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Let $X$ be a Hilbert space with orthonormal subset $\{ e_{n} \}_{n=1}^{\infty}$. Define $F : \mathcal{D}(F) \subset X \rightarrow \mathbb{C}$ by $F(x) = \sum_{n=1}^{\infty}n(x,e_n)$ on the domain $$\mathcal{D}(F)=\{ x \in X : \sum_{n=1}^{\infty}|n(x,e_n)| < \infty \}.$$ Then $\{ \frac{1}{n}e_{n} \}_{n=1}^{\infty}$ converges to $0$ and ...

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Claim: There exists a function $f : \mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Furthermore, for any $x \notin \mathbb{Q}$, one may choose $f$ so that $f(\alpha x) \ne \alpha f(x)$ for some $\alpha \in \mathbb{R}$. Proof of claim: To prove the existence of $f$, define an equivalence relation $\sim$ on ...

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No, Evans means that a supporting hyperplane at $(x, f(x))$ is the graph of the mapping $y \mapsto f(x) + r\cdot (y - x)$.

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This is not true: Consider the function $f(x)=\chi_{\mathbb{R}\setminus(-1,1)}(x)|x|^{-1/2}$. Then for $p\leq 2$ we have $\| f\|_{p}=\infty$ and for $p>2$ we get $$\| f\|_p = \left(\frac{4}{p-2}\right)^{1/p}.$$

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For example, take $$e^{(1)} = (1,0,0, \cdots )$$ Clearly, $e^{(1)}\in B$. Now suppose that for $t \in (0,1)$, you have $b=\{ b_j \}_j , d=\{ d_j \}_j$ with $b,d \in B$ such that $$e^{(1)} = t b+ ( 1- t )d$$ Then, you must have $1=tb_1+(1-t)d_1$ and $0=tb_j+(1-t)dj$ for $j>1$ ,which gives that $$b_j=d_j=\left\{ \begin{array}{cc} 1 & \text{if } ... 0 We know that the L^p norm is convex as a function of p by Minkowski's inequality. Then use the fact that a composition of convex functions is convex when the outer function is non-decreasing. 1 Riesz-Thorin interpolation (RTI) bounds the norms of linear maps acting between L^p spaces. Unlike Marcinkiewicz interpolation, RTI only works for strong type operators. Overview of interpolation: to understand how to obtain control on the expression ||Tf||_{L^q} for operator T & function f, one would divide f into two (or more) components ... 0 If A \in B(X,Y) then A^{\ast} \in B(Y^{\ast},X^{\ast}) is clearly weak-star continuous. Conversely, suppose B \in B(Y^{\ast},X^{\ast}) is weak-star continuous, we want to define A : X\to Y such that$$ B(f)(x) = f(Ax) \quad\forall f \in Y^{\ast}, x \in X $$Since B is weak-star continuous, for any x\in X fixed$$ \hat{x} \circ B : Y^{\ast} \to ...

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Here is an outline: in light of Bessel's inequality you have $$\sum_i |\langle f,\phi_i \rangle|^2 \le \|f\|^2 < \infty.$$ Use this to show that the sequence $$f_n = \sum_{i=1}^n \langle f,\phi_i \rangle \phi_i$$ is Cauchy in $H$, so that $f_n \to g$ for some $g \in H$. By definition you have $$g = \sum_i \langle f,\phi_i \rangle \phi_i.$$ Use the ...

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According to your definition this certainly can not be true. If $f_{max}\in C$ then so is $a(f_{max})$ for any $a\in \mathbb{R}$. But even if you fix that problem I'm pretty sure such a function can't exist.

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No, for the same reason that the result doesn't work for functions (just define an antiderivative, and so on). The result for functions is discussed in this question. Similarly for series and sequences, basically by using the integral test.

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Hint By construction, the support of $\phi \circ F^{-1}$ is $F(\text{supp }\phi)$.

