }\right) \) and \( f,{\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \) are \( \mu \) - measurable. Suppose \( {f}_{k} \rightarrow {f\mu } \) -a.e. and \( \left| {f}_{k}\right| \leq {g}_{k... | Proof: From \( \left| {f}_{k}\right| \leq {g}_{k} \) and \( {f}_{k} \rightarrow f,{g}_{k} \rightarrow {g\mu } \) -a.e., we have \( \left| f\right| \leq {g\mu } \) -a.e.\n\nChoose \( \widetilde{g} = {g\mu } \) -a.e. and \( \widetilde{g} \geq \left| f\right| \), note that \( \widetilde{g} \) is also \( \mu \) -measurable... | Yes |
Lemma 1.26 Let \( \mu \) be a measure on \( X \), for any \( \mu \) -measurable set \( \Omega \subseteq X \), let \( f \in {\mathcal{L}}^{1}\left( {\Omega ,\mu }\right) \) . Then for any \( \epsilon > 0 \), there is \( \delta > 0 \) such that for any \( \mu \) -measurable set \( A \subseteq \Omega \) satisfying \( \mu ... | Proof: Let\n\n\[ \n{f}_{n}\left( x\right) = \left\{ \begin{matrix} \left| {f\left( x\right) }\right| , & \text{ if }\left| {f\left( x\right) }\right| \leq n \\ 0, & \text{ otherwise } \end{matrix}\right. \]\n\nthen \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = \left| {f\left( x\right) }\rig... | Yes |
Lemma 1.27 For any \( a, b \geq 0, p, q > 1 \) and \( \frac{1}{p} + \frac{1}{q} = 1 \), we have\n\n\[ {ab} \leq \frac{{a}^{p}}{p} + \frac{{b}^{q}}{q} \] | Proof: If \( a \) or \( b \) is 0, then conclusion is trivial. Assume \( a, b > 0 \) in the rest of the proof.\n\nNote \( {\left( \ln x\right) }^{\prime \prime } < 0 \) for any \( x > 0 \), then \( \ln x \) is concave function. Note \( \frac{1}{p} + \frac{1}{q} = 1 \), then we get\n\n\[ \ln \left( {\frac{{a}^{p}}{p} + ... | Yes |
Lemma 1.28 (Hölder and Minkowski inequality) For any \( 1 \leq p, q \leq \infty \) and \( \frac{1}{p} + \frac{1}{q} = 1 \), we have\n\n\[ \parallel {fg}{\parallel }_{1} \leq \parallel f{\parallel }_{p} \cdot \parallel g{\parallel }_{q}\;\text{ and }\;\parallel f + g{\parallel }_{p} \leq \parallel f{\parallel }_{p} + \p... | Proof: If \( \parallel f{\parallel }_{p} = 0 \) or \( \infty \), or \( \parallel g{\parallel }_{q} = 0 \) or \( \infty \), then the first inequality is proved.\n\nIf \( p = 1, q = \infty \), from Lemma 1.24, we know that \( \left| {fg}\right| \leq \left| f\right| \cdot \parallel g{\parallel }_{\infty }\mu \) -a.e. From... | Yes |
Lemma 1.33 For any \( C \in {\mathcal{P}}_{1} \), there are disjoint sets \( {A}_{i} \times {B}_{i} \in {\mathcal{P}}_{0}, i = 1,\cdots \), such that \( C = { \cup }_{i = 1}^{\infty }\left( {{A}_{i} \times {B}_{i}}\right) \) . | Proof: Assume \( C = { \cup }_{i = 1}^{\infty }\left( {{C}_{i} \times {D}_{i}}\right) \), where \( {C}_{i} \times {D}_{i} \in {\mathcal{P}}_{0} \), then we have\n\n\[ C = { \cup }_{i = 1}^{\infty }{\Omega }_{i} \]\n\nwhere \( {\Omega }_{i} = \left( {{C}_{i} \times {D}_{i}}\right) - { \cup }_{j = 1}^{i - 1}\left( {{C}_{... | Yes |
Corollary 1.34 \( {\mathcal{P}}_{1} \subseteq \mathcal{F} \) . And if \( {S}_{i} \in {\mathcal{P}}_{1}, i = 1,\cdots, k \), then \( { \cap }_{i = 1}^{k}{S}_{i} \in {\mathcal{P}}_{1} \) . | Proof: Note for disjoint sets \( {A}_{i} \times {B}_{i} \in {\mathcal{P}}_{0}, i = 1,\cdots \) , \[ {\int }_{X}{\chi }_{{ \cup }_{i = 1}^{\infty }\left( {{A}_{i} \times {B}_{i}}\right) }\left( {x, y}\right) {d\mu }\left( x\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mu \left( {A}_{i}\right) \cdot {\chi }_{{B}_{i... | Yes |
Lemma 1.35 For each \( S \subseteq X \times Y \), we have\n\n\[ \left( {\mu \times v}\right) \left( S\right) = \inf \left\{ {\rho \left( R\right) \mid S \subseteq R \in {\mathcal{P}}_{1}}\right\} \] | Proof: From Lemma 1.33, we know that \( R = { \cup }_{i = 1}^{\infty }\left( {{A}_{i}^{\prime } \times {B}_{i}^{\prime }}\right) \), where \( \left( {{A}_{i}^{\prime } \times {B}_{i}^{\prime }}\right) \in {\mathcal{P}}_{0} \) are disjoint sets. Then \( S \subseteq { \cup }_{i = 1}^{\infty }\left( {{A}_{i}^{\prime } \ti... | Yes |
Proposition 1.36 If \( A \subseteq X \) is \( \mu \) -measurable and \( B \subseteq Y \) is \( v \) -measurable, then \( A \times B \) is \( \left( {\mu \times v}\right) \) -measurable and \( \left( {\mu \times v}\right) \left( {A \times B}\right) = \mu \left( A\right) \cdot v\left( B\right) \) . | Proof: Step (1). By the definition of product measure, we know that\n\n\[ \left( {\mu \times v}\right) \left( {A \times B}\right) \leq \mu \left( A\right) \cdot v\left( B\right) \]\n\nAssume \( A \times B \subseteq { \cup }_{i = 1}^{\infty }{A}_{i} \times {B}_{i} \), where \( {A}_{i} \times {B}_{i} \in {\mathcal{P}}_{0... | Yes |
