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Proposition 13.2. \( C \) is a zero set. | Proof. For any fixed \( \varepsilon > 0 \), we can find an \( n \) for which \( {C}^{n} \) has length less than \( \frac{1}{2}\varepsilon \) . These intervals are closed, so if we dilate the closed intervals by a factor of 2 to an open interval, this will cover the Cantor set. | No |
Evidently \( \left| {W\left( n\right) }\right| = {2}^{n} \) for any \( n \) . Now we remark that if \( \# S \leq {2}^{n} \), then there exists a surjection \( W\left( n\right) \rightarrow S \) . | So there exists an \( {n}_{1} \) so that for which we can construct a surjection \( W\left( {n}_{1}\right) \rightarrow {\mathcal{M}}_{1} \) ; we can label each of the pieces of \( {\mathcal{M}}_{1} \) by labels such that each label is is used. Now we pick an \( {n}_{2} \) sufficiently large and construct an extended su... | No |
Theorem 16.6 (Mean Value Theorem). Suppose \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is continuous and the restriction of \( f \) to \( \left( {a, b}\right) \) is differentiable. Then there exists \( {a\theta } \in \left( {a, b}\right) \) such that\n\n\[ f\left( b\right) - f\left( a\right) = {f}... | Proof. The intuition is secants. We refer the reader to the diagram in the book.\n\nDefine\n\n\[ \phi \left( x\right) = f\left( x\right) - S\left( {x - a}\right) . \]\n\nEvidently \( \phi \left( a\right) = f\left( a\right) \) and \( \phi \left( b\right) = f\left( a\right) \) . Furthermore, \( \phi \) is differentiable,... | No |
Theorem 16.7 (Ratio MVT). Let \( f, g : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be continuous and differentiable on \( \left( {a, b}\right) \) . Then there exists \( \theta \in \left( {a, b}\right) \) such that\n\n\[{\Delta f}{g}^{\prime }\left( \theta \right) = {f}^{\prime }\left( \theta \right) {\D... | Proof. Consider\n\n\[ \Phi \left( x\right) = {\Delta f}\left( {g\left( x\right) - g\left( a\right) }\right) - \left( {f\left( x\right) - f\left( a\right) }\right) {\Delta g}.\]\n\nClearly \( \Phi \) is \ | Yes |
Theorem 16.8 (L’Hospital’s Rule). Let \( f \) and \( g \) be functions differentiable on \( \left( {a, b}\right) \) . Suppose that \( f\left( x\right) \rightarrow 0 \) and \( g\left( x\right) \rightarrow 0 \) as \( x \rightarrow b \) . Furthermore, suppose \( g\left( x\right) \neq 0 \) and \( {g}^{\prime }\left( x\righ... | Here is an intuitive description of the proof. Consider a fixed \( x \) . Then we can find a \( t \) much closer to \( b \) than \( x \) . Because \( g \rightarrow 0 \), then \( g\left( t\right) \) is very close to zero, and negligible in comparison to the \( g\left( x\right) \neq 0 \) . (Note that \( g\left( x\right) ... | No |
Theorem 16.10. If \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is differentiable, then \( {f}^{\prime } \) is Darboux continuous. That is, if \( {f}^{\prime }\left( {x}_{1}\right) < \alpha < {f}^{\prime }\left( {x}_{2}\right) \), then \( {f}^{\prime }\left( \theta \right) = \alpha \) for some \( \alpha \) betw... | Proof. Fix a positive \( h > 0 \) . Then let \( S\left( x\right) \) be the slope of the secant joining \( \left( {x, f\left( x\right) }\right) \) and \( \left( {x + h, f\left( {x + h}\right) }\right) \) .\n\nFor some sufficiently small \( h > 0 \), we have \( S\left( {x}_{1}\right) < \alpha < S\left( {x}_{2}\right) \) ... | No |
Theorem 17.13. Fix \( x \) and \( f : \\left( {a, b}\\right) \\rightarrow \\mathbb{R} \), where \( f \) is rth order differentiable at \( x \) . Define \( R\\left( h\\right) = f\\left( {x + h}\\right) - P\\left( h\\right) \) . Then \( \\frac{R\\left( h\\right) }{{h}^{r}} \\rightarrow 0 \) as \( h \\rightarrow 0 \) if a... | Proof. Clearly \( R\\left( 0\\right) = 0 \) . Furthermore, by MVT, we know \( R\\left( h\\right) - R\\left( 0\\right) = {R}^{\\prime }\\left( {\\theta }_{1}\\right) \\cdot h \) for some \( {\\theta }_{1} \\in \\left( {0, h}\\right) \) .\n\nNow we can easily check \( {R}^{\\prime }\\left( 0\\right) = 0 \) by constructio... | No |
Proposition 18.4. If \( f \in \mathcal{R} \), then \( f \) is bounded. | Proof. Suppose not. Then \( \exists I \in \mathbb{R},{\delta .0} \) such that \( \parallel P\parallel < \delta \Rightarrow \left| {R\left( {f, P, T}\right) - I}\right| < {2013} \). Fix \( P \). Now there exists a \( k \) such that \( f\left( t\right) \) is unbounded as \( {x}_{k - 1} \leq t \leq {x}_{k} \); after all \... | No |
Proposition 18.5. The map \( \mathcal{R} \rightarrow \mathbb{R} \) by \( f \mapsto {\int }_{a}^{b}f\left( x\right) {dx} \) is bilinear. Furthermore, if \( f, g \in \mathcal{R} \) and \( f\left( x\right) \leq g\left( x\right) \) for all \( x \), then \( \int f \leq \int g \) . Finally, if \( f\left( x\right) \equiv c \)... | Proof. This is obvious. | No |
Proposition 18.7. For any partitions \( {P}_{1} \) and \( {P}_{2} \), we have \( L\left( {f,{P}_{1}}\right) \leq U\left( {f,{P}_{2}}\right) \) . | Proof. \( L\left( {f,{P}_{1}}\right) \leq L\left( {f,{P}_{1} \cup {P}_{2}}\right) \leq U\left( {f,{P}_{1} \cup {P}_{2}}\right) \leq U\left( {f,{P}_{2}}\right) \) and we’re done. We refer to \( {P}_{1} \cup {P}_{2} \) as the common refinement. | Yes |
Theorem 18.9. The following are equivalent for \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \left\lbrack {-M, M}\right\rbrack \) .\n\n(a) \( \underline{I} = \bar{I} \) . ( \( \underline{I} \leq \bar{I} \) is always true.)\n\n(b) \( \forall \varepsilon > 0\exists P \) such that \( U\left( {f, P}\right) - L\left( ... | Proof. First, let us prove (a) implies (b). By (a) we can find \( {P}_{1} \) and \( {P}_{2} \) such that \( \bar{I} - L\left( {f,{P}_{1}}\right) < \frac{1}{2}\varepsilon \) and \( U\left( {f,{P}_{2}}\right) - \bar{I} < \frac{1}{2}\varepsilon \) . Now let \( P = {P}_{1} \cup {P}_{2} \) . We find\n\n\[ L\left( {f,{P}_{1}... | Yes |
Theorem 19.1 (Riemann-Lebesgue). A function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is Riemann integrable if and only if \( f \) is bounded and its set of discontinuities is a zero set. | Proof. First, we show that if \( f \) is Riemann integrable, then it is bounded and its discontinuity set \( D \) is a zero set.\n\nLet \( \varepsilon > 0 \) be given. Define\n\n\[ \operatorname{osc}{xf} = \mathop{\limsup }\limits_{{t \rightarrow x}}f\left( t\right) - \mathop{\liminf }\limits_{{t \rightarrow x}}f\left(... | No |