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Remarks: The argument below proves the following. Let $1<r<\infty$ and $f\in L^r(\Omega, \Sigma, \mu)$ for some probability space $(\Omega, \Sigma, \mu)$. Then, the function $\phi:[1,r]\to$$\mathbb R defined by \phi(p)=\|f\|^{p}_{L^p} is convex. Choose \theta\in[0,1] and p,q\in[1,r]. Then, \theta p+(1-\theta)q\in[1,r]. Note that, by the ... 6 You have$$ \phi(p) = \int_0^1 \lvert f(x) \rvert^p \, dx $$p \mapsto x^p = e^{p\log{x}} is smooth for x \geqslant 0, so we can differentiate twice with respect to p,$$ \phi''(p) = \int_0^1 \lvert f(x) \rvert^p (\log{\lvert f(x) \rvert})^2 \, dx, $$which is obviously nonnegative. (This is consistent even for f possessing zeros if we make the ... 0 Jensen: take q>p, examine: L(q,p) = \frac {(\int |f|^q)^{\frac 1q}}{(\int |f|^p)^{\frac 1p}} By Jensen's inequality (\int |f|^p)^{\frac qp} ≤ (\int |f|^q) Because \frac qp≥1 \Rightarrow g(x) = x^{\frac qp} is convex Thus, L(q,p)^q≥ \frac {\int |f|^q}{\int |f|^q} = 1 \Rightarrow L(q,p)≥1 \Rightarrow p \rightarrow (\int ... 1 wouldn't it be "easier" to show A weaker statement is not always easier to prove. See Examples where it is easier to prove more than less. You are right: to derive that u is a minimizer it would suffice to show that$$I[u] \leq \limsup_{j\to\infty} I[u_{k_j}] \tag{1}$$Why does Evans talk about the stronger property$$I[u] \leq ... 0$c_0$has no extreme points. Let$x \in c_0$unit ball. Then,$x_n \to 0$. Choose$N$s.t.$|x_n| < 1/2$for$n\geq N$and use two sequences which match$x_n$for$n \leq N$and are$x_n + 2^{-n}$and$x_n - 2^{-n}$for$n > N$. Both of these sequences will be in the unit ball and their average will be$x$. You can try$\ell^\infty$on your own ... 0 Yes. For a bounded selfadjoint operator, its numerical range is the convex hull of its eigenvalues. So if the numerical range of$T$is$[a,b]$, this tells you that the max and min of the spectrum of$T$are$b$and$a$respectively, so for$T^{-1}$they are$a^{-1}$and$b^{-1}$, and the numerical range of$T^{-1}$is$[b^{-1},a^{-1}]$0 Let$p $be a projection with$p\leq\pi (e) $. Suppose$\pi (x_j)\to p $weakly. As$\pi (e)p=p $, we get that$\pi (ex_je)=\pi (e)\pi (x_j)\pi (e)\to p $. But$e $is minimal, so$ex_je=\lambda_je $for numbers$\lambda_j $. Then$p=\lim \lambda_j\pi ( e)\in\mathbb C\,\pi (e )$, and so either$p=0$or$p=\pi (e )$. Thus,$\pi (e) $is rank-one. If$A $is ... 2 In general, no. Take$U = (0,1) \subset \mathbb{R}^1$,$p=2$and$F(z,w,x) = z^4$. Then take $$w_n(x) = \begin{cases} n^{1/4} x, & 0 < x < n^{-1} \\ n^{-3/4}, & x \ge n^{-1} \end{cases}$$ so that$w_n' = n^{1/4} 1_{(0, n^{-1})}$. It's easy then to see that$w_n' \to 0$and$w_n \to 0$in$L^2$, so$w_n \to 0$in$W^{1,2}(U)$. Taking$w = ...

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the prove is clear with definition of algebra and topology . for example consider $X=\Bbb{N}$ and $$S=\{A\subset \Bbb{N} |A\, or A^c is finite\}$$ it is clear that S is algebra ,and S is not topology,because let $A_{n}$= $\{2n\}$ and ($\bigcup_{n=1}^{\infty}A_{n}$=even numbers) is not belong to $S$ because itself and complement of it is not finite.