Proposition 1.38 Assume \( \\left( {X,\\mu }\\right) \) and \( \\left( {Y,\\nu }\\right) \) are measure spaces, if \( S \) is \( \\left( {\\mu \\times \\nu }\\right) \) -measurable and \( \\left( {\\mu \\times \\nu }\\right) \\left( S\\right) < \\) \( \\infty \), then\n\n\[ \n{\\int }_{X \\times Y}{\\chi }_{S}d\\left( ... | Proof: From Corollary 1.37, we have \( {S}_{\\infty } \\in \\mathcal{F} \) such that \( S \\subseteq {S}_{\\infty } \) and \( \\left( {\\mu \\times v}\\right) \\left( S\\right) = \\rho \\left( {S}_{\\infty }\\right) = \\left( {\\mu \\times v}\\right) \\left( {S}_{\\infty }\\right) \). Note \( S \) is \( \\left( {\\mu \... | Yes |
Lemma 1.39 Let \( f : X \rightarrow \lbrack 0,\infty ) \) be a \( \mu \) -measurable function, then there exists \( \mu \) -measurable sets \( {\left\{ {A}_{k}\right\} }_{k = 1}^{\infty } \) in \( X \) such that \( f = \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{k}{\chi }_{{A}_{k}} \) . | Proof: Let \( {A}_{1} = \{ x \in X : f\left( x\right) \geq 1\} \), and define \( {A}_{k}, k \geq 2 \) inductively,\n\n\[ \n{A}_{k} = \left\{ {x \in X : f\left( x\right) \geq \frac{1}{k} + \mathop{\sum }\limits_{{j = 1}}^{{k - 1}}\frac{1}{j}{\chi }_{{A}_{j}}}\right\} \n\]\n\nLet \( {h}_{k} = f - \frac{1}{k} - \mathop{\s... | Yes |
Proposition 1.41 (Fubini’s Theorem) Assume \( \left( {X,\mu }\right) \) and \( \left( {Y, v}\right) \) are both \( \sigma \) -finite measure spaces, if \( f \) is \( \left( {\mu \times v}\right) \) -integrable, then\n\n\[ \n{\int }_{X \times Y}{fd}\left( {\mu \times v}\right) = {\int }_{Y}\left( {{\int }_{X}f\left( {x,... | Proof: We firstly assume that \( f \geq 0 \) . From Lemma 1.39, we can find \( \mu \times v \) -measurable sets \( {\left\{ {S}_{i}\right\} }_{i = 1}^{\infty } \) in \( X \times Y \) such that \( f = \mathop{\sum }\limits_{{i = 1}}^{\infty }\frac{1}{i}{\chi }_{{S}_{i}} \) .\n\nAssume \( X = { \cup }_{k = 1}^{\infty }{A... | Yes |
Lemma 2.3 For Radon measure \( v \) on \( {\mathbb{R}}^{n} \), assume \( {A}_{1} \subseteq {A}_{2} \subseteq \cdots \subseteq {A}_{m} \subseteq \cdots \), then\n\n\[ \mathop{\lim }\limits_{{m \rightarrow \infty }}v\left( {A}_{m}\right) = v\left( {\mathop{\bigcup }\limits_{{m = 1}}^{\infty }{A}_{m}}\right) \] | Proof: From \( v \) is Radon measure, there is open set \( {O}_{m} \) with \( {A}_{m} \subseteq {O}_{m} \subseteq {\mathbb{R}}^{n} \), such that\n\n\[ v\left( {A}_{m}\right) + \frac{1}{m} \geq v\left( {O}_{m}\right) \]\n\nDefine \( {U}_{m} = { \cap }_{j \geq m}{O}_{j} \), then \( {A}_{m} \subseteq {U}_{m} \subseteq {U}... | Yes |
Lemma 2.4 Let \( \mu \) be a Radon measure on \( {\mathbb{R}}^{n} \) , and \( A \subseteq {\mathbb{R}}^{n} \) is \( \mu \) -measurable, then \( \mu \llcorner A \) is a Radon measure. | Proof: It is easy to verify that \( \mu \llcorner A \) is a measure. Let \( v = \mu \llcorner A \), then for any \( K \subseteq \subseteq {\mathbb{R}}^{n} \), we have\n\n\[ v\left( K\right) = \mu \left( {K \cap A}\right) \leq \mu \left( K\right) < \infty \]\n\nFor any open \( \Omega \subseteq {\mathbb{R}}^{n} \), any \... | Yes |
Corollary 2.6 For any open interval \( \left( {a, b}\right) \subseteq \mathbb{R} \), we have \( {\mathcal{L}}^{1}\left( {a, b}\right) = b - a = \operatorname{diam}\left( {a, b}\right) \) . | Proof: From the definition of \( {\mathcal{L}}^{1} \), we get \( {\mathcal{L}}^{1}\left( {a, b}\right) \leq \operatorname{diam}\left( {a, b}\right) = b - a \) .\n\nOn the other hand, if \( \left( {a, b}\right) \subseteq { \cup }_{i = 1}^{\infty }\left( {{\widetilde{a}}_{i},{\widetilde{b}}_{i}}\right) \), where \( \left... | Yes |
Lemma 2.8 The Lebesgue measure \( {\mathcal{L}}^{n} \) is a Radon measure on \( {\mathbb{R}}^{n} \), where \( n \geq 1 \) . | Proof: From Lemma 2.7, we will prove the conclusion by induction. Assume \( {\mathcal{L}}^{n - 1} \) is a Radon measure,\n\nwe need to show that \( {\mathcal{L}}^{n} \) is also a Radon measure.\n\nAssume \( A = {\Pi }_{i = 1}^{n}\left( {{a}_{i},{b}_{i}}\right) \subseteq {\mathbb{R}}^{n} \), note \( {\Pi }_{i = 1}^{n - ... | Yes |
Lemma 2.11 For each \( \epsilon > 0, f \in {\mathcal{L}}^{p}\left( U\right) \), where \( U \) is open and bounded in \( {\mathbb{R}}^{n},1 \leq p \leq \infty \), we have \( {f}^{\epsilon } \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) . | Proof: We can define \( f\left( x\right) = 0 \) where \( x \in {\mathbb{R}}^{n} - U \) . For any \( x \in {\mathbb{R}}^{n}, h \in \mathbb{R},{e}_{i} = \left( {0,\cdot ,1,\cdots ,0}\right) \in {\mathbb{R}}^{n} \), there is some \( V \subseteq \subseteq {\mathbb{R}}^{n} \) such that\n\n\[ \n\frac{{f}^{\epsilon }\left( {x... | Yes |