Corollary 19.4. Continuous implies Riemann integrable. | Proof. Bounded because the domain is closed, and the discontinuity set is empty. You can prove this directly. Compactness implies that a continuous \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is uniformly continuous: for each \( \varepsilon > 0 \) there exists a \( \delta > 0 \) such that \( \left... | Yes |
Corollary 19.6. \( f, g \in \mathcal{R} \Rightarrow f \cdot g \in \mathcal{R} \) | Proof. \( D\left( f\right) \cup D\left( g\right) \) contains \( D\left( {fg}\right) \) | No |
Corollary 19.7. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \left\lbrack {-M, M}\right\rbrack \) is Riemann integrable, and \( h : \left\lbrack {-M, M}\right\rbrack \rightarrow \mathbb{R} \) is continuous, then \( h \circ f \) is Riemann integrable. | Proof. \( D\left( {h \circ f}\right) \subseteq D\left( f\right) \) .because \( h \) is continuous. Furthermore, \( h \) is bounded because it is continuous. | Yes |
Proposition 20.2. If \( \left\lbrack {c, d}\right\rbrack \overset{h}{ \rightarrow }\left\lbrack {a, b}\right\rbrack \overset{f}{ \rightarrow }\mathbb{R} \) has \( h \) a diffeomorphism and \( f \in \mathcal{R} \), then \( f \circ h \) is Riemann integrable. | Proof. It suffices to show \( {h}^{-1} \) satisfies the Lipschitz condition. We know \( {\left( {h}^{-1}\right) }^{\prime } \) is bounded by some constant \( L \) for all \( x \) . Now consider \( x, y \) . By the Mean Value theorem there exists \( \theta \) between \( x \) and \( y \) with\n\n\[ \left| {{h}^{-1}\left(... | Yes |
Proposition 20.5. If \( g \) has a jump discontinuity, then \( g \) does NOT have an antiderivative. | Proof. This follows from the fact that the derivatives have the intermediate value property. | No |
Example 20.6. The fairly nice function \( f : \left\lbrack {0,2}\right\rbrack \rightarrow \mathbb{R} \) by \( f = {\chi }_{\left\lbrack 0,1\right\rbrack } + {\chi }_{(1,2\rbrack } \) has no antiderivative. Meanwhile, the much more unfortunate function | How do we find this antiderivative? We basically want \( G\left( x\right) = {\int }_{0}^{x}g\left( t\right) {dt} \) but it’s not obvious how to do this. The key is that for any \( \alpha > 0 \) we have\n\n\[ \n{\left. {t}^{2}\cos \left( 1/t\right) \right| }_{\alpha }^{x} = {\int }_{\alpha }^{x}\sin \frac{1}{t}{dt} + {\... | No |
Example 21.2. Consider \( f\left( t\right) = \frac{1}{{t}^{2}} \) . We have | \[ {\int }_{1}^{b}\frac{1}{{t}^{2}}{dt} = 1 - \frac{1}{b} \] so the improper integral \( {\int }_{1}^{\infty } = 1 \) . | Yes |
Proposition 21.13. If \( \sum {a}_{n} \) converges and \( {b}_{n} \uparrow b \), then \( \sum {a}_{n}{b}_{n} \) converges. | One way to see this is\n\n\[ \n{a}_{m}{b}_{m} + {a}_{m + 1}{b}_{m} + \cdots + {a}_{n - 1}{b}_{m} + {a}_{n}{b}_{m} \]\n\n\[ \n{a}_{m + 1}\left( {{b}_{m + 1} - {b}_{m}}\right) + \cdots + {a}_{n - 1}\left( {{b}_{m + 1} - {b}_{m}}\right) + {a}_{n}\left( {{b}_{m + 1} - {b}_{m}}\right) \]\n\n\[ \n\because + \text{:}\n\n\[ \n... | No |
Proposition 21.15 (Root Test). If \( \rho < 1 \) then \( \sum {a}_{k} \) converges absolutely; if \( \rho > 1 \) then \( \sum {a}_{k} \) diverges. The test is inconclusive if \( \rho = 1 \) . | Proof. If \( \mathop{\limsup }\limits_{{k \rightarrow \infty }}{\left| {a}_{k}\right| }^{1/k} = \rho \), then for \( k \) large we have \( \left| {a}_{k}\right| < {\rho }^{k} \) and comparison to the geometric series works. If \( \rho > 1 \), then there are infinitely many terms with magnitude greater than one, so it c... | Yes |
Theorem 22.4. Suppose \( {f}_{n} \rightrightarrows f \) and \( {x}_{0} \in \left\lbrack {a, b}\right\rbrack \) . If \( {f}_{n} \) is continuous at \( {x}_{0} \) for infinitely many \( n \), then so is \( f \) . | Proof. Let \( \varepsilon > 0 \) be given. We wish to show that there is a \( \delta > 0 \) such that \( \left| {x - {x}_{0}}\right| < \) \( \delta \Rightarrow \left| {f\left( x\right) - f\left( {x}_{0}\right) }\right| < \varepsilon . \n\nBy uniform convergence, there is some large \( N \) such that \( \left| {{f}_{N}\... | Yes |
Theorem 22.10. \( {C}_{b} \) is a complete metric space. | Proof. Let \( \left( {f}_{n}\right) \) be a Cauchy sequence. Evidently the sequence \( {\left( {f}_{n}\left( x\right) \right) }_{n \in \mathbb{N}} \) is Cauchy for each \( x \), because \( \left| {{f}_{n}\left( x\right) - {f}_{m}\left( x\right) }\right| \leq d\left( {{f}_{n},{f}_{m}}\right) \). Since this occurs in the... | Yes |
Corollary 22.11. \( {C}_{0} \) is complete. So is \( \mathcal{R} \) . | Proof. We claim they are closed subsets of the complete space \( {C}_{b} \) ; in fact \( {C}^{0} \subset \mathcal{R} \subset {C}_{b} \) . That \( {C}^{0} \) is closed follows from our earlier theorem. Hence, we only consider the second statement, showing that \( \mathcal{R} \) is closed. Suppose \( {f}_{n} \in \mathcal... | Yes |
Proposition 22.12. If \( {f}_{n} \rightrightarrows f \) and each \( {f}_{n} \) is Riemann integrable, then\n\n\[{\int }_{a}^{b}{f}_{n}\left( t\right) {dt} \rightarrow {\int }_{a}^{b}f\left( t\right) {dt}\] | Proof. Just write the inequalities\n\n\[ \left| {{\int }_{a}^{b}\left( {{f}_{n}\left( t\right) - f\left( t\right) }\right) {dt}}\right| \leq {\int }_{a}^{b}\left| {{f}_{n}\left( t\right) - f\left( t\right) }\right| {dt} \leq {\int }_{a}^{b}d\left( {{f}_{n}, f}\right) {dt} = \left( {b - a}\right) d\left( {{f}_{n}, f}\ri... | Yes |
Theorem 22.14. Suppose \( {f}_{n} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is continuously differentiable and \( {f}_{n} \rightrightarrows f \) . Suppose further \( {f}_{n}^{\prime } \rightrightarrows g \) for some function \( g \) . Then \( f \) is differentiable and \( {f}^{\prime } = g \) . | Proof. Uses the Fundamental Theorem of Calculus. We know that\n\n\[ \n{f}_{n}\left( x\right) = {f}_{n}\left( a\right) + {\int }_{a}^{x} + {\int }_{a}^{x}{f}_{n}^{\prime }\left( t\right) {dt}.\n\]\n\nConsidering the behavior as \( n \rightarrow \infty \), we obtain\n\n\[ \nf\left( x\right) = f\left( a\right) + {\int }_{... | Yes |
Theorem 22.16 (Weierstrass M-Test). Let \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{M}_{k} \) be a convergent sequence of nonnegative reals. If \( \left( {f}_{n}\right) \) is a sequence of functions with \( \begin{Vmatrix}{f}_{k}\end{Vmatrix} \leq {M}_{k} \) for all \( k \), then \( \sum {f}_{k} \) converges uniformly... | Proof. Just apply the Cauchy condition. | No |
If \( \left| x\right| < R \) then the series converges to a function \( f : \left( {-R, R}\right) \rightarrow \mathbb{R} \) . Furthermore, \( f \) is \( {C}^{1} \), and converges uniformly on closed intervals. Moreover, \( {f}^{\prime } \) is also given by a canonical power series | \[ \mathop{\sum }\limits_{{k = 1}}^{\infty }k{c}_{k}{x}^{k - 1} \] | No |
