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Extreme point of the unit ball in H(U)

There is a famous result due to Szego that characterizes the extreme points of the closed unit ball in $H^{\infty}(\mathbb{D})$ (where recall that $H^{\infty}(\mathbb{D})$ denotes the space of all ...
Wjvr46's user avatar
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1 vote

Resolvent of maximal-monotone operator

Existence follows from Rockafellar's surjectivity theorem: Suppose that $X$ is reflexive and $A \colon X \to 2^{X^*}$ is a maximal monotone operator. Then $$R(A+λ \mathcal F ) =X^*, \quad \forall \...
Evangelopoulos Foivos's user avatar
2 votes

Can an unbounded operator and its adjoint both have full domains?

This is not possible. Any adjointable operator is closed. Indeed, let $x_n \to x$, $Tx_n \to y$. Then for any $z \in H$, we have, $$\langle z, y \rangle = \lim_n \langle z, Tx_n \rangle = \lim_n \...
David Gao's user avatar
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3 votes
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For a given $k \in \mathbb{N}$, define the mapping $A: X \to X$ by the rule $ (Af)(x) = x^k f(x). $

To show that $A$ is linear means to show that $$ A(\alpha f+\beta g)=\alpha Af+\beta Ag $$ or, equivalently, that $$ x^k(\alpha f(x)+\beta g(x))=\alpha x^kf(x)+\beta x^kg(x). $$ This identity clearly ...
John B's user avatar
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0 votes

Kernel dimension is preserved under uniform convergence of bounded operators on a Hilbert space

The conclusion does not follow the opposite way. Assume $T$ is the operator defined by $Te_n=2^{-n}e_n,$ where $\{e_n\}_{n=1}^\infty$ is an orthonormal basis. For fixed $m$ choose a sequence of ...
Ryszard Szwarc's user avatar
1 vote
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I want to prove that $T$ belongs to $B(X)$, meaning it is a bounded linear operator.

I think that $f \in X^*$ should be a bounded linear operator as well. I'm assuming it is a operator of this kind: $f:X\to\mathbb{K}$ where $\mathbb{K}$ is a field equal to $\mathbb{R}$ or $\mathbb{C}$...
Alejandro Sánchez Yalí's user avatar
2 votes
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If $X$ is a normed space and $x \neq 0$ and $f(y)x = 0$ for every $f \in X^*$, then $y = 0$.

Hint: By Hahn-Banach there exists $f \in X^*$ such that $f(y) = \|y\|$
Evangelopoulos Foivos's user avatar
1 vote
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Supremum Norm on Space of Continuous Functions

The issue is that there are unbounded continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(0)=0$, so there are elements of $C$ which are unbounded. As you stated in your question, the ...
Dean Miller's user avatar
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1 vote

Kernel dimension is preserved under uniform convergence of bounded operators on a Hilbert space

Fix any $T$ such that $\dim \ker T=k$, and let $C_n=\frac1n\,T$. Then $\dim\ker C_n=k$ for all $n$, but $C_n\to0$, so the kernel of the limit is infinite-dimensional. Worse, you can prescribe $\dim \...
Martin Argerami's user avatar
1 vote

Closeness of the operator $(Tx )(n)= nx(n).$

Let $\ell^1$ denote the set of sequences $z=(z(1),z(2),z(3),\ldots)$ such that $\sum_{k=1}^{\infty}|z(k)|<\infty$ equipped with the norm $\|z\|:=\sum_{k=1}^{\infty}|z(k)|$. Let $T$ be defined by $(...
Steen82's user avatar
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1 vote

Closeness of the operator $(Tx )(n)= nx(n).$

I think your arguments are problematic, because you seem to make some confusion with this "$n$" in $x(n)$ or $x^m(n)$. Remember that when people talk, for example, about the function $f(x) = ...
André Caldas's user avatar
2 votes
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Continuity of $T(f) = \int (h\circ f)\operatorname df$

The answer is that if we consider the particularization of the $n$-dimensional BV norm for $n=1$, i.e. $$\DeclareMathOperator{\Dm}{d\!} \Vert f\Vert_{BV} \triangleq \Vert f\Vert_{L_1}+\operatorname{TV}...
Daniele Tampieri's user avatar
1 vote
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Infinite linear combination: conditions for not escaping the source space

No. Here is a counterexample: Let $(\Omega, P)$ be s.t. $\Omega = \mathbb{N}$ and $P = \sum_{n = 1}^\infty 2^{-n} \delta_n$ (effectively the counting measure - I’m adding the $2^{-n}$ weights to ...
David Gao's user avatar
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1 vote
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Sum of topological interior and topological interior of the sum

The latter is, in fact, still a subset of the former, though this is less trivial. The observation here is that, if $B$ is a TVS, $U \subset B$ is open, and $V \subset B$ is an arbitrary subset, then $...
David Gao's user avatar
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0 votes

Closed range of $T^*T$ for Hilbert space map.