Lemma 2.12 Let \( \Omega \subseteq {\mathbb{R}}^{n} \) be an open set and \( K \subseteq \Omega \) be compact, then there exists a function \( {J}_{K} \in {C}_{c}^{\infty }\left( \Omega \right) \) such that \( 0 \leq {J}_{K}\left( x\right) \leq 1 \) for all \( x \in \Omega \) and \( {J}_{K}\left( x\right) = 1 \) for \(... | Proof: Since \( K \) is compact, there is \( d > 0 \) such that \( {K}^{ + } \mathrel{\text{:=}} \{ x : d\left( {x, K}\right) \leq {2d}\} \subseteq \Omega \), define \( {K}_{ + } \mathrel{\text{:=}} \{ x \) : \( d\left( {x, K}\right) \leq d\} \), then we have\n\n\[ K \subseteq {K}_{ + } \subseteq {K}^{ + } \subseteq \O... | Yes |
Corollary 2.14 For any function \( h \in {\mathcal{L}}^{1}\left( {{\mathbb{R}}^{n},{\mathcal{L}}^{n}}\right) \), there exists a sequence of Riemannian integrable functions \( {h}_{i} \) on \( {\mathbb{R}}^{n} \), such that \( \mathop{\lim }\limits_{{i \rightarrow \infty }}{\begin{Vmatrix}{h}_{i} - h\end{Vmatrix}}_{1} =... | Proof: From the theory of Rimannian integral, we know any function \( \psi \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) is Riemannian integrable, hence the conclusion follows from Theorem 2.13. | No |
Lemma 2.15 (Vitali) Let \( A \subseteq {\mathbb{R}}^{n} \) with \( {\mathcal{L}}^{n}\left( A\right) < \infty \), assume \( \mathcal{F} \) is any collection of closed balls satisfying the following: for any \( \epsilon > 0, x \in A \) there is a closed ball \( B\left( {a, r}\right) \in \mathcal{F} \) such that\n\n\[ x \... | Proof: Choose open set \( B \) such that\n\n\[ A \subseteq B\;\text{ and }\;{\mathcal{L}}^{n}\left( B\right) \leq {\mathcal{L}}^{n}\left( A\right) + 1 < \infty \]\n\nFrom the definition of \( \mathcal{F} \), if we define \( \widetilde{\mathcal{F}} \) to be the collection of all closed balls \( B\left( {a, r}\right) \in... | Yes |
Lemma 2.18 Every absolutely continuous function defined on \( \left\lbrack {a, b}\right\rbrack \) is the difference of two increasing functions on \( \left\lbrack {a, b}\right\rbrack \) . | Proof: We define\n\n\[ g\left( x\right) = \mathop{\sup }\limits_{{a = {x}_{0} < {x}_{1} < \cdots < {x}_{k} = x}}\mathop{\sum }\limits_{{i = 1}}^{k}{\left\lbrack f\left( {x}_{i}\right) - f\left( {x}_{i - 1}\right) \right\rbrack }^{ + }\;\text{ and }\;h\left( x\right) = \mathop{\sup }\limits_{{a = {x}_{0} < {x}_{1} < \cd... | Yes |
Corollary 2.19 Every absolutely continuous function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is differentiable \( {\mathcal{L}}^{1} \) -a.e. and \( {f}^{\prime } \in \) \( {\mathcal{L}}^{1}\left( {\left\lbrack {a, b}\right\rbrack ,{\mathcal{L}}^{1}}\right) \) . | Proof: The conclusion follows from Lemma 2.18 and Proposition 2.16. | No |
Proposition 2.20 If \( f \in {\mathcal{L}}^{1}\left\lbrack {a, b}\right\rbrack \), then \( F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \) is absolutely continuous function on \( \left\lbrack {a, b}\right\rbrack \) . | Proof: For any \( \epsilon > 0 \), from Lemma 1.26, there exists \( \delta > 0 \) such that for any \( \Omega \subseteq \left\lbrack {a, b}\right\rbrack \) with \( {\mathcal{L}}^{1}\left( \Omega \right) < \delta \) , we have\n\n\[ \n{\int }_{\Omega }f\left( t\right) d{\mathcal{L}}^{1} \leq \epsilon \n\]\n\nIf \( A = { ... | Yes |
Lemma 2.21 If \( f \in {\mathcal{L}}^{1}\left\lbrack {a, b}\right\rbrack \) and \( {\int }_{a}^{x}f\left( t\right) {dt} = 0 \) for all \( x \in \left\lbrack {a, b}\right\rbrack \), then \( f\left( t\right) = 0{\mathcal{L}}^{1} \) -a.e. in \( \left\lbrack {a, b}\right\rbrack \) . | Proof: By contradiction. Let \( {E}_{1} = \{ x \in \left( {a, b}\right) : f\left( x\right) > 0\} \) and \( {E}_{2} = \{ x \in \left( {a, b}\right) : f\left( x\right) < 0\} \), from \( {\int }_{a}^{b}f = 0 \), if \( f\left( t\right) = 0{\mathcal{L}}^{1} \) -a.e. does not hold, we have that \( {\mathcal{L}}^{1}\left( {E}... | Yes |
Lemma 2.22 For \( f \in {\mathcal{L}}^{1}\left\lbrack {a, b}\right\rbrack \) and \( F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \), we have \( {F}^{\prime }\left( x\right) = f\left( x\right) \) for almost all \( x \in \left\lbrack {a, b}\right\rbrack \) . | Proof: Without loss of generality, we assume \( f \geq 0 \) . \n\nStep (1). If \( f \leq K \) for some \( K \), then set \n\n\[ \n{g}_{n}\left( x\right) = \frac{F\left( {x + \frac{1}{n}}\right) - F\left( x\right) }{\frac{1}{n}} = n{\int }_{x}^{x + \frac{1}{n}}f\left( t\right) {dt} \n\] \n\nwe get \( {g}_{n} \leq K \) .... | Yes |
Lemma 2.23 If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is absolutely continuous function, and \( {f}^{\prime } = 0{\mathcal{L}}^{1} \) -a.e., then \( f \) is constant function. | Proof: Choose any \( c \in (a, b\rbrack \), let \( E = \left\{ {x \in \left( {a, c}\right) : {f}^{\prime }\left( x\right) = 0}\right\} \), from the assumption, we have\n\n\[{\mathcal{L}}^{1}\left( E\right) = c - a\]\n\n(2.20)\n\nChoose any \( \epsilon > 0 \), there exists \( {\eta }_{0} > 0 \), such that for every fini... | Yes |