Theorem 23.3. Suppose \( {f}_{n} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) has the property that each \( {f}_{n} \) is differentiable and there exists \( L \) such that \( \left| {{f}_{n}^{\prime }\left( x\right) }\right| \leq L \) . Then \( \left( {f}_{n}\right) \) is equicontinuous. | Proof. Let \( \varepsilon > 0 \) be given. Choose \( \delta < \frac{\varepsilon }{L} \) . Now cite the Mean Value Theorem and win:\n\n\[ \left| {{f}_{n}\left( x\right) - {f}_{n}\left( y\right) }\right| = {f}^{\prime }\left( \theta \right) \left| {x - y}\right| \leq {L\delta } \leq \varepsilon . \] | Yes |
Theorem 23.6. A set \( \mathcal{F} \subseteq {C}^{0} \) is compact if and only if \( \mathcal{F} \) is closed, bounded, and equicontinuous. | Proof. One direction is trivial. The other direction is \( 2/3 \) trivial; we only need to prove \( \mathcal{F} \) is continuous given compactness. Consider an open subcover\n\n\[ {M}_{\varepsilon /3}\left( f\right) : f \in \mathcal{F} \]\n\nwhere \( {M}_{\varepsilon /3} \) is the \( \varepsilon /3 \) neighborhood. Hne... | Yes |
Lemma 25.2. Assume the two properties above. For all \( {p}_{1},{p}_{2} \in M \) and \( {c}_{1},{c}_{2} \in \mathbb{R} \), we can find a function \( f \in \mathcal{A} \) with \( {p}_{1} \mapsto {c}_{1} \) and \( {p}_{2} \mapsto {c}_{2} \) . | Proof. This is a matter of cooking up functions.\n\nLet \( {g}_{i} \in \mathcal{A} \) not vanish at \( {p}_{i} \) for \( i = 1,2 \) and let \( h \in \mathcal{A} \) separate \( {p}_{1} \) and \( {p}_{2} \) . For convenience, define \( g = {g}_{1}^{2} + {g}_{2}^{2} \) and note that \( g\left( {p}_{1}\right), g\left( {p}_... | Yes |
Lemma 25.5. If \( \mathcal{A} \) is a function algebra and \( f \) is in \( \overline{\mathcal{A}} \), then so is \( \left| f\right| \) . | Proof. There exists a large \( B \in \mathbb{R} \) such that \( f \) takes values in \( \left\lbrack {-B, B}\right\rbrack \) (because \( M \) is compact). The Weierstrass Approximation Theorem produces a polynomial\n\n\[ P\left( y\right) = {a}_{0} + {a}_{1}y + {a}_{2}{y}^{2} + \cdots + {a}_{n}{y}^{n} \]\n\nfor which \(... | Yes |
Theorem 26.2 (Banach Contraction Theorem). If \( M \) is a complete metric space, there exists a unique point \( p \in M \) such that \( {fp} = p \) ; that is, there exists a unique fixed point. | Proof. Choose any \( {x}_{0} \in M \) arbitrarily and let \( {x}_{n} = {f}^{n}\left( {x}_{0}\right) \) for every positive integer \( n \) . We claim \( \left( {x}_{n}\right) \) is Cauchy. After all, \( d\left( {{x}_{n},{x}_{n + 1}}\right) \leq {k}^{n}\left( {{x}_{0},{x}_{1}}\right) \) by a simple induction, and hence \... | Yes |
Theorem 26.4 (Picard). Let \( U \subset {\mathbb{R}}^{m} \) be open. Suppose that \( f : U \rightarrow {\mathbb{R}}^{m} \) satisfies a Lipschitz condition: there exists \( L \in \mathbb{R} \) with \( \left| {f\left( x\right) - f\left( y\right) }\right| \leq L\left| {x - y}\right| \) for every \( x, y \in U \) . Then th... | Proof. We can instead search for a continuous curve \( \gamma \left( t\right) \) (not necessarily differentiable) with\n\n\[ \gamma \left( t\right) = {x}_{0} + {\int }_{0}^{t}f\left( {\gamma \left( s\right) }\right) {ds}. \]\n\nfor all \( t \) . This is sufficient, since \( \gamma \left( 0\right) = 0 \) and \( {\gamma ... | Yes |
\( \mathbb{R} \smallsetminus \mathbb{Q} \) is thick in \( \mathbb{R} \) | For each rational number \( \alpha \) define\n\n\[ G\left( \alpha \right) = \mathbb{R} \smallsetminus \{ \alpha \}\]\n\nEach \( G\left( \alpha \right) \) is open and dense in \( \mathbb{R} \), and the intersection of these countably many \( G \)'s is precisely \( \mathbb{R} \smallsetminus \mathbb{Q} \). | Yes |
Theorem 27.6 (Baire’s Theorem). If \( M \) is a complete metric space and \( S \subset M \) is thick, then \( S \) is dense in \( M \) . | Proof. Given \( {p}_{0} \in M \) and \( {\varepsilon }_{0} > 0 \), we wish to show that \( S \cap {M}_{{\varepsilon }_{0}}\left( {p}_{0}\right) \neq \varnothing \) . Set \( S = \) \( { \cap }_{n = 1}^{\infty }{G}_{n} \), where each \( n \) is open and dense.\n\nBy density of \( {G}_{1} \), we can furnish an \( {\vareps... | Yes |
Theorem 27.7. There exists a sequence \( {R}_{n} \) of open dense sets in \( {C}^{0}\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \) such that \( \forall F \in \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{R}_{n}, F \) is nowhere differentiable. | Proof. This proof will use the Weierstrass Approximation Theorem.\n\nDefine\n\n\[ R\left( n\right) \in \left\{ {f \in {C}^{0} : \forall x \in \left\lbrack {a, b - \frac{1}{n}}\right\rbrack \exists h > 0\text{ with }\left| \frac{f\left( {x + h}\right) - f\left( x\right) }{h}\right| > n}\right\} .\n\]\n\nThe point is tha... | Yes |
Theorem 1.1 (Helly, Hahn-Banach analytic form). Let \( p : E \rightarrow \mathbb{R} \) be a function satisfying \( {}^{1} \n\n(1)\n\n\[ \np\left( {\lambda x}\right) = {\lambda p}\left( x\right) \;\forall x \in E\;\text{ and }\;\forall \lambda > 0, \]\n\n(2)\n\n\[ \np\left( {x + y}\right) \leq p\left( x\right) + p\left(... | The proof of Theorem 1.1 depends on Zorn's lemma, which is a celebrated and very useful property of ordered sets. Before stating Zorn's lemma we must clarify some notions. Let \( P \) be a set with a (partial) order relation \( \leq \) . We say that a subset \( Q \subset P \) is totally ordered if for any pair \( \left... | Yes |
Proposition 1.5. The hyperplane \( H = \left\lbrack {f = \alpha }\right\rbrack \) is closed if and only if \( f \) is continuous. | Proof. It is clear that if \( f \) is continuous then \( H \) is closed. Conversely, let us assume that \( H \) is closed. The complement \( {H}^{c} \) of \( H \) is open and nonempty (since \( f \) does not vanish identically). Let \( {x}_{0} \in {H}^{c} \), so that \( f\left( {x}_{0}\right) \neq \alpha \), for exampl... | Yes |
Lemma 1.2. Let \( C \subset E \) be an open convex set with \( 0 \in C \) . For every \( x \in E \) set\n\n\[ p\left( x\right) = \inf \left\{ {\alpha > 0;{\alpha }^{-1}x \in C}\right\} \]\n\n( \( p \) is called the gauge of \( C \) or the Minkowski functional of \( C \) ).\n\nThen \( p \) satisfies (1),(2), and the fol... | Proof of Lemma 1.2. It is obvious that (1) holds.\n\nProof of (9). Let \( r > 0 \) be such that \( B\left( {0, r}\right) \subset C \) ; we clearly have\n\n\[ p\left( x\right) \leq \frac{1}{r}\parallel x\parallel \;\forall x \in E. \]\n\nProof of (10). First, suppose that \( x \in C \) ; since \( C \) is open, it follow... | Yes |