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If $X$ is only topological, "bounded" has no meaning. If $X$ is a metric space, the answer is "yes", as the closure consists of points that are within $\epsilon$ of points in the set.

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Let $S$ denote the class of sets under consideration. All you need to show is that if $A_n \in S$ for all $n \in \mathbb{N}$ then $A:= \bigcup A_n$ is in $S$ too. So define $$B_n:= \bigcup_{k \leq n} A_k.$$ The $B_k$ form an increasing sequence, and they increase to $A$ which is thus in $S$ (by the monotone class property).

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Let $F$ be a locally free sheaf on an $n$-dimensional complex manifold $X$. The obvious is to consider the complex $$O_X(F) \to A_X^{0,0,r}(F) \to A_X^{0,1,r-1}(F)\to \dots \to A_X^{0,n,r-n}(F) \to 0$$ where $A_X^{p,q,s}(F)$ denotes the space of $(p,q)$ forms on $X$ with values in $F$ of class $C^s$, and all differentials are $\overline \partial$. ...

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In a newer edition of the book, the proof has been rewritten. In effect, $S \subset (a,b]$ or $S \subseteq (a,b]$ has replaced the old definition. Sorry for wasting people's time.

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All norms in $\mathbb{R}^n$ are equivalent. There is a constant $C>1$ (depending on $n$ and $q$) such that $$\frac{1}{C}\,\Bigl(\sum_{k=1}^n|x_n|^q\Bigr)^{1/q}\le\Bigl(\sum_{k=1}^n|x_n|^2\Bigr)^{1/2}\le C\,\Bigl(\sum_{k=1}^n|x_n|^q\Bigr)^{1/q}$$ Note that the last line of your argument is incorrect. $\|Dw\|^q_{L^q(U)}$ is NOT equal to $(\int_U ... 0 Define: $$f: \mathbb{R}x \rightarrow \mathbb{R}: ax \mapsto a\|x\|^2$$ by the Hahn-Banach theorem there exists a$g \in X^*$s.t.$\|g\| = \|f\|$and$g = f$on$\mathbb{R}x$. from this we get: $$\|g\| = \|f\| = \sup_{\|ax\| \leq 1} |f(x)| = \sup_{\|ax\| \leq 1} (|a|\|x\| \|x\|) = \|x\|$$ and $$g(x) = f(x) = f(1\cdot x) = 1\cdot \|x\|^2 = \|x\|^2$$ ... 2 It's a simple calculation error. When substituting to convert$\int_a^b \lvert f(x)\rvert^2\,dx$into the integral over$\lvert g(t)\rvert^2, we have $$x = \frac{b-a}{2\pi}t + \frac{a+b}{2},$$ and hence obtain $$dx = \frac{b-a}{2\pi}\,dt.$$ So $$\int_a^b\lvert f(x)\rvert^2\,dx = \int_{-\pi}^\pi\left\lvert ... 0 First let us assume that B is bounded. Let (x_n) be any sequence in B and (a_n)\in c_0. Then \|x_n\|\leq M for all n and given \epsilon >0 there exist a natural number N such that |a_n|<\frac{\epsilon}{M} for all n\geq N. So, for n\geq N we have \|a_nx_n\| = |a_n|\|x_n\|<\frac{\epsilon}{M}M = \epsilon. ... 0 Your 1 is not correct. On a ball, weak ^*= wot. Hahn-Banach says that the weak and norm closures agree on a bounded convex set. 2 T is bounded for all \beta > -d. The strategy is to show that T is bounded both as a function L^1(d\mu) \rightarrow L^1(d\mu) and as a function L^\infty \rightarrow L^\infty; the Riesz-Thorin theorem then shows that it is bounded L^p(d\mu) \rightarrow L^p(d\mu) for any p. We just need the following estimate; everything else is standard. ... 