Proposition 2.24 (Newton-Leibniz Formula) If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is absolutely continuous, then for any \( x \in \left\lbrack {a, b}\right\rbrack \), we have \( {\int }_{a}^{x}{f}^{\prime }\left( t\right) {dt} = f\left( x\right) - f\left( a\right) \) . | Proof: From Corollary 2.19, we have \( {f}^{\prime } \in {\mathcal{L}}^{1}\left\lbrack {a, b}\right\rbrack \) . Define \( g\left( x\right) = {\int }_{a}^{x}{f}^{\prime }\left( t\right) {dt} \), from Proposition 2.20 and \( {f}^{\prime } \in {\mathcal{L}}^{1}\left\lbrack {a, b}\right\rbrack \), we know that \( g\left( x... | Yes |
For \( A \subseteq {\mathbb{R}}^{n} \), assume \( f : A \rightarrow {\mathbb{R}}^{m} \) is Lipschitz, then there exists a Lipschitz map \( \bar{f} : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) such that \[ {\left. \bar{f}\right| }_{A} = {\left. f\right| }_{A}\;\text{ and }\;\operatorname{Lip}\left( \bar{f}\right) ... | Proof: First assume \( m = 1 \), then for any \( x \in {\mathbb{R}}^{n} \), we define \[ \bar{f}\left( x\right) = \mathop{\inf }\limits_{{a \in A}}\{ f\left( a\right) + \operatorname{Lip}\left( f\right) \left| {x - a}\right| \} \] it is easy to see that \( \bar{f} = f \) on \( A \) . If \( x, y \in {\mathbb{R}}^{n} \) ... | Yes |
Lemma 3.3 Assume \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is a Lipschitz function, for \( v \in \partial B\left( 1\right) \subseteq {\mathbb{R}}^{n} \), let \( {A}_{v} = \left\{ {x \in {\mathbb{R}}^{n}}\right. \) : \( {D}_{v}f \) does not exist \( \} \), then \( {\mathcal{L}}^{n}\left( {A}_{v}\right) = 0,{D}_{... | Proof: Step (1). Because \( f \) is continuous, from Lemma 1.11 and Lemma 1.12, we know\n\n\[ \n{\bar{D}}_{v}f\left( x\right) \mathrel{\text{:=}} \overline{\mathop{\lim }\limits_{{t \rightarrow 0}}}\frac{f\left( {x + {tv}}\right) - f\left( x\right) }{t} = \mathop{\lim }\limits_{{k \rightarrow \infty }}\mathop{\sup }\li... | Yes |
Lemma 3.4 For \( f \in {\mathcal{L}}_{loc}^{1}\left( {{\mathbb{R}}^{n},{\mathcal{L}}^{n}}\right) \), if \( {\int }_{{\mathbb{R}}^{n}}f \cdot g = 0 \) for any \( g \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \), then \( f = 0{\mathcal{L}}^{n} \) -a.e. | Proof: For any \( \epsilon > 0, m > 0 \), let \( E = \{ x \in B\left( m\right) : f\left( x\right) > \epsilon \} \), where \( B\left( m\right) \) is the open ball centered at origin with radius \( m \) in \( {\mathbb{R}}^{n} \) . From Lemma 2.8, for any \( \delta > 0 \), there exists open \( \widetilde{A} \), such that\... | Yes |
Lemma 3.5 If \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is a Lipschitz function, then for any \( v \in \partial B\left( 1\right) \subseteq {\mathbb{R}}^{n} \) , \[ {D}_{v}f\left( x\right) = v \cdot \operatorname{grad}f\left( x\right) ,\;{\mathcal{L}}^{n} - \text{ a.e. }x \] | Proof: Let \( \xi \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \), for \( k \in {\mathbb{Z}}^{ + } \), we have \[ {\int }_{{\mathbb{R}}^{n}}\frac{f\left( {x + \frac{1}{k}v}\right) - f\left( x\right) }{\frac{1}{k}}\xi \left( x\right) {dx} = - {\int }_{{\mathbb{R}}^{n}}f\left( x\right) \cdot \frac{\xi \left( x\rig... | Yes |
Corollary 3.7 Let \( f : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) be locally Lipschitz, then \( {Df} : A \rightarrow {M}_{m \times n} \) is Borel measurable map, where \( {M}_{m \times n} \) is the set of all \( m \times n \) -matrices and \( A = \left\{ {x \in {\mathbb{R}}^{n} : {Df}\left( x\right) \text{exist... | Proof: It follows from Theorem 3.6 directly. | Yes |
Lemma 3.8 For continuous map \( f : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \), if \( A \subseteq {\mathbb{R}}^{m} \) is a Borel set, then \( {f}^{-1}\left( A\right) \) is Borel set in \( {\mathbb{R}}^{n} \). | Proof: Let \( \mathcal{F} = \left\{ {A \subseteq {\mathbb{R}}^{m} : {f}^{-1}\left( A\right) }\right. \) is Borel set in \( \left. {\mathbb{R}}^{n}\right\} \) . Then from the continuity of \( f \), we know that every open set of \( {\mathbb{R}}^{m} \) belongs to \( \mathcal{F} \) .\n\nNote \( \varnothing ,{\mathbb{R}}^{... | Yes |
Lemma 3.11 For \( A \subseteq {\mathbb{R}}^{n} \) with \( \operatorname{diam}A < \infty \), define \( {A}^{ * } = {S}_{{e}_{n}} \circ {S}_{{e}_{n - 1}} \circ \cdots \circ {S}_{{e}_{1}}\left( A\right) \) . Then for any \( x \in {A}^{ * } \), we have \( - x \in {A}^{ * } \) . | Proof: We define \( {A}_{k} = {S}_{{e}_{k}}\left( {A}_{k - 1}\right) \) by induction, where \( {A}_{1} = {S}_{{e}_{1}}\left( A\right) \), then \( {A}^{ * } = {A}_{n} \). From the definition of \( {A}_{1} \), we know that for any \( \left( {{x}_{1},\cdots ,{x}_{n}}\right) \in {A}_{1} \), then \( \left( {-{x}_{1},{x}_{2}... | Yes |