Lemma 1.3. Let \( C \subset E \) be a nonempty open convex set and let \( {x}_{0} \in E \) with \( {x}_{0} \notin C \) . Then there exists \( f \in {E}^{ \star } \) such that \( f\left( x\right) < f\left( {x}_{0}\right) \;\forall x \in C \) . In particular, the hyperplane \( \left\lbrack {f = f\left( {x}_{0}\right) }\r... | Proof of Lemma 1.3. After a translation we may always assume that \( 0 \in C \) . We may thus introduce the gauge \( p \) of \( C \) (see Lemma 1.2). Consider the linear subspace \( G = \mathbb{R}{x}_{0} \) and the linear functional \( g : G \rightarrow \mathbb{R} \) defined by\n\n\[ g\left( {t{x}_{0}}\right) = t,\;t \... | Yes |
Proposition 1.9. Let \( M \subset E \) be a linear subspace. Then\n\n\[ \left( {{M}^{ \bot }{)}^{ \bot } = \bar{M}}\right) \text{.} \]\n\nLet \( N \subset {E}^{ \star } \) be a linear subspace. Then\n\n\[ {\left( {N}^{ \bot }\right) }^{ \bot } \supset \bar{N} \] | Proof. It is clear that \( M \subset {\left( {M}^{ \bot }\right) }^{ \bot } \), and since \( {\left( {M}^{ \bot }\right) }^{ \bot } \) is closed we have \( \bar{M} \subset \) \( {\left( {M}^{ \bot }\right) }^{ \bot } \) . Conversely, let us show that \( {\left( {M}^{ \bot }\right) }^{ \bot } \subset \bar{M} \) . Suppos... | Yes |
Proposition 1.10. Assume that \( \varphi : E \rightarrow ( - \infty , + \infty \rbrack \) is convex l.s.c. and \( \varphi ≢ + \infty \) . Then \( {\varphi }^{ \star } ≢ + \infty \), and in particular, \( \varphi \) is bounded below by an affine continuous function. | Proof. Let \( {x}_{0} \in D\left( \varphi \right) \) and let \( {\lambda }_{0} < \varphi \left( {x}_{0}\right) \) . We apply Theorem 1.7 (Hahn-Banach, second geometric form) in the space \( E \times \mathbb{R} \) with \( A = \operatorname{epi}\varphi \) and \( B = \left\{ \left\lbrack {{x}_{0},{\lambda }_{0}}\right\rbr... | Yes |
Consider \( \varphi \left( x\right) = \parallel x\parallel \) . It is easy to check that | \[ {\varphi }^{ \star }\left( f\right) = \left\{ \begin{array}{ll} 0 & \text{ if }\parallel f\parallel \leq 1 \\ + \infty & \text{ if }\parallel f\parallel > 1 \end{array}\right. \] It follows that \[ {\varphi }^{\star \star }\left( x\right) = \mathop{\sup }\limits_{\substack{{f \in {E}^{ \star }} \\ {\parallel f\paral... | No |
Given a nonempty set \( K \subset E \), we set \[ {I}_{K}\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x \in K \\ + \infty & \text{ if }x \notin K \end{array}\right. \] The function \( {I}_{K} \) is called the indicator function of \( K \) (and should not be confused with the characteristic function, \( {\... | It is easy to see that if \( K = M \) is a linear subspace then \( {\left( {I}_{M}\right) }^{ \star } = {I}_{{M}^{ \bot }} \) and \( {\left( {I}_{M}\right) }^{\star \star } = \) \( {I}_{{\left( {M}^{ \bot }\right) }^{ \bot }} \) . Assuming that \( M \) is a closed linear space and writing that \( {\left( {I}_{M}\right)... | Yes |
Lemma 1.4. Let \( C \subset E \) be a convex set, then Int \( C \) is convex. \( {}^{7} \) If, in addition, Int \( C \neq \varnothing \), then | For the proof of Lemma 1.4, see, e.g., Exercise 1.7. | No |
Let \( K \) be a nonempty convex set. We claim that for every \( {x}_{0} \in E \) we have\n\n\[ \operatorname{dist}\left( {{x}_{0}, K}\right) = \mathop{\inf }\limits_{{x \in K}}\begin{Vmatrix}{x - {x}_{0}}\end{Vmatrix} = \mathop{\max }\limits_{\substack{{f \in {E}^{ \star }} \\ {\parallel f\parallel \leq 1} }}\left\{ {... | Indeed, we have\n\n\[ \mathop{\inf }\limits_{{x \in K}}\begin{Vmatrix}{x - {x}_{0}}\end{Vmatrix} = \mathop{\inf }\limits_{{x \in E}}\{ \varphi \left( x\right) + \psi \left( x\right) \} \]\n\nwith \( \varphi \left( x\right) = \begin{Vmatrix}{x - {x}_{0}}\end{Vmatrix} \) and \( \psi \left( x\right) = {I}_{K}\left( x\righ... | No |
Let \( \varphi : E \rightarrow \mathbb{R} \) be convex and continuous and let \( M \subset E \) be a linear subspace. Then we have\n\n\[ \mathop{\inf }\limits_{{x \in M}}\varphi \left( x\right) = - \mathop{\min }\limits_{{f \in {M}^{ \bot }}}{\varphi }^{ \star }\left( f\right) \] | It suffices to apply Theorem 1.12 with \( \psi = {I}_{M} \) . | No |
Corollary 2.3. Let \( E \) and \( F \) be two Banach spaces. Let \( \left( {T}_{n}\right) \) be a sequence of continuous linear operators from \( E \) into \( F \) such that for every \( x \in E,{T}_{n}x \) converges (as \( n \rightarrow \infty \) ) to a limit denoted by \( {Tx} \) . Then we have\n\n(a) \( \mathop{\sup... | Proof. (a) follows directly from Theorem 2.2, and thus there exists a constant \( c \) such that\n\n\[ \begin{Vmatrix}{{T}_{n}x}\end{Vmatrix} \leq c\parallel x\parallel \;\forall n,\;\forall x \in E. \]\n\nAt the limit we find\n\n\[ \parallel {Tx}\parallel \leq c\parallel x\parallel \;\forall x \in E. \]\n\nSince \( T ... | Yes |
Corollary 2.5. Let \( G \) be a Banach space and let \( {B}^{ \star } \) be a subset of \( {G}^{ \star } \) . Assume that\n\nfor every \( x \in G \) the set \( \left\langle {{B}^{ \star }, x}\right\rangle = \left\{ {\langle f, x\rangle ;f \in {B}^{ \star }}\right\} \) is bounded (in \( \mathbb{R} \) ).\n\nThen\n\n(6)\n... | Proof. Use Theorem 2.2 with \( E = G, F = \mathbb{R} \), and \( I = {B}^{ \star } \) . For every \( b \in {B}^{ \star } \) set\n\n\[ \n{T}_{b}\left( x\right) = \langle b, x\rangle \;\left( {x \in G = E}\right) . \n\]\n\nWe find that there exists a constant \( c \) such that\n\n\[ \n\left| {\langle b, x\rangle }\right| ... | Yes |
Corollary 2.8. Let \( E \) be a vector space provided with two norms, \( {\begin{Vmatrix}\end{Vmatrix}}_{1} \) and \( \parallel {\parallel }_{2} \) . Assume that \( E \) is a Banach space for both norms and that there exists a constant \( C \geq 0 \) such that\n\n\[ \parallel x{\parallel }_{2} \leq C\parallel x{\parall... | Proof of Corollary 2.8. Apply Corollary 2.7 with\n\n\[ E = \left( {E,\parallel {\parallel }_{1}}\right), F = \left( {E,\parallel {\parallel }_{2}}\right) ,\text{ and }T = I. \] | Yes |
Proposition 2.14. Let \( G \) and \( L \) be two closed subspaces in \( E \). Then\n\n(16)\n\n\[ G \cap L = {\left( {G}^{ \bot } + {L}^{ \bot }\right) }^{ \bot } \] | Proof of (16). It is clear that \( G \cap L \subset {\left( {G}^{ \bot } + {L}^{ \bot }\right) }^{ \bot } \) ; indeed, if \( x \in G \cap L \) and \( f \in {G}^{ \bot } + {L}^{ \bot } \) then \( \langle f, x\rangle = 0 \) . Conversely, we have \( {G}^{ \bot } \subset {G}^{ \bot } + {L}^{ \bot } \) and thus \( {\left( {... | Yes |
