# Tag Info

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The conclusion of the theorem you want to prove holds if and only if $X^*$ has the Radon–Nikodym property with respect to the Lebesgue measure on $(0,1)$. This is precisely what you need to recover Rademacher's theorem about differentiation of Lipschitz maps from $(0,1)$ to $X^*$. Separability in your proof is redundant but it makes it easier as you may ...

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One need not to know the whole dual space of $C[0,1]$ to see the lack of reflexivity. Simply note that $C[0,1]^*$, opposed to $C[0,1]$, is non-separable because for each $t\in [0,1]$ the map $$\langle f, \delta_t\rangle = f(t)\quad (f\in C[0,1])$$ is a norm-one linear functional on $C[0,1]$ and $\|\delta_t - \delta_s\| =2$ for distinct $s,t$. To see this, ...

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No it is not. The dual of $C[0,1]$ is the space $\mathfrak{M}([0,1])$ of complex (or signed) regular Borel measures on $[0,1]$. The dual of $\mathfrak{M}([0,1])$ is the set $\mathcal L^\infty([0,1])$ of bounded Borel functions on $[0,1]$. For example, $\chi_{\{0\}}$, the function which vanishes everywhere, except at $x=0$, where it is equal to $1$, ...

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Consider the subspace $V_\infty$ of $\mathscr l^2(\mathbb N)$ where only a finite amount of terms in a series is non-zero. This is an infinite dimensional normed vector space. Define also the subspace $V_n$ where only the first $n$ terms of a series are non-zero. $V_n \cong \mathbb R ^n$ with the standard norm. As such there is a sequence of compacta ...

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No, the last assumption does not follows from the first two one. To see this consider operators $T_n f = f\left(x^n \right).$