2 As user40276 has mentioned, using Taylor’s Theorem, we can write$$ \forall \mathbf{x} \in \mathbb{R}^{n}: \quad f(\mathbf{x}) = f(\mathbf{a}) + \sum_{i = 1}^{n} \frac{\partial f}{\partial x_{i}} \Bigg|_{\mathbf{x} = \mathbf{a}} \cdot (x_{i} - a_{i}) + R(\mathbf{x}), $$where R: ... 0 The defining property of the adjoint is that for all x\in X and y^*\in Y^* then$$(Tx,y^*)=(x,T^*y^*)$$We have that$$\|T^*y^*\|=\sup_{x\neq0}\frac{(x,T^*y^*)}{\|x\|}=\sup_{x\neq0}\frac{(Tx,y^*)}{\|x\|}=\sup_{x\neq0}\left(\frac{Tx}{\|x\|},y*\right)=\sup_{\|y\|=1}(y,y^*)=\|y^*\|$$The second-to-last equality is because \frac{Tx}{\|x\|} returns all ... 1 Because the unit ball of the C^*-algebra generated by A sits in between the unit ball of A and the unit ball of the weak closure of A; so if the former is dense in the latter, so is the unit ball of C^*(A). Because A is non-degenerate, the identity is in it weak closure. So if the unit ball of A, together with the identity, is dense in the ... 0 If x\geq0, let f be a state on B(H_2); then the functional f\circ\phi is unital and contractive. It is well-known that a unital contractive functional is positive (I know of proofs in Paulsen and Davidson's books). So f(\phi(x))\geq0 for all states on B(H_2), which means that \phi(x)\geq0. Finally, this can be done for each amplification of ... 4 How about this: define T : \ell^1 \to \ell^1 by$$T(a_1, a_2, a_3, \cdots, a_n, \cdots) = (0, (1-1/2)a_2, (1-1/3)a_3, \cdots, (1-1/n)a_n, \cdots)$$Then$$||T|| = \sup_{||x|| = 1} ||Tx|| = \sup_n (1-1/n) = 1$$However, for all non-zero x \in \ell^1, ||Tx|| < ||x|| = ||T|| \cdot ||x||. 0 I do not think that this result is true, you need that K is closed w.r.t. the weak-* topology. The following is a counterexample: Take X = [0,1] and m the Lebesgue measure. K is the closed convex hull of the Dirac measures. Then, m \not\in K (see below) but you cannot separate m from K with a continuous function. It remains to show m \not\in ... 1 Let H be Hilbert space. By Riesz representation theorem for any bounded linear functional F\in H' there exists a unique element g\in H such that$$\forall h\in H\quad f(h)=(g,h)_H.$$We will denote this isomorphism by i:H'\to H Apparently \|f\|_{H'}= \|g\|_H. Now take a scalar product in H' by$$(f_1,f_2)_{H'} = (i(f_1),i(f_2))_{H}.$$All you ... 1 Short answer: y_n=x_n is a convex combination of x_1,\dots,x_n. Clearly, this need not converge strongly. Longer answer: for any particular choice of coefficients one can cook up a sequence (x_n) where the strong convergence of convex combinations fails. Here's an idea. For any sequence a_n\to \infty we have \sin a_n t\to 0 weakly in L^2[0,1]. ... 0 One way is by showing that the inverse image of a(n) basic interval (a,b) \subset \mathbb R under d is open in X \times X. Take a point (u,v) in V:=d^{-1}(a,b) and show there is an open ball B_1 \times B_2 \subset V . Use open balls B(u,r_1), B(v,r_2) with the right radii r_1, r_2 and show that the product is contained in V. For #2, you ... 