Corollary 2.15. Let \( G \) and \( L \) be two closed subspaces in \( E \) . Then\n\n(18)\n\n\[{\left( G \cap L\right) }^{ \bot } \supset \overline{{G}^{ \bot } + {L}^{ \bot }}\]\n\n(19)\n\n\[{\left( {G}^{ \bot } \cap {L}^{ \bot }\right) }^{ \bot } = \overline{G + L}\] | Proof. Use Propositions 1.9 and 2.14. | No |
Proposition 2.17. Let \( A : D\left( A\right) \subset E \rightarrow F \) be a densely defined unbounded linear operator. Then \( {A}^{ \star } \) is closed, i.e., \( G\left( {A}^{ \star }\right) \) is closed in \( {F}^{ \star } \times {E}^{ \star } \) . | Proof. Let \( {v}_{n} \in D\left( {A}^{ \star }\right) \) be such that \( {v}_{n} \rightarrow v \) in \( {F}^{ \star } \) and \( {A}^{ \star }{v}_{n} \rightarrow f \) in \( {E}^{ \star } \) . One has to check that (a) \( v \in D\left( {A}^{ \star }\right) \) and (b) \( {A}^{ \star }v = f \) .\n\nWe have\n\n\[ \left\lan... | Yes |
Corollary 2.18. Let \( A : D\\left( A\\right) \\subset E \\rightarrow F \) be an unbounded linear operator that is densely defined and closed. Then\n\n(i)\n\n\[ N\\left( A\\right) = R{\\left( {A}^{ \\star }\\right) }^{ \\bot },\]\n\n(ii)\n\n\[ N\\left( {A}^{ \\star }\\right) = R{\\left( A\\right) }^{ \\bot }\n\n(iii)\n... | Proof. Note that (iii) and (iv) follow directly from (i) and (ii) combined with Proposition 1.9. There is a simple and direct proof of (i) and (ii) (see Exercise 2.18). However, it is instructive to relate these facts to Proposition 2.14 by the following device. Consider the space \( X = E \\times F \), so that \( {X}^... | No |
Problem 1. Construct a topology on \( X \) that makes all the maps \( {\left( {\varphi }_{i}\right) }_{i \in I} \) continuous. If possible, find a topology \( \mathcal{T} \) that is the most economical in the sense that it has the fewest open sets. | Note that if we equip \( X \) with the discrete topology (i.e., every subset of \( X \) is open), then every map \( {\varphi }_{i} \) is continuous; of course, this topology is far from being the \ | No |
Given a set \( X \) and a family \( {\left( {U}_{\lambda }\right) }_{\lambda \in \Lambda } \) of subsets in \( X \), construct the cheapest topology \( \mathcal{T} \) on \( X \) in which \( {U}_{\lambda } \) is open for all \( \lambda \in \Lambda \). | In other words, we must find the cheapest family \( \mathcal{F} \) of subsets of \( X \) that is stable \( {}^{1} \) by \( { \cap }_{\text{finite }} \) and \( { \cup }_{\text{arbitrary }} \) and with the property that \( {U}_{\lambda } \in \mathcal{F} \) for every \( \lambda \in \Lambda \) . The construction goes as fo... | No |
Lemma 3.1. The family \( \mathcal{F} \) is stable under \( { \cap }_{\text{finite }} \) . | The proof of Lemma 3.1 - a delightful exercise in set theory - is left to the reader; see e.g., G. Folland [2]. It is now obvious that the above construction gives the cheapest topology with the required property. | No |
Proposition 3.3. The weak topology \( \\sigma \\left( {E,{E}^{ \\star }}\\right) \) is Hausdorff. | Proof. Given \( {x}_{1},{x}_{2} \\in E \) with \( {x}_{1} \\neq {x}_{2} \) we have to find two open sets \( {O}_{1} \) and \( {O}_{2} \) for the weak topology \( \\sigma \\left( {E,{E}^{ \\star }}\\right) \) such that \( {x}_{1} \\in {O}_{1},{x}_{2} \\in {O}_{2} \), and \( {O}_{1} \\cap {O}_{2} = \\varnothing \) . By H... | Yes |
The unit sphere \( S = \{ x \in E;\parallel x\parallel = 1\} \), with \( E \) infinite-dimensional, is never closed in the weak topology \( \sigma \left( {E,{E}^{ \star }}\right) \) . More precisely, we have\n\n\[{\bar{S}}^{\sigma \left( {E,{E}^{ \star }}\right) } = {B}_{E}\]\n\nwhere \( {\bar{S}}^{\sigma \left( {E,{E}... | First let us check that every \( {x}_{0} \in E \) with \( \begin{Vmatrix}{x}_{0}\end{Vmatrix} < 1 \) belongs to \( {\bar{S}}^{\sigma \left( {E,{E}^{ \star }}\right) } \) . Indeed, let \( V \) be a neighborhood of \( {x}_{0} \) in \( \sigma \left( {E,{E}^{ \star }}\right) \) . We have to prove that \( V \cap S \neq \var... | Yes |
The unit ball \( U = \{ x \in E;\parallel x\parallel < 1\} \), with \( E \) infinite-dimensional, is never open in the weak topology \( \sigma \left( {E,{E}^{ \star }}\right) \) . | Suppose, by contradiction, that \( U \) is weakly open. Then its complement \( {U}^{c} = \{ x \in E;\parallel x\parallel \geq 1\} \) is weakly closed. It follows that \( S = {B}_{E} \cap {U}^{c} \) is also weakly closed; this contradicts Example 1. | Yes |
Corollary 3.8 (Mazur). Assume \( \left( {x}_{n}\right) \) converges weakly to \( x \) . Then there exists a sequence \( \left( {y}_{n}\right) \) made up of convex combinations of the \( {x}_{n} \) ’s that converges strongly to \( x \) . | Proof. Let \( C = \operatorname{conv}\left( {{ \cup }_{p = 1}^{\infty }\left\{ {x}_{p}\right\} }\right) \) denote the convex hull of the \( {x}_{n} \) ’s. Since \( x \) belongs to the weak closure of \( { \cup }_{p = 1}^{\infty }\left\{ {x}_{p}\right\} \) it belongs a fortiori to the weak closure of \( C \) . By Theore... | Yes |
Theorem 3.10. Let \( E \) and \( F \) be two Banach spaces and let \( T \) be a linear operator from \( E \) into \( F \) . Assume that \( T \) is continuous in the strong topologies. Then \( T \) is continuous from \( E \) weak \( \sigma \left( {E,{E}^{ \star }}\right) \) into \( F \) weak \( \sigma \left( {F,{F}^{ \s... | Proof. In view of Proposition 3.2 it suffices to check that for every \( f \in {F}^{ \star } \) the map \( x \mapsto \langle f,{Tx}\rangle \) is continuous from \( E \) weak \( \sigma \left( {E,{E}^{ \star }}\right) \) into \( \mathbb{R} \) . But the map \( x \mapsto \langle f,{Tx}\rangle \) is a continuous linear func... | Yes |
Proposition 3.12. Let \( {f}_{0} \in {E}^{ \star } \) ; given a finite set \( \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{k}}\right\} \) in \( E \) and \( \varepsilon > 0 \) , consider\n\n\[ \nV = V\left( {{x}_{1},{x}_{2},\ldots ,{x}_{k};\varepsilon }\right) = \left\{ {f \in {E}^{ \star };\left| \left\langle {f - {f}_{0},{x}... | Proof. Same as the proof of Proposition 3.4. | No |
Lemma 3.2. Let \( X \) be a vector space and let \( \varphi ,{\varphi }_{1},{\varphi }_{2},\ldots ,{\varphi }_{k} \) be \( \left( {k + 1}\right) \) linear functionals on \( X \) such that\n\n\[ \left\lbrack {{\varphi }_{i}\left( v\right) = 0\;\forall i = 1,2,\ldots, k}\right\rbrack \Rightarrow \left\lbrack {\varphi \le... | Proof of Lemma 3.2. Consider the map \( F : X \rightarrow {\mathbb{R}}^{k + 1} \) defined by\n\n\[ F\left( u\right) = \left\lbrack {\varphi \left( u\right) ,{\varphi }_{1}\left( u\right) ,{\varphi }_{2}\left( u\right) ,\ldots ,{\varphi }_{k}\left( u\right) }\right\rbrack .\n\]\n\nIt follows from assumption (2) that \( ... | Yes |