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By definition weak* convergence of $f_k$ to $f$ in $X=L^{\infty}(\mathbb R)$ means: $$lim_{k \to \infty}\langle f^*,f_k\rangle=\langle f^*,f \rangle \space (\forall f^*\in X^*)$$ This is a convergent sequence of reals, hence we have $$sup_{k \in \mathbb N}\Vert \langle f^*,f_k \rangle \Vert = sup_{k \in \mathbb N} \Vert \iota(f_k)(f^*)\Vert < \infty ... 2 Of course this is quite simple, as Jonas showed. Here's a fun way to look at it. Let K=\{x_n'\}\cup\{x'\}. Then K is a compact metric space. Regard x_n and x as functions from K to \Bbb C. Uniform Boundedness shows that ||x_n|| is bounded, and this shows that our family of functions from K to \Bbb C is equicontinuous. And x_n\to x ... 1 Note that$$ |x_n'(x_n)-x'(x)|\le|x_n'(x_n)-x'(x_n)+x'(x_n-x)|\le\|x_n'-x'\|\cdot\|x_n\|+|x'(x_n-x)| $$and the sequence \|x_n\| is bounded because it converges. 1 The proof follows from the following theorem from basic calculus: Let \{a_n\}_{n\in\mathbb N} be a sequence of nonnegative numbers with the property that a_{n+m}\leq a_n \cdot a_m for all n,m\in\mathbb N. Then the limit \lim\limits_{n\to\infty}\sqrt[n]{a_n} exists and it is equal to \inf\limits_{n\in\mathbb N}\sqrt[n]{a_n}. Theorem about ... 1 A(\lambda x_1,\ldots,\lambda x_m) = \lambda^{m-j}\overline{\lambda}^jA(x_1,\ldots,x_m)= \lambda^{m-k}\overline{\lambda}^k A(x_1,\ldots,x_m) From the above, since \lambda \neq 0, we have:$$\left[ \left(\frac{\lambda}{\overline{\lambda}}\right)^{k-j} - 1\right] A(x_1,\ldots,x_m) = 0$$Thus: \left[ \left(\frac{\lambda}{|\lambda|}\right)^{2(k-j)} - ... 0 I find it easier to prove (b) first; using the hint that was given for (a) appears more relevant to (b). (b) If no such C exists, there is a sequence (x_n,y_n) such that$$|B(x_n,y_n)| = 1\quad\text{ and }\quad \|x_n\|\|y_n\|\to 0 \tag{1}$$By replacing (x_n,y_n) with (sx_n, s^{-1}y_n) we can arrange \|x_n\|=\|y_n\| without changing (1). ... 0 EDIT (see below) Following Robert Israel's idea here's another example, which is even Hilbertian: on the space L^2(0,1) consider, again, the polynomials and the linear operator T defined as$$T(x^{2k})=0\quad > T(x^{2k+1})=x^{2k+1}, $$which agrees with the identity on the dense subspace consisting of odd-degree polynomials and is ... 3 Yes, f(B(x_{0}; \epsilon)) is open. It does not follow that B(x_{0}; \epsilon) = f^{-1}((B_{\mathbb{F}}(f(x_{0}); \epsilon'). It doesn't follow even in a finite-dimensional space. To get an idea what's happening here, say X=\Bbb R^2 with the euclidean norm. Say B is the unit ball of X and define f(x,y)=x. Then f(B) is the open interval ... 3 You are correct up until B(x;\epsilon) = f^{-1}(B_\mathbb{F}(f(x_0);\epsilon')). All you can conclude from the statement before that is B(x;\epsilon) \subseteq f^{-1}(B_\mathbb{F}(f(x_0);\epsilon')), which does not imply that B(x;\epsilon) is weakly open. In general, an open ball for the norm topology in an infinite-dimensional Banach space has empty ... 1 It implies that C\|x\|\leqslant \|Tx\| holds true which means that T is injective and has closed range. Note that it is enough to check this only on a dense subspace and {\rm span}\{\delta_x\colon x\in X\} is such a subspace. 1 Let M=\|u\|_{\infty}. Since G' is continuous, there exists a constant K such that |G'(s)|\leq K for all s\in [-M,M]. Thus |G'\circ u|\leq K; it follows that |G'\circ u|^p is bounded by K^p. The desired conclusion follows from Hölder's inequality as you explained with a little correction (you have to change parenthesis by norms): ... 1 Consider the Banach space X = C[0,1] of continuous functions on [0,1], with the operator A of multiplication by x (i.e. Af(x) = x f(x)). This has spectrum [0,1]. Let E be the subspace of X consisting of polynomials. This is invariant under A. However, A - \lambda I is never surjective as an operator from E to E, e.g. there is no ... -1 X=R^2=Vect\{e_1,e_2\}, A(e_1)=e_1, A(e_2)=2e_2, E=R^2-Re_2. The spectrum of A restricted to E is 1 and the spectrum of A is \{1,2\}. 0 The answer can be found in "Calculus without derivatives" of Prof. Jean-Paul Penot. I refer to Lemma 3.94 p.251. The answer is the following : an equivalent norm on the dual X^* is the dual norm of an equivalent norm on X if and only if it is weak-* lower semicontinuous. The major idea is the following. Let us consider the notations introduced in the ... 1 Y is the kernel of the surjective (and continuous) map$$C^1([0,1]^n) \to \mathbb R, f \mapsto f(0).$$In particular, Y is closed (as a kernel) and of codimension 1 (since C^1([0,1]^n)/Y \cong \mathbb R is one-dimensional). 1 The spectrum of a is a compact set that does not contain 0. So there is a disk D around 0 with D\cap\sigma(a)=\emptyset. Thus, on \sigma(a), f:t\longmapsto 1/t is continuous, so f\in C(\sigma(a)). Then f(a)\in C^*(a) via the Gelfand transform. Or, even easier, you could check that f is analytic on \sigma(a), and so a^{-1} belongs ... 2 That \Gamma(y) should be closed doesn't follow from the fact that \Gamma is a closed subalgebra. Example. Let \mathbb N^+ = \mathbb N \setminus \{0\} be the set of positive integers. Consider X = \ell^2(\mathbb N^+) and let R : X \to X be the following weighted right shift:$$ (Rx)(n) = \begin{cases} \quad 0, & \quad\text{if $n = 1$}; ...

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Lemma (Exercise 3.29, Brezis). Let $E$ be a normed vector space with a uniformly convex norm and fix $p > 1$. If $x$, $y \in \overline{N(0, M)} =: B$ are at least $\epsilon > 0$ apart, then there is some $\delta$ such that$$\left\|{{x + y}\over2}\right\|^p \le {{\|x\|^p + \|y\|^p}\over2} - \delta.$$ Suppose not for the sake of contradiction. Then ...

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Reflexivity is superfluous. You can use the simple fact that if $T:X\rightarrow X$ is a compact map on a Banach space $X$, with $\dim(X)=\infty$, then $0\in \sigma(T)$.