Proposition 3.12 (Isodiametric inequality) For any \( A \subseteq {\mathbb{R}}^{n} \), we have \( {\mathcal{L}}^{n}\left( A\right) \leq {\omega }_{n}{\left( \frac{\operatorname{diam}A}{2}\right) }^{n} \), where \( {\omega }_{n} \) is the volume of unit ball in \( {\mathbb{R}}^{n} \) . | Proof: From Lemma 3.11, for any \( x \in {\left( \bar{A}\right) }^{ * } \), we have \( - x \in {\left( \bar{A}\right) }^{ * } \), hence\n\n\[ 2\left| x\right| = \left| {x - \left( {-x}\right) }\right| \leq \operatorname{diam}{\left( \bar{A}\right) }^{ * } \]\n\nwhich implies \( {\left( \bar{A}\right) }^{ * } \subseteq ... | Yes |
Proposition 3.16 \( {\mathcal{H}}^{n} = {\mathcal{L}}^{n} \) on \( {\mathbb{R}}^{n} \) . | Proof: Assume \( A \subseteq {\mathbb{R}}^{n} \) . \n\nStep (1). For any \( \delta > 0 \), choose \( {C}_{j} \subseteq {\mathbb{R}}^{n} \) such that \( A \subseteq { \cup }_{j = 1}^{\infty }{C}_{j} \) and diam \( {C}_{j} \leq \delta \), from Proposition 3.12, \n\n\[ \n{\mathcal{L}}^{n}\left( A\right) \leq \mathop{\sum ... | Yes |
Lemma 3.17 Let \( A \subseteq {\mathbb{R}}^{n}, f : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) is Lipschitz and \( s \in \left( {0,\infty }\right) \), then\n\n\[{\mathcal{H}}^{s}\left( {f\left( A\right) }\right) \leq {\left( \operatorname{Lip}f\right) }^{s} \cdot {\mathcal{H}}^{s}\left( A\right)\]\n\nFurthermore ... | Proof: Step (1). Fix \( \delta > 0 \), choose \( {C}_{i} \subseteq {\mathbb{R}}^{n} \) such that\n\n\[ \operatorname{diam}{C}_{i} \leq \delta ,\;A \subseteq { \cup }_{i = 1}^{\infty }{C}_{i} \]\n\nthen diam \( f\left( {C}_{i}\right) \leq \operatorname{Lip}f \cdot \operatorname{diam}{C}_{i} \leq \operatorname{Lip}f \cdo... | Yes |
Corollary 3.18 For affine isometry \( L : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}, A \subseteq {\mathbb{R}}^{n} \), we have \( {\mathcal{H}}^{s}\left( {L\left( A\right) }\right) = {\mathcal{H}}^{s}\left( A\right) \) . | Proof: For affine isometry \( L \), we have that Lip \( L = 1 \) . Apply Lemma 3.17 on \( L,{L}^{-1} \), the conclusion follows. | Yes |
Proposition 3.22 \( L : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is a linear function and \( A \subseteq {\mathbb{R}}^{n} \) is \( {\mathcal{L}}^{n} \) -measurable with \( {\mathcal{L}}^{n}\left( A\right) < \infty \), then \( \varphi \left( y\right) = {\mathcal{H}}^{n - 1}\left( {A \cap {L}^{-1}\{ y\} }\right) \) is ... | Proof: \( \operatorname{Step}\left( \mathbf{1}\right) \) . If \( \dim \left( {L\left( {\mathbb{R}}^{n}\right) }\right) < 1 \), then \( L\left( {\mathbb{R}}^{n}\right) = a \in \mathbb{R} \) and \( \left| {\nabla L}\right| = 0 \), we have\n\n\[ \varphi \left( b\right) = 0,\;\forall b \neq a \]\n\n\[ \varphi \left( a\righ... | Yes |
Theorem 3.24 If \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is Lipschitz, then for \( g \in {\mathcal{L}}^{1}\left( {{\mathbb{R}}^{n},{\mathcal{L}}^{n}}\right) \), we have \( {\left. g\right| }_{{f}^{-1}\left( y\right) } \in {\mathcal{L}}^{1}\left( {{f}^{-1}\left( y\right) ,{\mathcal{H}}^{n - 1}}\right) \) for \(... | Proof: Note \( g = {g}^{ + } - {g}^{ - } \), to prove the conclusion we can assume \( g \geq 0 \) . From Lemma 1.39, we can write \( g = \mathop{\sum }\limits_{{i = 1}}^{\infty }\frac{1}{i}{\chi }_{{A}_{i}} \), where \( {A}_{i} \) are \( {\mathcal{L}}^{n} \) -measurable sets. From \( g \in {\mathcal{L}}^{1} \), we know... | Yes |
Lemma 4.1 The balls \( {\left\{ B\left( {a}_{j},\frac{{r}_{j}}{3}\right) \right\} }_{j = 1}^{\infty } \) are disjoint, and \( A \subseteq { \cup }_{j = 1}^{J}{B}_{j} \) . | Proof: Note for \( j > i,{a}_{j} \notin {B}_{i} \), then\n\n\[ \left| {{a}_{j} - {a}_{i}}\right| > {r}_{i} = \frac{{r}_{i}}{3} + \frac{2}{3}{r}_{i} \geq \frac{{r}_{i}}{3} + \frac{2}{3} \cdot \frac{3}{4}{r}_{j} \geq \frac{{r}_{i}}{3} + \frac{{r}_{j}}{3} \]\n\n(4.1)\n\nhence \( B\left( {{a}_{i},\frac{{r}_{i}}{3}}\right) ... | Yes |
Lemma 4.3 (Besicovitch’s Covering Theorem) There exists a positive integer \( C\left( n\right) \) depending only on \( n \) such that the following holds. If \( \mathfrak{F} \) is any collection of closed balls in \( {\mathbb{R}}^{n} \) with \( \sup \{ \operatorname{diam}\left( B\right) : B \in \mathfrak{F}\} < \infty ... | Proof: Step (1). If \( A \) is bounded, then define \( \sigma : \{ 1,2,\cdots \} \rightarrow \left\{ {1,\cdots ,{M}_{n}}\right\} \) as follows:\n\n(i) \( \sigma \left( i\right) = i \) for \( 1 \leq i \leq {M}_{n} \).\n\n(ii) For \( k \geq {M}_{n} \), define \( \sigma \left( {k + 1}\right) \) inductively as follows. By ... | Yes |