Lemma 3.3 (Helly). Let \( E \) be a Banach space. Let \( {f}_{1},{f}_{2},\ldots ,{f}_{k} \) be given in \( {E}^{ \star } \) and let \( {\gamma }_{1},{\gamma }_{2},\ldots ,{\gamma }_{k} \) be given in \( \mathbb{R} \) . The following properties are equivalent:\n\n(i) \( \forall \varepsilon > 0\exists {x}_{\varepsilon } ... | Proof. (i) \( \Rightarrow \) (ii). Fix \( {\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{k} \) in \( \mathbb{R} \) and let \( S = \mathop{\sum }\limits_{{i = 1}}^{k}\left| {\beta }_{i}\right| \) . It follows from (i)\n\nthat\n\[ \left| {\mathop{\sum }\limits_{{i = 1}}^{k}{\beta }_{i}\left\langle {{f}_{i},{x}_{\varepsilon ... | Yes |
Lemma 3.4 (Goldstine). Let \( E \) be any Banach space. Then \( J\left( {B}_{E}\right) \) is dense in \( {B}_{{E}^{\star \star }} \) with respect to the topology \( \sigma \left( {{E}^{\star \star },{E}^{ \star }}\right) \), and consequently \( J\left( E\right) \) is dense in \( {E}^{\star \star } \) in the topology \(... | Proof. Let \( \xi \in {B}_{{E}^{\star \star }} \) and let \( V \) be a neighborhood of \( \xi \) for the topology \( \sigma \left( {{E}^{\star \star },{E}^{ \star }}\right) \). We must prove that \( V \cap J\left( {B}_{E}\right) \neq \varnothing \). As usual, we may assume that \( V \) is of the form\n\n\[ V = \left\{ ... | Yes |
Corollary 3.21. A Banach space \( E \) is reflexive if and only if its dual space \( {E}^{ \star } \) is reflexive. | Proof. E reflexive \( \Rightarrow {E}^{ \star } \) reflexive. The idea of the proof is simple, since, roughly speaking, we have that \( {E}^{\star \star } = E \Rightarrow {E}^{\star \star \star } = {E}^{ \star } \) . More precisely, let \( J \) be the canonical isomorphism from \( E \) into \( {E}^{\star \star } \) . L... | Yes |
Theorem 3.24. Let \( E \) and \( F \) be two reflexive Banach spaces. Let \( A : D\left( A\right) \subset E \rightarrow \) \( F \) be an unbounded linear operator that is densely defined and closed. Then \( D\left( {A}^{ \star }\right) \) is dense in \( {F}^{ \star } \) . Thus \( {A}^{\star \star } \) is well defined \... | Proof.\n\n1. \( D\left( {A}^{ \star }\right) \) is dense in \( {F}^{ \star } \) . Let \( \varphi \) be a continuous linear functional on \( {F}^{ \star } \) that vanishes on \( D\left( {A}^{ \star }\right) \) . In view of Corollary 1.8 it suffices to prove that \( \varphi \equiv 0 \) on \( {F}^{ \star } \) . Since \( F... | Yes |
Proposition 3.25. Let \( E \) be a separable metric space and let \( F \subset E \) be any subset. Then \( F \) is also separable. | Proof. Let \( \left( {u}_{n}\right) \) be a countable dense subset of \( E \) . Let \( \left( {r}_{m}\right) \) be any sequence of positive numbers such that \( {r}_{m} \rightarrow 0 \) . Choose any point \( {a}_{m, n} \in B\left( {{u}_{n},{r}_{m}}\right) \cap F \) whenever this set is nonempty. The set \( \left( {a}_{... | Yes |
Theorem 3.26. Let \( E \) be a Banach space such that \( {E}^{ \star } \) is separable. Then \( E \) is separable. | Proof. Let \( {\left( {f}_{n}\right) }_{n \geq 1} \) be countable and dense in \( {E}^{ \star } \) . Since\n\n\[\n\begin{Vmatrix}{f}_{n}\end{Vmatrix} = \mathop{\sup }\limits_{\substack{{x \in E} \\ {\parallel x\parallel \leq 1} }}\left\langle {{f}_{n}, x}\right\rangle\n\]\n\nwe can find some \( {x}_{n} \in E \) such th... | Yes |
Corollary 3.27. Let \( E \) be a Banach space. | Proof. We already know (Corollary 3.21 and Theorem 3.26) that \[ \left\lbrack {{E}^{ \star }\text{reflexive and separable}}\right\rbrack \Rightarrow \left\lbrack {E\text{reflexive and separable}}\right\rbrack \text{.} \] Conversely, if \( E \) is reflexive and separable, so is \( {E}^{\star \star } = J\left( E\right) \... | Yes |
Corollary 3.30. Let \( E \) be a separable Banach space and let \( \left( {f}_{n}\right) \) be a bounded sequence in \( {E}^{ \star } \) . Then there exists a subsequence \( \left( {f}_{{n}_{k}}\right) \) that converges in the weak* topology \( \sigma \left( {{E}^{ \star }, E}\right) \) . | Proof. Without loss of generality we may assume that \( \begin{Vmatrix}{f}_{n}\end{Vmatrix} \leq 1 \) for all \( n \) . The set \( {B}_{{E}^{ \star }} \) is compact and metrizable for the topology \( \sigma \left( {{E}^{ \star }, E}\right) \) (by Theorems 3.16 and 3.28). The conclusion follows. | No |
Let \( E = {\mathbb{R}}^{2} \) . The norm \( \parallel x{\parallel }_{2} = {\left\lbrack {\left| {x}_{1}\right| }^{2} + {\left| {x}_{2}\right| }^{2}\right\rbrack }^{1/2} \) is uniformly convex, while the norm \( \parallel x{\parallel }_{1} = \left| {x}_{1}\right| + \left| {x}_{2}\right| \) and the norm \( \parallel x{\... | This can be easily seen by staring at the unit balls, as shown in Figure 3. | No |
Theorem 3.31 (Milman-Pettis). Every uniformly convex Banach space is reflexive. | Proof. Let \( \xi \in {E}^{\star \star } \) with \( \parallel \xi \parallel = 1 \) . We have to show that \( \xi \in J\left( {B}_{E}\right) \) . Since \( J\left( {B}_{E}\right) \) is closed in \( {E}^{\star \star } \) in the strong topology, it suffices to prove that\n\n(7)\n\n\[ \forall \varepsilon > 0\;\exists x \in ... | Yes |
Proposition 3.32. Assume that \( E \) is a uniformly convex Banach space. Let \( \left( {x}_{n}\right) \) be a sequence in \( E \) such that \( {x}_{n} \rightharpoonup x \) weakly \( \sigma \left( {E,{E}^{ \star }}\right) \) and\n\n\[ \lim \sup \begin{Vmatrix}{x}_{n}\end{Vmatrix} \leq \parallel x\parallel \]\n\nThen \(... | Proof. We may always assume that \( x \neq 0 \) (otherwise the conclusion is obvious). Set\n\n\[ {\lambda }_{n} = \max \left( {\begin{Vmatrix}{x}_{n}\end{Vmatrix},\parallel x\parallel }\right) ,{y}_{n} = {\lambda }_{n}^{-1}{x}_{n}\text{, and }y = \parallel x{\parallel }^{-1}x, \]\n\nso that \( {\lambda }_{n} \rightarro... | Yes |
Theorem 4.3 (density). The space \( {C}_{c}\left( {\mathbb{R}}^{N}\right) \) is dense in \( {L}^{1}\left( {\mathbb{R}}^{N}\right) \) ; i.e., | \[ \forall f \in {L}^{1}\left( {\mathbb{R}}^{N}\right) \;\forall \varepsilon > 0\;\exists {f}_{1} \in {C}_{c}\left( {\mathbb{R}}^{N}\right) \text{ such that }{\begin{Vmatrix}f - {f}_{1}\end{Vmatrix}}_{1} \leq \varepsilon . \] | Yes |
