New answers tagged

0

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 ...


3

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, ...


2

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 $1$, ...


1

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 ...


2

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


0

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 $n = 1$}; ...


2

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 ...


1

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 $0\in \sigma(T)$.


1

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 $\mathbb{B}(\mathbb{H})$, ...


2

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.


1

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 $[0,1]\times[1,2]$. ...


1

$\newcommand{\ip}[2]{\left\langle #1,#2\right\rangle}$ $\newcommand{\norm}[1]{\left|\left| #1\right|\right|}$ I think your confusion is due to notation used in Pythagorean theorem: we don't actually require any norm-property of $||\cdot||$ induced by the inner product in proving it. It's only after we recognise $||\cdot||$ as a norm that it has its usual ...


2

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 ...


2

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) = ...


3

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 $[0,1]$. ...


3

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 ...


1

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).


1

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 $A'\tilde r = \tilde x$. ...


1

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 ...


2

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 ...


1

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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