Corollary 4.4 Let \( \mu \) be a Radon measure on \( {\mathbb{R}}^{n} \) and \( \mathcal{F} \) is any collection of closed balls. Let \( A = \{ a \) : \( B\left( {a, r}\right) \in \mathcal{F}\} \) and assume \( \mu \left( A\right) < \infty \), also for each \( a \in A \), we have\n\n\[ \inf \{ r : B\left( {a, r}\right)... | Proof: Let \( {\mathcal{F}}_{1} = \{ B \in \mathcal{F} : \operatorname{diam}B \leq 1, B \subseteq U\} \) . From Lemma 4.3, there are families \( {\mathcal{G}}_{1},\cdots ,{\mathcal{G}}_{C\left( n\right) } \) of disjoint balls in \( {\mathcal{F}}_{1} \) such that \( \left( {A \cap U}\right) \subseteq \mathop{\bigcup }\l... | Yes |
Lemma 4.5 For any \( \alpha \in \left( {0,\infty }\right) \), (i) . If \( A \subseteq \left\{ {x \in {\mathbb{R}}^{n} \mid {\underline{D}}_{\mu }v\left( x\right) \leq \alpha }\right\} \), then \( v\left( A\right) \leq {\alpha \mu }\left( A\right) \). (ii) . If \( A \subseteq \left\{ {x \in {\mathbb{R}}^{n} \mid {\bar{D... | Proof: Step (1). If \( v\left( A\right) < \infty \) . For any \( \epsilon > 0 \), let \( U \) be open such that \[ A \subseteq U,\;\mu \left( A\right) + \epsilon \geq \mu \left( U\right) \] Set \( \mathcal{F} = \{ B = B\left( {a, r}\right) : a \in A, B \subseteq U, v\left( B\right) \leq \left( {\alpha + \epsilon }\righ... | Yes |
Lemma 4.6 If \( v < < \mu \) are Radon measures, \( A \) is \( \mu \) -measurable, then \( A \) is \( v \) -measurable. | Proof: Firstly we assume that \( \mu \left( A\right) < \infty \) . From \( \mu \) is Radon measure, we can find open set \( {\Omega }_{k} \subseteq {\mathbb{R}}^{n} \) such that\n\n\[ A \subseteq {\Omega }_{k}\;\text{ and }\;\mu \left( {\Omega }_{k}\right) \leq \mu \left( A\right) + \frac{1}{k} \]\n\nLet \( \Omega = { ... | Yes |
Corollary 4.8 If \( f \geq 0 \) is \( \mu \) -measurable and \( v < < \mu \), then \( {\int }_{{\mathbb{R}}^{n}}{fdv} = {\int }_{{\mathbb{R}}^{n}}f \cdot {D}_{\mu }{vd\mu } \) . | Proof: Apply Lemma 1.39, we can write \( f = \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{k}{\chi }_{{E}_{k}} \), where \( {E}_{k} \) is \( \mu \) -measurable, apply Proposition 4.7 , the conclusion follows. | No |
Lemma 4.9 For \( f \in {\mathcal{L}}_{loc}^{1}\left( {{\mathbb{R}}^{n},\mu }\right) \), if \( {\int }_{{\mathbb{R}}^{n}}f \cdot {\chi }_{A}{d\mu } = 0 \) for any open set \( A \subseteq {\mathbb{R}}^{n} \), then \( f = {0\mu } \) -a.e. | Proof: For any \( \epsilon > 0, m > 0 \), let \( E = \{ x \in B\left( m\right) : f\left( x\right) > \epsilon \} \), where \( B\left( m\right) \) is the open ball centered at origin with radius \( m \) in \( {\mathbb{R}}^{n} \). From Lemma 1.11, we know that \( E \) is \( \mu \) -measurable. From \( \mu \) is Radon meas... | Yes |
Proposition 4.11 (Lebesgue-Besicovitch Differentiation Theorem) Let \( \mu \) be a Radon measure on \( {\mathbb{R}}^{n} \) and \( f \in {\mathcal{L}}_{loc}^{1}\left( {{\mathbb{R}}^{n},\mu }\right) \), then for \( \mu \) -a.e. \( x \in {\mathbb{R}}^{n} \), we have\n\n\[ \n\mathop{\lim }\limits_{{r \rightarrow 0}}{f}_{B\... | Proof: By \( f = {f}^{ + } - {f}^{ - } \), we can assume \( f \geq 0 \) in the rest of the proof. Define\n\n\[ \nv\left( A\right) = \inf \left\{ {{\int }_{\Omega }{fd\mu } : A \subseteq \Omega \subseteq {\mathbb{R}}^{n},\Omega \text{ is open }}\right\} \n\]\n\nthen from Lemma 4.10, \( v \) is a Radon measure. From Lemm... | Yes |
Corollary 4.12 Let \( \mu \) be a Radon measure on \( {\mathbb{R}}^{n}, f \in {\mathcal{L}}_{loc}^{p}\left( {{\mathbb{R}}^{n},\mu }\right) \) where \( 1 \leq p < \infty \), then\n\n\[ \mathop{\lim }\limits_{{r \rightarrow 0}}{\int }_{B\left( {x, r}\right) }{\left| f\left( y\right) - f\left( x\right) \right| }^{p}{d\mu ... | Proof: Let \( {\left\{ {r}_{i}\right\} }_{i = 1}^{\infty } = \mathbb{Q} \subseteq \mathbb{R} \) . Note \( {\left| f\left( y\right) - {r}_{i}\right| }^{p} \in {\mathcal{L}}_{loc}^{1}\left( {{\mathbb{R}}^{n},\mu }\right) \), by Proposition 4.11, for each \( i \in {\mathbb{Z}}^{ + } \), \n\n\[ \mathop{\lim }\limits_{{r \r... | Yes |
Lemma 4.14 (Partition of Unity) Let \( {\left\{ {V}_{i}\right\} }_{i = 1}^{m} \) be open sets of \( {\mathbb{R}}^{n}, K \) is compact and \( K \subseteq \mathop{\bigcup }\limits_{{i = 1}}^{m}{V}_{i} \), then there are smooth function \( {h}_{i} \geq 0 \) such that \( \operatorname{spt}\left( {h}_{i}\right) \subseteq {V... | Proof: For any \( x \in K \), there is a neighborhood \( {W}_{k} \) with compact closure \( \overline{{W}_{k}} \subseteq {V}_{i} \) for some \( i \) . From the compactness of \( K \), there are \( {x}_{1},\cdots ,{x}_{k} \) such that \( K \subseteq { \cup }_{i = 1}^{k}{W}_{{x}_{i}} \) .\n\nWe define the set \( {J}_{i} ... | Yes |