Theorem 4.5 (Fubini). Assume that \( F \in {L}^{1}\left( {{\Omega }_{1} \times {\Omega }_{2}}\right) \) . Then for a.e. \( x \in {\Omega }_{1} \) , \( F\left( {x, y}\right) \in {L}_{y}^{1}\left( {\Omega }_{2}\right) \) and \( {\int }_{{\Omega }_{2}}F\left( {x, y}\right) d{\mu }_{2} \in {L}_{x}^{1}\left( {\Omega }_{1}\r... | \[ {\int }_{{\Omega }_{1}}d{\mu }_{1}{\int }_{{\Omega }_{2}}F\left( {x, y}\right) d{\mu }_{2} = {\int }_{{\Omega }_{2}}d{\mu }_{2}{\int }_{{\Omega }_{1}}F\left( {x, y}\right) d{\mu }_{1} = {\iint }_{{\Omega }_{1} \times {\Omega }_{2}}F\left( {x, y}\right) d{\mu }_{1}d{\mu }_{2}. \] | Yes |
Theorem 4.7. \( {L}^{p} \) is a vector space and \( \parallel {\parallel }_{p} \) is a norm for any \( p,1 \leq p \leq \infty \) . | Proof. The cases \( p = 1 \) and \( p = \infty \) are clear. Therefore we assume \( 1 < p < \infty \) and let \( f, g \in {L}^{p} \) . We have\n\n\[ \n{\left| f\left( x\right) + g\left( x\right) \right| }^{p} \leq {\left( \left| f\left( x\right) \right| + \left| g\left( x\right) \right| \right) }^{p} \leq {2}^{p}\left(... | Yes |
Theorem 4.9. Let \( \left( {f}_{n}\right) \) be a sequence in \( {L}^{p} \) and let \( f \in {L}^{p} \) be such that \( {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{p} \) \( \rightarrow 0 \) . Then, there exist a subsequence \( \left( {f}_{{n}_{k}}\right) \) and a function \( h \in {L}^{p} \) such that (a) \( {f}_{{n}_{k... | Proof. The conclusion is obvious when \( p = \infty \) . Thus we assume \( 1 \leq p < \infty \) . Since \( \left( {f}_{n}\right) \) is a Cauchy sequence we may go back to the proof of Theorem 4.8 and consider a subsequence \( \left( {f}_{{n}_{k}}\right) \) -denoted by \( \left( {f}_{k}\right) \) -satisfying (6), such t... | Yes |
Theorem 4.12. The space \( {C}_{c}\left( {\mathbb{R}}^{N}\right) \) is dense in \( {L}^{p}\left( {\mathbb{R}}^{N}\right) \) for any \( p,1 \leq p < \infty \) . | Proof. First, we claim that given \( f \in {L}^{p}\left( {\mathbb{R}}^{N}\right) \) and \( \varepsilon > 0 \) there exist a function \( g \in {L}^{\infty }\left( {\mathbb{R}}^{N}\right) \) and a compact set \( K \) in \( {\mathbb{R}}^{N} \) such that \( g = 0 \) outside \( K \) and\n\n(11)\n\n\[ \parallel f - g{\parall... | Yes |
Theorem 4.13. Assume that \( \Omega \) is a separable measure space. Then \( {L}^{p}\left( \Omega \right) \) is separable for any \( p,1 \leq p < \infty \) . | Proof of Theorem 4.13 when \( \Omega = {\mathbb{R}}^{N} \) . Let \( \mathcal{R} \) denote the countable family of sets in \( {\mathbb{R}}^{N} \) of the form \( R = \mathop{\prod }\limits_{{k = 1}}^{N}\left( {{a}_{k},{b}_{k}}\right) \) with \( {a}_{k},{b}_{k} \in \mathbb{Q} \) . Let \( \mathcal{E} \) denote the vector s... | No |
Lemma 4.2. Let \( E \) be a Banach space. Assume that there exists a family \( {\left( {O}_{i}\right) }_{i \in I} \) such that\n\n(i) for each \( i \in I,{O}_{i} \) is a nonempty open subset of \( E \) ,\n\n(ii) \( {O}_{i} \cap {O}_{j} = \varnothing \) if \( i \neq j \) ,\n\n(iii) I is uncountable.\n\nThen \( E \) is n... | Proof of Lemma 4.2. Suppose, by contradiction, that \( E \) is separable. Let \( {\left( {u}_{n}\right) }_{n \in \mathbb{N}} \) denote a dense countable set in \( E \) . For each \( i \in I \), the set \( {O}_{i} \cap {\left( {u}_{n}\right) }_{n \in \mathbb{N}} \neq \varnothing \) and we may choose \( n\left( i\right) ... | Yes |
Proposition 4.16. Let \( f \in {L}^{1}\left( {\mathbb{R}}^{N}\right), g \in {L}^{p}\left( {\mathbb{R}}^{N}\right) \) and \( h \in {L}^{{p}^{\prime }}\left( {\mathbb{R}}^{N}\right) \) . Then we have\n\n\[ \n{\int }_{{\mathbb{R}}^{N}}\left( {f \star g}\right) h = {\int }_{{\mathbb{R}}^{N}}g\left( {\breve{f} \star h}\righ... | Proof. The function \( F\left( {x, y}\right) = f\left( {x - y}\right) g\left( y\right) h\left( x\right) \) belongs to \( {L}^{1}\left( {{\mathbb{R}}^{N} \times {\mathbb{R}}^{N}}\right) \) since\n\n\[ \n\int \left| {h\left( x\right) }\right| {dx}\int \left| {f\left( {x - y}\right) }\right| \left| {g\left( y\right) }\rig... | Yes |
Proposition 4.17 (and definition of the support). Let \( f : {\mathbb{R}}^{N} \rightarrow \mathbb{R} \) be any function. Consider the family \( {\left( {\omega }_{i}\right) }_{i \in I} \) of all open sets on \( {\mathbb{R}}^{N} \) such that for each \( i \in I, f = 0 \) a.e. on \( {\omega }_{i} \) . Set \( \omega = \ma... | Proof of Proposition 4.17. Since the set \( I \) need not be countable it is not clear that \( f = 0 \) a.e. on \( \omega \) . However we may recover the countable case as follows. There is a countable family \( \left( {O}_{n}\right) \) of open sets in \( {\mathbb{R}}^{N} \) such that every open set on \( {\mathbb{R}}^... | Yes |
Proposition 4.19. Let \( f \in {C}_{c}\left( {\mathbb{R}}^{N}\right) \) and \( g \in {L}_{\mathrm{{loc}}}^{1}\left( {\mathbb{R}}^{N}\right) \) . Then \( \left( {f \star g}\right) \left( x\right) \) is well defined for every \( x \in {\mathbb{R}}^{N} \), and, moreover, \( \left( {f \star g}\right) \in C\left( {\mathbb{R... | Proof. Note that for every \( x \in {\mathbb{R}}^{N} \) the function \( y \mapsto f\left( {x - y}\right) g\left( y\right) \) is integrable on \( {\mathbb{R}}^{N} \) and therefore \( \left( {f \star g}\right) \left( x\right) \) is defined for every \( x \in {\mathbb{R}}^{N} \). Let \( {x}_{n} \rightarrow x \) and let \(... | Yes |
Proposition 4.21. Assume \( f \in C\left( {\mathbb{R}}^{N}\right) \) . Then \( \left( {{\rho }_{n} \star f}\right) \underset{n \rightarrow \infty }{ \rightarrow }f \) uniformly on compact sets of \( {\mathbb{R}}^{N} \) . | Proof. \( {}^{4} \) Let \( K \subset {\mathbb{R}}^{N} \) be a fixed compact set. Given \( \varepsilon > 0 \) there exists \( \delta > 0 \) (depending on \( K \) and \( \varepsilon \) ) such that\n\n\[ \left| {f\left( {x - y}\right) - f\left( x\right) }\right| < \varepsilon \;\forall x \in K,\;\forall y \in B\left( {0,\... | Yes |
Corollary 4.24. Let \( \Omega \subset {\mathbb{R}}^{N} \) be an open set and let \( u \in {L}_{\text{loc }}^{1}\left( \Omega \right) \) be such that\n\n\[ \int {uf} = 0\;\forall f \in {C}_{c}^{\infty }\left( \Omega \right) \]\n\nThen \( u = 0 \) a.e. on \( \Omega \) . | Proof. Let \( g \in {L}^{\infty }\left( {\mathbb{R}}^{N}\right) \) be a function such that supp \( g \) is a compact set contained in \( \Omega \) . Set \( {g}_{n} = {\rho }_{n} \star g \), so that \( {g}_{n} \in {C}_{c}^{\infty }\left( \Omega \right) \) provided \( n \) is large enough. Therefore we have\n\n(19)\n\n\[... | Yes |