Upon reflection, I found a more simple argument than the answers posted. I’m still enormously grateful for the help. As noted in the question, we know $\ker(T^*T) = \ker(T)$, and $\text{image}(T^*T) = ...
Joe's user avatar
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2 votes
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A false proof of Dini theorem

The mistake, as most often happens, it's in the part of the proof where you gloss over stuff. The problem is in the last sentence: you cannot guarantee that if $N>i_k$, then $A_{N,\epsilon}\subset ...
Martin Argerami's user avatar
0 votes

Brezis's Exercise 1.10

One way to see this is to consider first the vector space $\mathbb R[I]$ with basis $I$ (the subspace of the space of $\mathbb R$-valued functions on $I$ spanned by the indicator functions of the ...
krm2233's user avatar
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1 vote
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What is the connection between double commutant theorem and group representation theory?

They are not the same result. The von Neumann double commutant theorem applies to infinite-dimensional Hilbert spaces and infinite-dimensional operator algebras acting on them, which are not ...
Qiaochu Yuan's user avatar
2 votes
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Suppose $\phi(a) \geq 0$ for every state $\phi$ of $\mathcal{A}$. Can we conclude that $a\geq 0$.

Yes, we can. Assume first that $A = C^\ast(a)$. As $a$ is self-adjoint, we have $C^\ast(a) = C_0(\Omega)$ for some locally compact Hausdorff $\Omega$ (in fact, $\Omega = \sigma(a) \setminus \{0\}$). ...
David Gao's user avatar
  • 7,830
0 votes

Computing operator norm of convolution on bounded interval

Fix $x,$ $0\le x\le 1.$ Then by the Riesz-Markov theorem we have $$\sup_{\|f\|_\infty \le 1}|(f*c)(x)|=\sup_{\|f\|_\infty \le 1}\left |\int_0^1 f(t)c(x-t)\,dt \right |\\ \underset{\rm R-M }{=}\int_0^1 ...
Ryszard Szwarc's user avatar
2 votes
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Is every proper subspace of T. V. S. closed?

No. For example, if you consider the space $(C([a,b]),||\cdot||_{\infty})$, this normed space has a dense proper subspace (which can't be closed): the polynomials, as a consequence of Weierstrass ...
Valere's user avatar
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1 vote
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Estimating integrals and measures over Hilbert space using finite dimensional projections

Interchanging the limit and the integral is fine in this case. Indeed, using the definition of Bochner integrals, by approximating with simple functions, one easily observes that Bochner integrals can ...
David Gao's user avatar
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1 vote
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Interpreting Green's function in Evans' Partial Differential Equations

Let $x\in U$. Since $\delta_x$ is the convolution identity at point $x$, we have $$ u(x) = u*\delta_x = \int_Uu(y)\delta_x(x-y)dy$$ then since $-\Delta_x G(x,y) = \delta_x (x-y)$: $$ u(x) = -\int_Uu(y)...
Lucas Cândido's user avatar
2 votes
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Inequality for spectral families of self-adjoint operators

This inequality fails even for self-adjoint matrices $A$ and $B$. In terms of functional calculus, what you are asking is the following: If $A\leq B$, does one have $1_{[n,\infty)}(A)\leq 1_{[n,\infty)...
MaoWao's user avatar
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7 votes
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Is the best degree $n$ polynomial approximation an interpolation on $L^2[0,1]$?

Let $\deg p\le n$ and $p$ minimizes $\|f-q\|_2^2$ for all polynomials $\deg q\le n.$ Then $f-p\perp q$ for any polynomial $\deg q\le n.$ Assume $f-p$ changes sign only at $x_1,\ldots ,x_k$ and $k\le ...
Ryszard Szwarc's user avatar
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Finding a Closed Form Expression for a Distribution Defined by an Integral Involving Sine and Bessel Functions

Mathematica and WolframAlpha give the result $$D(t,x)=\mathcal{M}_{\omega}[\sin(\omega t)\, J_0(\omega x)](3)=\int_0^{\infty} \sin(\omega t)\, J_0(\omega x)\, \omega^{3-1} \, d\omega\\=-\frac{4 \sqrt{\...
Steven Clark's user avatar
  • 7,611
0 votes

$\beta(B)\subseteq C(X)$ is closed

We want to show $\beta(B)$ is closed in the topology on C(X) induced by the sup-norm. Let’s use the sequential characterisation of closed sets in metric spaces to prove this result. That is suppose we ...
Wjvr46's user avatar
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1 vote
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Intersection of spheres in normed space

The correct statement should be the spheres $S(x_1, r_1)$ and $S(x_2, r_2)$ intersect iff $|r_1 - r_2| \leq \|x_1 - x_2\| \leq r_1 + r_2$, under the assumption that the real dimension of $X$ is at ...
David Gao's user avatar
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0 votes

Weak derivative of $|x|^{-\lambda}$.