Lemma 4.15 Assume \( V \) is a positive subspace of \( {C}_{c}\left( {{\mathbb{R}}^{n},{\mathbb{R}}^{m}}\right) \), the map \( L : V \rightarrow \mathbb{R} \) is uniformly bounded on compact sets and satisfies \( L\left( {{af} + {bg}}\right) = {aL}\left( f\right) + {bL}\left( g\right) ,\forall f, g \in V, a, b \geq 0 \... | Proof: Step (1). For simplicity, in the proof we use the notation \( \mu \) instead of \( {\mu }_{L} \) . It is obvious that \( \mu \left( \varnothing \right) = 0 \) by \( L\left( 0\right) = 0 \), where 0 is the zero function.\n\nFor any \( \epsilon > 0 \), from the definition of \( \mu \left( A\right) \) for \( A \sub... | Yes |
Corollary 4.22 For any \( K \subseteq \subseteq {\mathbb{R}}^{n} \) and \( f \in {C}_{c}\left( K\right) \), there is \( {\left\{ {f}_{i}\right\} }_{i = 1}^{\infty } \subseteq {C}_{c}^{\infty }\left( K\right) \) such that\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}\mathop{\sup }\limits_{{x \in {\mathbb{R}}^{n}... | Proof: Let \( \Omega = \{ x : f\left( x\right) \neq 0\} \), then \( \Omega \) is open and \( \Omega \subseteq K \) . Define \( {K}_{i} = \left\{ {x : \left| {f\left( x\right) }\right| \geq {2}^{-i}}\right\} \), then \( {K}_{i} \subseteq \subseteq \Omega \) , from Lemma 2.12 we can find \( {J}_{i} \in {C}_{c}^{\infty }\... | Yes |
Lemma 4.27 Let \( {\left\{ {\mu }_{k}\right\} }_{k = 1}^{\infty },\mu \) be Radon measures on \( {\mathbb{R}}^{n} \), if \( {\mu }_{k} \rightharpoonup \mu \), then for any open set \( U \subseteq {\mathbb{R}}^{n} \) we have\n\n\[ \mu \left( U\right) \leq \mathop{\lim }\limits_{{k \rightarrow \infty }}{\mu }_{k}\left( U... | Proof: Let \( U \subseteq {\mathbb{R}}^{n} \) be open, for any compact set \( K \subseteq U \), from Lemma 2.12, we can find \( {J}_{K} \in {C}_{c}\left( U\right) \) such that \( 0 \leq {J}_{K} \leq 1 \) and \( {\left. {J}_{K}\right| }_{K} = 1 \), then\n\n\[ \mu \left( K\right) \leq {\int }_{{\mathbb{R}}^{n}}{J}_{K}{d\... | Yes |
Lemma 4.29 If \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is locally Lipschitz, then for any non-zero \( v \in {\mathbb{R}}^{n} \), we have\n\n\[ \n{\int }_{\Omega }{D}_{v}f\left( x\right) \cdot {gdx} = - {\int }_{\Omega }f \cdot {D}_{v}{gdx},\;\forall g \in {C}_{c}^{\infty }\left( \Omega \right) \n\] | Proof: Let \( K = \operatorname{spt}\left( g\right) \subseteq \subseteq \widetilde{\Omega } \), where \( \widetilde{\Omega } \) is a bounded open set. Note for \( x \in K \) and \( 0 < t < 1 \), we have\n\n\[ \n\mathop{\sup }\limits_{{x \in K}}\left| {\frac{f\left( {x + {tv}}\right) - f\left( x\right) }{t}g\left( x\rig... | Yes |
Lemma 4.31 Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be convex, then \( f \) is locally Lipschitz; if \( {Df}\left( x\right) \) exists we have:\n\n\[ f\left( y\right) \geq f\left( x\right) + {Df}\left( x\right) \left( {y - x}\right) \] | Proof: Step (1). Let \( \Omega = {\left\lbrack -m, m\right\rbrack }^{n} \) with vertex \( {\left\{ {v}_{k}\right\} }_{k = 1}^{{2}^{n}} \), then for any \( x \in \Omega \), we can find \( {\left\{ {\lambda }_{k}\right\} }_{k = 1}^{{2}^{n}} \) such that\n\n\[ x = \mathop{\sum }\limits_{{k = 1}}^{{2}^{n}}{\lambda }_{k}{v}... | Yes |
Lemma 4.35 If \( L : {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \rightarrow \mathbb{R} \) is linear and \( L\left( {{C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \cap {C}^{ + }\left( {\mathbb{R}}^{n}\right) }\right) \subseteq \overline{{\mathbb{R}}^{ + }} \) . Then there is a Radon measure \( \mu \) on \( {\math... | Proof: For any \( K \subseteq \subseteq {\mathbb{R}}^{n} \), choose \( {J}_{K} \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) with \( {\left. {J}_{K}\right| }_{K} = 1,0 \leq {J}_{K} \leq 1 \) . Then for any \( f \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) with \( \operatorname{spt}\left( f\right) \su... | Yes |
Theorem 4.39 Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be convex, then for \( {\mathcal{L}}^{n} \) -a.e. \( x \), we have\n\n\[ \mathop{\lim }\limits_{{y \rightarrow x}}\frac{\left| f\left( y\right) - f\left( x\right) - Df\left( x\right) \left( y - x\right) - \frac{1}{2}{\left( y - x\right) }^{T}{D}^{2}f\le... | Proof: From Corollary 4.12 and Lemma 4.13, we know that there is a set \( A \subseteq {\mathbb{R}}^{n} \) with \( {\mathcal{L}}^{n}\left( {{\mathbb{R}}^{n} - A}\right) = 0 \) , such that \( {Df}\left( x\right) \) exists for any \( x \in A \) and the following holds:\n\n\[ \mathop{\lim }\limits_{{t \rightarrow 0}}{\int ... | Yes |
For \( f \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \), then for \( r > 0 \), we have\n\n\[ \left| {{\mathcal{A}}_{r}\left( f\right) \left( x\right) - {\mathcal{A}}_{r}\left( f\right) \left( y\right) }\right| \leq C\left( n\right) {r}^{-n}\parallel f{\parallel }_{{\mathcal{L}}^{p}\left( {\mathbb{R}}^{n}\right)... | Proof: Let \( {\Omega }_{r}\left( {x, y}\right) = \left( {B\left( {x, r}\right) - B\left( {y, r}\right) }\right) \cup \left( {B\left( {y, r}\right) - B\left( {x, r}\right) }\right) \), note \( {\mathcal{L}}^{n}\left( {{\Omega }_{r}\left( {x, y}\right) }\right) \leq {\varphi }_{r}\left( \left| {x - y}\right| \right) \) ... | Yes |