Corollary 4.27. Let \( \mathcal{F} \) be a bounded set in \( {L}^{p}\left( {\mathbb{R}}^{N}\right) \) with \( 1 \leq p < \infty \) . Assume (22) and also\n\n(27)\n\n\[ \left\{ \begin{array}{l} \forall \varepsilon > 0\exists \Omega \subset {\mathbb{R}}^{N},\text{ bounded, measurable such that } \\ \parallel f{\parallel ... | Proof. Given \( \varepsilon > 0 \) we fix \( \Omega \subset {\mathbb{R}}^{N} \) bounded measurable such that (27) holds. By Theorem 4.26 we know that \( {\mathcal{F}}_{\mid \Omega } \) has compact closure in \( {L}^{p}\left( \Omega \right) \) . Hence we may cover \( {\mathcal{F}}_{\mid \Omega } \) with a finite number ... | Yes |
Corollary 4.28. Let \( G \) be a fixed function in \( {L}^{1}\left( {\mathbb{R}}^{N}\right) \) and let\n\n\[ \mathcal{F} = G \star \mathcal{B} \]\n\nwhere \( \mathcal{B} \) is a bounded set in \( {L}^{p}\left( {\mathbb{R}}^{N}\right) \) with \( 1 \leq p < \infty \) . Then \( {\mathcal{F}}_{\mid \Omega } \) has compact ... | Proof. Clearly \( \mathcal{F} \) is bounded in \( {L}^{p}\left( {\mathbb{R}}^{N}\right) \) . On the other hand, if we write \( f = G \star u \) with \( u \in \mathcal{B} \) we have\n\n\[ {\begin{Vmatrix}{\tau }_{h}f - f\end{Vmatrix}}_{p} = {\begin{Vmatrix}\left( {\tau }_{h}G - G\right) \star u\end{Vmatrix}}_{p} \leq C{... | No |
Lemma 4.3. Let \( G \in {L}^{q}\left( {\mathbb{R}}^{N}\right) \) with \( 1 \leq q < \infty \) . Then \[ \mathop{\lim }\limits_{{h \rightarrow 0}}{\begin{Vmatrix}{\tau }_{h}G - G\end{Vmatrix}}_{q} = 0 \] | Proof. Given \( \varepsilon > 0 \), there exists (by Theorem 4.12) a function \( {G}_{1} \in {C}_{c}\left( {\mathbb{R}}^{N}\right) \) such that \( {\begin{Vmatrix}G - {G}_{1}\end{Vmatrix}}_{q} < \varepsilon \) . We write \[ {\begin{Vmatrix}{\tau }_{h}G - G\end{Vmatrix}}_{q} \leq {\begin{Vmatrix}{\tau }_{h}G - {\tau }_{... | Yes |
Theorem 4.33 (Young). Assume \( f \in {L}^{p}\left( {\mathbb{R}}^{N}\right) \) and \( g \in {L}^{q}\left( {\mathbb{R}}^{N}\right) \) with \( 1 \leq p \leq \infty \) , \( 1 \leq q \leq \infty \) and \( \frac{1}{r} = \frac{1}{p} + \frac{1}{q} - 1 \geq 0 \) . Then \( f \star g \in {L}^{r}\left( {\mathbb{R}}^{N}\right) \) ... | For a proof see, e.g., Exercise 4.30. | No |
Proposition 5.3. Let \( K \subset H \) be a nonempty closed convex set. Then \( {P}_{K} \) does not increase distance, i.e., \[ \left| {{P}_{K}{f}_{1} - {P}_{K}{f}_{2}}\right| \leq \left| {{f}_{1} - {f}_{2}}\right| \;\forall {f}_{1},{f}_{2} \in H. \] | Proof. Set \( {u}_{1} = {P}_{K}{f}_{1} \) and \( {u}_{2} = {P}_{K}{f}_{2} \) . We have (6) \[ \left( {{f}_{1} - {u}_{1}, v - {u}_{1}}\right) \leq 0\;\forall v \in K \] (7) \[ \left( {{f}_{2} - {u}_{2}, v - {u}_{2}}\right) \leq 0\;\forall v \in K. \] Choosing \( v = {u}_{2} \) in (6) and \( v = {u}_{1} \) in (5) and add... | Yes |
Assume that \( M \subset H \) is a closed linear subspace. Let \( f \in H \) . Then \( u = {P}_{M}f \) is characterized by\n\n\[ u \in M\\text{ and }\\left( {f - u, v}\\right) = 0\\;\\forall v \in M. \] | Proof. By (3) we have\n\n\[ \\left( {f - u, v - u}\\right) \\leq 0\\;\\forall v \in M \]\n\nand thus\n\n\[ \\left( {f - u,{tv} - u}\\right) \\leq 0\\;\\forall v \in M,\\;\\forall t \in \\mathbb{R}. \]\n\nIt follows that (8) holds.\n\nConversely, if \( u \) satisfies (8) we have\n\n\[ \\left( {f - u, v - u}\\right) = 0\... | Yes |
Theorem 5.6 (Stampacchia). Assume that \( a\left( {u, v}\right) \) is a continuous coercive bilinear form on \( H \) . Let \( K \subset H \) be a nonempty closed and convex subset. Then, given any \( \varphi \in {H}^{ \star } \), there exists a unique element \( u \in K \) such that\n\n\[ a\left( {u, v - u}\right) \geq... | Proof of Theorem 5.6. From the Riesz-Fréchet representation theorem (Theorem 5.5) we know that there exists a unique \( f \in H \) such that\n\n\[ \langle \varphi, v\rangle = \left( {f, v}\right) \;\forall v \in H. \]\n\nOn the other hand, if we fix \( u \in H \), the map \( v \mapsto a\left( {u, v}\right) \) is a cont... | Yes |
Lemma 5.1. Assume that \( \left( {v}_{n}\right) \) is any sequence in \( H \) such that\n\n(21)\n\n\[ \left( {{v}_{m},{v}_{n}}\right) = 0\;\forall m \neq n, \]\n\n(22)\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }{\left| {v}_{k}\right| }^{2} < \infty \]\n\nSet\n\n---\n\n\( {}^{3} \) The linear space spanned by the \(... | Proof of Lemma 5.1. Note that for \( m > n \) we have\n\n\[ {\left| {S}_{m} - {S}_{n}\right| }^{2} = \mathop{\sum }\limits_{{k = n + 1}}^{m}{\left| {v}_{k}\right| }^{2}. \]\n\nIt follows that \( {S}_{n} \) is a Cauchy sequence and thus \( S = \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n} \) exists. On the othe... | Yes |
Theorem 5.16 (Minty-Browder). Let \( E \) be a reflexive Banach space. Let \( A : E \rightarrow \) \( {E}^{ \star } \) be a continuous nonlinear map such that\n\n\[ \n\left\langle {A{v}_{1} - A{v}_{2},{v}_{1} - {v}_{2}}\right\rangle > 0\;\forall {v}_{1},{v}_{2} \in E,\;{v}_{1} \neq {v}_{2},\n\]\n\nand\n\n\[ \n\mathop{\... | The interested reader will find in F. Browder [1] and J.-L. Lions [3] a proof of Theorem 5.16 as well as many extensions and applications; see also Problem 31. | No |
Theorem 6.1. The set \( \mathcal{K}\left( {E, F}\right) \) is a closed linear subspace of \( \mathcal{L}\left( {E, F}\right) \) (in the topology associated to the norm \( \parallel {\parallel }_{\mathcal{L}\left( {E, F}\right) } \) ). | Proof. Clearly the sum of two compact operators is a compact operator. Suppose that \( \left( {T}_{n}\right) \) is a sequence of compact operators and \( T \) is a bounded operator such that \( {\begin{Vmatrix}{T}_{n} - T\end{Vmatrix}}_{\mathcal{L}\left( {E, F}\right) } \rightarrow 0 \) . We claim that \( T \) is a com... | Yes |
Proposition 6.3. Let \( E, F \), and \( G \) be three Banach spaces. Let \( T \in \mathcal{L}\left( {E, F}\right) \) and \( S \in \mathcal{K}\left( {F, G}\right) \) [resp. \( T \in \mathcal{K}\left( {E, F}\right) \) and \( \left. {S \in \mathcal{L}\left( {F, G}\right) }\right\rbrack \) . Then \( S \circ T \in \mathcal{... | The proof is obvious. | No |
Theorem 6.4 (Schauder). If \( T \in \mathcal{K}\left( {E, F}\right) \), then \( {T}^{ \star } \in \mathcal{K}\left( {{F}^{ \star },{E}^{ \star }}\right) \) . And conversely. | Proof. We have to show that \( {T}^{ \star }\left( {B}_{{F}^{ \star }}\right) \) has compact closure in \( {E}^{ \star } \) . Let \( \left( {v}_{n}\right) \) be a sequence in \( {B}_{{F}^{ \star }} \) . We claim that \( \left( {{T}^{ \star }\left( {v}_{n}\right) }\right) \) has a convergent subsequence. Set \( K = \) \... | Yes |
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