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For every normed space $X$, the dual $X^*$ is $1$-complemented in $X^{***}$. Indeed, let $i:X\to X^{**}$ be the canonical embedding; then its adjoint $i^*$ is a projection of norm $1$ of $X^{***}$ to $X^*$. Simply put, it takes a functional $\phi:X^{**}\to \mathbb{C}$ and composes it with $i$. In particular, the above applies to $\mathbb{B}(\mathbb{H})$, ...

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I don't see the point of reflexivity here, $inf\{\|T(x)\|, \|x\|=1\}>0$ implies that the spectrum of $T$ does not contain zero. Since $T$ is compact, $X$ must be finite dimensional so $S$ is compact.

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Yes. Say $f_t\in L^2[0,1]$ for $t\in[1,2]$ and $f_t$ depends continuously on $t$. Then $K=\{f_t:t\in[1,2]\}$ is a compact subset of $L^2[0,1]$. Define $T_n:L^2[0,1]\to L^2[0,1]$ by, say, letting $T_n=f*\phi_n$, where $\phi_n$ is an approximate identity. It's easy to see that for each $n$ the function $g_n(s,t)=T_nf_t(s)$ is continuous on $[0,1]\times[1,2]$. ...

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Only a partial answer: If you take the norm $\|x\|'=\|x\|+\alpha |x|$ as gerw said in the comments, it is easy to see that since $\|.\|\sim |.|$, i.e $\exists \underline c,\overline c>0: \underline c\|x\|\leq |x|\leq \overline c \|x\|,\forall x\in E\,\,$ you will have $$\|x\|\leq \|x\|+\alpha|x|\leq (1+\alpha \overline c)\|x\|,\forall x\in E$$. It is ...

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By assumption $p\neq q$. Case 1: $p,q\in(1,+\infty)$. Assume we have a surjection $T:\ell_p\to\ell_q$, then $T^*:\ell_{q'}\to \ell_{p'}$ is an embedding. Since $p',q'\in(1,+\infty)$, we get a contradiction because by corollary of Pitt's theorem theses spaces are totally incomparable. Thus for this case a desired surjection doesn't exists. Case 2: $q=1, ... 0 In the case$V_i = \Bbb R$for all$i$, and the codomain being$\Bbb R^n$, for some$n$, i.e.,$f: \Bbb R^d \to \Bbb R^n$,$x = (x_1,...,x_d) \mapsto f(x) = (f_1(x),...,f_n(x))$, if$a \in \Bbb R^d$is a point at which$f$is differentiable, then$Df(a)$is simply the Jacobian matrix of$f$at$a$. That's, $$Df(a) = \left( \frac{\partial f_i}{\partial ... 1 You are correct. We can note that our space is isometric to the subspace of l^2(\mathbb{N}) comprised of sequences with finitely many non-zero entries, by the isometry \sum_{n=0}^N a_n z^n\mapsto (a_0, a_1, \ldots, a_N, 0, 0, \ldots). (Note that, indeed, \left\langle \sum_{n=0}^N a_n z^n, \sum_{n=0}^M b_n z^n\right\rangle = ... 0 No, this seems not to be possible. Take E = L^2(0,1) and consider$$x_i = \chi_{((i-1)/n,n)}.$$Then,$$\sum_{i=1}^n \|x_i\|_{L^2(0,1)} = \sqrt{n}$$while$$\|\sum_{i=1}^n x_i\|_{L^2(0,1)} = 1.$$It is even worse with L^p(0,1), p > 2. And with p = \infty, you need the constant n. 1 You are definitely on the right track and nearing a complete proof! I'm going to write \mathcal{A}_p for what you're denoting by \sum_p A_n (since I find the placement of the p rather unsettling in your notation ;)). If you're comfortable with the idea of a direct product of vector spaces, you can think of \mathcal{A}_p (before assigning it a ... 1 Let (a_n)_j be a Cauchy sequence of sequences. Now, the "obvious" limit choice would be the sequence (a_n) where a_n = \lim_{j \to \infty} (a_n)_j. To show this, you have to prove that all those (a_n) exist for any j, and then that that is the actual limit. For the first part, it's enough to observe that$$ \|(a_n)_j - (a_n)_k\|_n^p \leq ... 1 You are confused: Let$H$be a Hilbert Space, let$B=\{u_j\}_{j=1}^\infty$be a countable orthonormal basis. So we know that if a set is a complete orthonormal basis, the set of all finite linear combinations is dense in$H$. It is true. Now, since the set of all finite linear combinations is dense let$x\in H$, we have ... 2 A very elegant proof can be based on tensor product representation $$C([0,1],E)= C([0,1]) \tilde \otimes_\varepsilon E.$$ For every dense subspace$L$of$C([0,1])$the tensor product$L\otimes E$is dense. 1 I already know that the unit ball in$X$(denoted$B$) is compact in the$\|.\|$topology. So I just need to have the estimate $$\| . \|_{\alpha} \leq C \|.\|$$ for some$C$to conclude that$B$is compact in the$\|.\|_{\alpha}$topology. Now$X$is closed in$C[0,1]$, the inclusion$i : C^\alpha[0,1] \to C[0,1]$is continuous so that$i^{-1}(X) = ...