Lemma 5.4 For \( 1 \leq r < s < t \), we have\n\n\[ \parallel f{\parallel }_{{\mathcal{L}}^{s}} \leq \parallel f{\parallel }_{{\mathcal{L}}^{r}}^{\theta } \cdot \parallel f{\parallel }_{{\mathcal{L}}^{t}}^{1 - \theta } \]\n\nwhere \( \theta \in \left( {0,1}\right) \) satisfies \( \frac{1}{s} = \frac{\theta }{r} + \frac... | Proof: Note\n\n\[ \parallel f{\parallel }_{{\mathcal{L}}^{s}\left( U\right) } \leq {\left( \int {\left| f\right| }^{\theta s} \cdot {\left| f\right| }^{\left( {1 - \theta }\right) s}\right) }^{\frac{1}{s}} \leq {\left( \int {\left| f\right| }^{{\theta s} \cdot \frac{r}{\theta s}}\right) }^{\frac{\theta s}{r} \cdot \fra... | Yes |
Lemma 5.8 For any \( f, g \in {W}^{1, p}\left( U\right) \), we have\n\n\[ \parallel f + g{\parallel }_{{W}^{1, p}\left( U\right) } \leq \parallel f{\parallel }_{{W}^{1, p}\left( U\right) } + \parallel g{\parallel }_{{W}^{1, p}\left( U\right) } \] | Proof: Using Lemma 1.28, we get\n\n\[ \parallel f + g{\parallel }_{{W}^{1, p}\left( U\right) } = {\left( {\int }_{U}{\left| f + g\right| }^{p} + {\left| Df + Dg\right| }^{p}\right) }^{\frac{1}{p}} = {\left( \parallel f + g{\parallel }_{{\mathcal{L}}^{p}\left( U\right) }^{p} + \parallel Df + Dg{\parallel }_{{\mathcal{L}... | Yes |
Lemma 5.9 If \( 1 < p \leq \infty \), the following are equivalent:\n\n(a) . \( f \in {W}^{1, p}\left( U\right) \) ;\n\n(b) . For \( f \in {\mathcal{L}}^{p}\left( U\right) \), where \( \frac{1}{p} + \frac{1}{q} = 1 \), and the following holds:\n\n\[{\int }_{U}f \cdot \operatorname{div}\left( \phi \right) {dx} \leq C\pa... | Proof: If (a) holds, we have \( f \in {W}^{1, p}\left( U\right) \), then\n\n\[ \int f \cdot \operatorname{div}\left( \phi \right) = - \int {Df} \cdot \phi \leq \parallel {Df}{\parallel }_{{\mathcal{L}}^{p}} \cdot \parallel \phi {\parallel }_{{\mathcal{L}}^{q}} \leq C\parallel \phi {\parallel }_{{\mathcal{L}}^{q}} \]\n\... | Yes |
Lemma 5.11 The following are equivalent:\n\n(a) . \( f \in \mathrm{{BV}}\left( U\right) \) ;\n\n(b) . \( f \in {\mathcal{L}}^{1}\left( U\right) \) and there is a Radon measure \( \mu \) on \( U \) with \( \mu \left( U\right) < \infty \) and \( \mu \) -meuasurable function \( \sigma : U \rightarrow {\mathbb{R}}^{n} \) w... | Proof: If (b) holds, then\n\n\[ \int f \cdot \operatorname{div}\left( \phi \right) = - \int \phi \cdot {\sigma d\mu } \leq \parallel \phi {\parallel }_{{\mathcal{L}}^{\infty }} \cdot \mu \left( U\right) \leq C\parallel \phi {\parallel }_{{\mathcal{L}}^{\infty }} \]\n\nIf (a) holds, then define \( {L}_{f}\left( \phi \ri... | Yes |
Lemma 5.13 If \( {f}_{k} \in \mathrm{{BV}}\left( U\right), f \in {\mathcal{L}}^{1}\left( U\right) ,\mathop{\lim }\limits_{{k \rightarrow \infty }}{\begin{Vmatrix}{f}_{k} - f\end{Vmatrix}}_{{\mathcal{L}}^{1}\left( U\right) } = 0 \) and \( \mathop{\lim }\limits_{{k \rightarrow \infty }}\begin{Vmatrix}{D{f}_{k}}\end{Vmatr... | Proof: From the definition of \( \parallel {Df}\parallel \) and Lemma 5.11, we in fact have\n\n\[ \n{\int }_{U}f\operatorname{div}\left( \phi \right) {dx} = \mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{U}{f}_{k}\operatorname{div}\left( \phi \right) {dx} = - \mathop{\lim }\limits_{{k \rightarrow \infty }}{\in... | Yes |
Proposition 5.14 \( \\left( {{W}^{1, p}\\left( U\\right) ,\\parallel \\cdot {\\parallel }_{{W}^{1, p}\\left( U\\right) }}\\right) \) and \( \\left( {\\mathrm{{BV}}\\left( U\\right) ,\\parallel \\cdot {\\parallel }_{\\mathrm{{BV}}\\left( U\\right) }}\\right) \) are complete spaces. | Proof: Step (1). For any Cauchy sequence \( {\\left\{ {f}_{k}\\right\} }_{k = 1}^{\\infty } \\subseteq \\left( {{W}^{1, p}\\left( U\\right) ,\\parallel \\cdot {\\parallel }_{{W}^{1, p}\\left( U\\right) }}\\right) \), from Theorem 1.30, there is \( f \\in {\\mathcal{L}}^{p}\\left( U\\right), g \\in {\\mathcal{L}}^{p}\\l... | Yes |
Proposition 5.16 (a) . If \( f \in {W}^{1, p}\left( U\right) \), then there are \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \subseteq {W}^{1, p}\left( U\right) \cap {C}^{\infty }\left( U\right) \) such that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}\parallel {f}_{k} - \) \( {\left. f\right.\parallel }_{{W}^... | Proof: Step (1). Set \( {\Omega }_{0} = \varnothing \) and\n\n\[{\Omega }_{k} = {U}_{{k}^{-1}} \cap \overset{ \circ }{B}\left( k\right) ,\;\forall k \in {\mathbb{Z}}^{ + }\]\n\ndefine \( {V}_{k} = {\Omega }_{k + 2} - \overline{{\Omega }_{k - 1}} \) . Then \( {W}_{k} \mathrel{\text{:=}} \overline{{\Omega }_{k + 1}} - {\... | Yes |
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