I'm pretty sure $r^{d-\lambda-1}$ is integrable on $(0, R)$ if $d-1-\lambda<0$, as long as $-1<d-\lambda-1$. Edit: Also, if $\lambda < d$, then $f(x) = |x|^{- \lambda} \in L^1_\mathrm{loc}(\...
Sebastián's user avatar
1 vote
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Does the closure of product of two ideals satisfy $\overline{I_1I_2}=\overline{I_1}\ \overline{I_2}$.

It suffices to show $\bar{I_1}\bar{I_2}$ is closed. Thus, it suffices to prove that, if $I_1, I_2 \subset A$ are two closed ideals, then $I_1I_2$ is closed. Recall that any closed ideal in a $C^\ast$-...
David Gao's user avatar
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2 votes
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Prove that for every $n \in \mathbb{N}$, the mapping $A_n$ is a bounded linear operator from $C[0,1]$ to $C[0,1]$ and calculate its norm.

I guess you mean to say "Therefore, $A_n(f)$ must be continuous for any $f\in C[0, 1]$". Your argument for this is fine: Obvoiusly $A_n(f)$ is continuous over $[0, \frac{2n}{2n+3}]$ and $[\...
Just a user's user avatar
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2 votes

$L^{\infty}$ (uniform) decay of Dirichlet heat equation $u_t=\Delta u$

Let $u_1, u_2, \dots \in C^{\infty}(\Omega)$ be an orthonormal basis of $L^2$ consisting of eigenfunctions of $-\Delta$ on $\Omega$. We have $$u(t) = e^{-t\Delta}u_0,$$ that is, $$u(t, x) = \sum_{j = ...
Kakashi's user avatar
  • 2,079
2 votes
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Uniqueness of solution of the operator equation $AXB=BXA$ on $B(H)$

There is a sufficient condition: $A$ and $B$ are injective (that's a necessary condition due to the comment of @Frederik vom Ende) and ${\rm Im}\,A\cap {\rm Im}\, B=\{0\}.$ If $AXB=BXA$ then $AXB=0=...
Ryszard Szwarc's user avatar
0 votes

Integral operator on $L^p$ is compact

Here is another solution to this old problem. Assume that $(X,\Sigma,\mu)$ is $\sigma$-finite and countably generated. That last assumption means that there is a countable collection $\mathcal{C}$ of ...
Mittens's user avatar
  • 39.9k
2 votes
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Proving a linear functional belongs to $L^{4/3}(0,T; H^{-1})$

To estimate $\|g\|_{H^{-1}}$ they (silently) took the supremum of the upper bounds of $\langle g,\phi\rangle$ over all $\phi$ with $\|\phi\|_{H^1} \le 1$, that is how the $\phi$-related terms ...
daw's user avatar
  • 49.9k
1 vote

$L^{\infty}$ (uniform) decay of Dirichlet heat equation $u_t=\Delta u$

WLOG we may assume that $0\in\Omega$ and consider the function $v(x,t)=e^{-bt}\displaystyle\prod\limits_{k=1}^N\cos cx_k$, where $b=c^2N>0$. It's easy to prove that $v_t=\Delta v$. Next, consider a ...
thing's user avatar
  • 1,700
2 votes

Defining operator as a series

Claim: $A$ is compact if, and only if, $\mu_i\xrightarrow{i\to \infty}0$ I am only showing the reverse direction here. So let $(\mu_i)_{i\in\mathbb N}$ be a zero sequence. Define: $$A_kx:=\sum_{i=1}^...
Konstruktor's user avatar
1 vote

Completeness of $\sin(kx)_{k=0}^{\infty}$ in $L_1[1, 4]$

We have $\sin n(\pi+x)=(-1)^n\sin nx.$ Let $$g(x)= \begin{cases} 1 & 2\pi -4\le x\le 4\\ 0 & {\rm otherwise} \end{cases}$$ Then $$\int\limits_1^4g(x)\sin nx\,dx =\int\limits_{2\pi-4}^4\sin ...
Ryszard Szwarc's user avatar
0 votes