0 If A^* is defined as A\left(f\right)\left(x\right)=f\left(A\left(x\right)\right) for all x\in X and all f\in Y', then we have$$\|A^*\|_{\mathcal{L}\left(Y';X'\right)} =\sup_{\|f\|_{Y'}=1}\|A^*\left(f\right)\|_{X'} =\sup_{\|f\|_{Y'}=1}\sup_{\|x\|_{X}=1}|f\left(A\left(x\right)\right)|$$... 1 Notice that if you are considering L:C[a,b]\to \mathbb{R}, then L is clearly a linear functional. Now note that for any h \in C[0,1] $$|L(h)| =\biggl{|}\int_\limits{a}^{b}h(x)f(x)dx\biggr{|}\leq (b-a)\sup_{a\leq x\leq b}|h(x)f(x)| \leq (b-a)\sup_{a\leq x\leq b}|h(x)|\sup_{a\leq x\leq b}|f(x)|,$$ i.e., |L(h)| ... 0 By the other thread:$$WN\subseteq NW\quad WN^*\subseteq N^*W$$And formally it holds:$$\psi\in\mathcal{D}(N^*):\quad A^*\psi=WN^*\psi$$So a calculation gives:$$AA^*\psi=NWWN^*\psi=WNN^*W\psi=WN^*NW\psi=A^*A\psi\quad(\psi\in\mathcal{D}N^*)$$But the domain was dense:$$\overline{\mathcal{D}(N^*)}=\mathcal{H}\implies AA^*=A^*A$$That finishes the ... 0 Meanwhile I got it... Existence It is a well-know result:$$W\in\mathcal{B}(\mathcal{H}):\quad W=W^*\geq0$$So for the square root:$$\sqrt{W}\in\mathcal{B}(\mathcal{H}):\quad\sqrt{W}=\sqrt{W}^*\geq0$$But the square root maps to:$$\sqrt{W}\mathcal{R}(\sqrt{W})=\mathcal{R}(W)=\mathcal{D}(1+N^*N)=\mathcal{D}(N^*N)\supseteq\mathcal{D}(N)$$Also it has ... 0 It seems there is no such formula. Vaguely speaking even for simple cases this problem is NP-hard. For details see this paper 1 \boldsymbol{\langle x,y\rangle=0\implies\|x+\alpha y\|^2\ge\|x\|^2}$$ \begin{align} \|x+\alpha y\|^2 &=\|x\|^2+2\mathrm{Re}\left(\langle x,\alpha y\rangle\right)+|\alpha|^2\|y\|^2\\ &=\|x\|^2+2\mathrm{Re}\left(\overline{\alpha}\langle x,y\rangle\right)+|\alpha|^2\|y\|^2\\ &=\|x\|^2+|\alpha|^2\|y\|^2\\ &\ge\|x\|^2 \end{align} $$... 2 Spectral theory in infinite-dimensional spaces is quite a bit more complicated than in the finite-dimensional case. In particular, we have to distinguish between the spectrum \sigma(A) of an operator and its eigenvalues. Let A be a linear operator on a Banach space X over the scalar field C. We have$$ \sigma(A) = \{ \lambda \in C: (\lambda I - A) ... 2 The multiplication operator(Mf)(x)=xf(x)$on$L^{2}[0,1]$is a classical example of an operator with no eigenvalues, but its spectrum is$[0,1]$.$M$has no eigenvalue because$Mf=\lambda f$gives$(x-\lambda)f=0$, which forces$f(x)=0$a.e.. To see that$[0,1]\in\sigma(M)$, note that the constant function$1$is not in the range of$(M-\lambda I)$for ... 0 They're not necessarily both spectral values.$I$has spectrum$\{1\}$and$0$has spectrum$\{0\}$. Every$x \in X$can be written as$x = (I-T)x+Tx$. And,$y=(I-T)x$,$z=Tx$satisfy$Ty=0$and$(T-I)z=0$. If$T$does not have a non-zero vector with eigenvalue$0$, then$(I-T)x=0$for all$x$, which means$T=I$. If$T$does not have a non-zero vector with ... 2 If$(x,y) \ne 0$then$\|y\|\ne 0\$, and the following is an orthogonal decomposition: $$x = \left[x-\frac{(x,y)}{(y,y)}y\right]+\frac{(x,y)}{(y,y)}y.$$ Hence, $$\|x\|^{2} = \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2}+\frac{|(x,y)|^{2}}{\|y\|^{2}} > \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2}.$$

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