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Seems like the answer is yes. Take $f\in C([0,1],E)$. First extend $f$ to a function in $C(\Bbb R, E)$, say by making $f$ constant on $[1,\infty)$ and constant on $(-\infty,0]$. Now say $\phi_n$ is a smooth (real-valued) approximate identity; then the convolution $f*\phi_n$ should be differentiable and it should be that $f*\phi_n\to f$ uniformly on $[0,1]$. ...

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You can just take the coefficients of the polynomials in $E$ to have $E$-valued polynomials. Trying to mimic the Stone-Weierstraß proof for $E$-valued functions may however be tricky. But from uniform continuity of continuous functions on $[0,1]$ you immediately get that you can uniformly approximate all functions in $C([0,1],E)$ by piecewise affine ...

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Hint Consider a Cauchy sequence $\{x_n\}$. See what happens with $$\left\{\frac1{\|x_n\|}x_n\right\}$$ (Note that you also have to consider sequences with zeros).

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Let $Y$ be the closure of the range of $A$. We define $B : X \to Y$ by $B x = A x$ for all $x \in X$. Let us show that $B' : \tilde Y \to \tilde X$ is invertible. Since the range of $B$ is dense, $B'$ is injective. It remains to show that $B$ is surjective. For any $\tilde x \in \tilde X$, there is $\tilde r \in \tilde X$, such that $A'\tilde r = \tilde x$. ...

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I think you can find the spectrum directly. First, $$T((a_j))_i=\sum_{j=2}^{\infty}a_j e_1+\sum_{i=2}^{\infty}a_{i-1} e_i=\sum_{j=2}^{\infty}a_j e_1+a_{i-1}$$ Then $$(T-\lambda I)((a_j))_1=\sum_{j=2}^{\infty}a_j -\lambda a_1,$$ $$(T-\lambda I)((a_j))_i=a_{i-1}-\lambda a_i$$ So if $\lambda$ is an eigenvalue, the second equation implies ...

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You don't need such heavy machinery as Hahn-Banach. This is a completely elementary fact: Let $J\colon E\times F\to F\times E,\,J(x,y)=(-y,x)$. It is a (more or less) immediate consequence of the definiton of $A^\ast$ that $G(A^\ast)=(J G(A))^\perp$, where $G(T)$ denotes the graph of the operator $T$. Then we have $x\in N(A)$ iff $(x,0)\in G(A)$ iff ...

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If the projection is continuous, then the range of $P$ is closed, as it is the kernel of the continuous projection $I-P$. However not all projections on a Banach space are continuous (as opposed to what happens in the case of a finite dimensional space). So, in general, this is not true. Take for instance the subspace generated by $\{(1,0,\dots), (0,1,0, ... 4 Convexity: To show$\psi$is convex we only need to show$\varphi^*$is convex. Let$f,g\in E^*, \lambda\in [0,1]$. We want to show $$\varphi^*(\lambda f+(1-\lambda)g)\leq \lambda \varphi^*(f)+(1-\lambda)\varphi^*(g)$$ To prove this let$x\in E$with$\varphi(x)<+\infty$. Then \begin{eqnarray} \langle \lambda f+(1-\lambda)g,x\rangle-\varphi(x) & ... 2 You can bound$(f\ast g)_n$below by $$\sum_{m = 1}^{n-1} m^{-\phi} (n-m)^{-\psi} \geqslant \sum_{m = 1}^{n-1} (n-1)^{-\phi}(n-1)^{-\psi} = (n-1)^{1 - \phi - \psi}.$$ So a necessary condition for$f\ast g \in \ell_{\infty}$is$\phi + \psi \geqslant 1$. That is also sufficient, as can be seen by splitting the sum at$n/2\$: \begin{align} \sum_{m = 1}^{n-1} ...

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