Convergence in $L^{p}$ and convergence of derivative

This is not true, since we can take a non-zero $f\in C_c^{\infty}(\mathbb{R})$ and define $f_n(x)=n^{1/2}f(nx)$. Then, it follows that $$\lVert f_n \rVert_1=n^{-1/2}\lVert f \rVert_1 \to 0, $$ while $$...
Hidde's user avatar
  • 601
1 vote
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Operator continuous with range in weak topology iff continuous with range in the norm topology

Let's take a closer look at this application of the uniform boundedness principle. Recall the statement (taken from Wikipedia, lightly simplified): Uniform Boundedness Principle — Let $X$ be a Banach ...
Theo Bendit's user avatar
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1 vote
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On the compactness of the square of a finite double norm integration operator $T$ on $L^1 (\mu)$.

The proof that I know relies on the Eberlein-Smulain theorem, and to be more precise, on the Dunford-Pettois theorem. The issue is that to determine compactness of operators on Banach spaces, it is ...
Mittens's user avatar
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0 votes
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How to interpret $L^2$ norm for functions from $[0,T]\to\mathbb{R}^n$?

In general, when you say that $\alpha \in L^2(0,T,A)$, this means that $$ \int_0^T \parallel \alpha(t)\parallel_A^2 dt < \infty $$ So, based on the norm defined on $A$, you can check whether the ...
Lucio Rosi's user avatar
1 vote

image of an operator is closed iff its image of adjoint is closed

The proof follows directly from the polar decomposition $T=U|T|$ where $U$ is the partial isometry which maps isometrically $\overline{{\rm Im}\, |T|}$ onto $\overline{{\rm Im}\, T}$ and $T$ vanishes ...
Ryszard Szwarc's user avatar
2 votes
Accepted

$L_p$ inequality for measurable sets

By triangle inequality $$ \int |\chi_A - \chi_B| = \mu(A\setminus B) + \mu(B \setminus A) = \mu(A) - \mu(A \cap B) + \mu(B)-\mu(A\cap B) \le 2\epsilon $$ Then by Hoelder inequality $$ \|\chi_A - \...
daw's user avatar
  • 49.9k
2 votes

$L^\infty(\Omega)$ is dense in $L^{p,\infty}(\Omega)$ if $\Omega$ is compact

Actually, the closure of $L^\infty$ in $L^{p,\infty}$ is $$ L^{p,\infty}_0:=\left\{ f:\lim_{t\to\infty}t^p\mu\{x:\lvert f(x)\rvert>t\}=0 \right\}. $$ Indeed, if $f$ belongs to the closure of $L^\...
Davide Giraudo's user avatar
1 vote

$L^\infty(\Omega)$ is dense in $L^{p,\infty}(\Omega)$ if $\Omega$ is compact

This density is not true. Let $\Omega=(0,+1)$. Let $p\in (0,\infty)$ and define $$ f(x) = x^{-\frac1p}. $$ Then $$ \mu(\{x:\ |f(x)| > t\} ) = t^{-p} $$ for $t\ge1$, so that $\|f\|_{L^{p,\infty}} = ...
daw's user avatar
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0 votes

Questions about the proof of Poisson's formula for half-space in Page 38 of Evans' book

To elaborate on @Martin's answer: Notation: $f_{x_i}$ is $\frac{\partial f}{\partial x_i}$. $\overrightarrow{e}_i=(0,\cdots,0,1,0,\cdots,0)$ (the $i-$th coordinate being $1$, other being $0$). For ...
Asigan's user avatar
  • 1,910
2 votes

Example of nonnuclear Montel space

For Köthe sequence spaces $\lambda^p(A)$, i.e., countable intersections of weighted sequence spaces $$\ell^p(a_n)=\{(x_k)_k\in \mathbb R^{\mathbb N}: (a_n(k)x_k)_k\in\ell^p\},$$ all relevant ...
Jochen's user avatar
  • 12.3k
2 votes

Closed range of $T^*T$ for Hilbert space map.

For a positive operator $S:A\to A$ the range of $S^2$ is closed iff the range of $S$ is closed. Indeed $\ker S=\ker S^2$ Hence $V:=\overline{{\rm Im}\,S}=\overline{{\rm Im}\,S^2}.$ If the range of $S^...
Ryszard Szwarc's user avatar
2 votes

Closed range of $T^*T$ for Hilbert space map.

First suppose that $T$ is injective and has dense range. To show that $R(T)$ is closed, suppose that $Tx_n \to y$ in $H$. Then $T^* Tx_n \to T^* y \in R(T^*T)$ so that $T^*y =T^*Tx$ for some $x \in ...
Evangelopoulos Foivos's user avatar

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