# Tag Info

## New answers tagged fa.functional-analysis

6

The answer is no. E.g., let $f(x):=\min(1,|x|)x$ for real $x$. Then $f$ is Lipschitz and strictly monotone. Let then $g$ be any function in $BV$ such that $g(1/n)=(-1)^n/n^2$ for all natural $n$; it is easy to see that such a function exists. (For instance, let $g=0$ on $(-\infty,0]\cup(1,\infty)$ and let $g$ be monotonic on $[\frac1{n+1},\frac1n]$ for each ...

1

Sure one can! Just note that $f$ can be written as the "convolution" of $-g$ with an appropriate integrable kernel, the Green function $G(x)$ for the flat torus: $$f(x) = -\int_\Omega g(y) G(x - y) dy.$$ The Green function for a flat torus is defined as follows. Pick a smooth, symmetric, compactly supported function $\phi$ such that the ...

0


2

Let me write out the equation $J''(u)w = g$ for $g\in H^{-1}$ and $w\in H^1_0(\Omega)$. This is equivalent to $$\int_\Omega \nabla w \cdot \nabla v - f''(u)wv = g(v) \quad \forall v\in H^1_0(\Omega).$$ This is the weak formulation of $$-\Delta w - f''(u)w = g$$ plus boundary conditions. To show existence of solutions, you need some assumptions on $f''$ to ...

0

The answer is yes. Suppose $g$ is orthogonal to the image of $S$, and let $v$ be the solution of the Dirichlet problem $\Delta v=g\delta(\partial D_1)$ on $D_2$, where $\delta(\partial D_1)$ is a delta function localized on $\partial D_1$. We find $$\int_{\partial D_1} gu\,dS=\int_{D_2}v\Delta u\,dx.$$ Now suppose this is true for every $u$ for which $\Delta ... 1 The answer is yes: take any smooth function$g_0$on$\partial D_1$and solve the Dirichlet problem $$\begin{cases} \Delta g = 0 & \text{ on } D_1\\ g = g_0 & \text{ on } \partial D_1. \end{cases}$$ Now extend$g$to a smooth function on$\mathbb{R}^n$. Multiply by a smooth cutoff function$\eta$which is$1$on$D_1$and compactly supported on$...

1

This paper concerns with your problem.

4

We probably don't need another answer, but here it is anyway, especially since Giorgio raised the question about projections. Lemma: There is an ONB $\{ v_j\}$ of $\mathbb C^n$ such that $|\langle e_1, v_j\rangle |= n^{-1/2}$ for all $j=1,2, \ldots , n$. Proof: This is trivial: it works if we replace $e_1$ by $w=(n^{-1/2}, \ldots, n^{-1/2})$ and use the ...

4

Let $$f_n(x) = \sup_{r \in {\mathbb Q} \cap [\frac{1}{n+1},\frac{1}{n})} \frac{|B(x,r)\setminus E|}{|B(x,r)|}\,,$$ so that $f_n(x) \to 0$ for a.e. $x \in E$. By Egorov's theorem [1], for every $\epsilon>0$ there is a subset $\tilde{E} \subset E$ with $|E \setminus \tilde{E}| <\epsilon$, such that $f_n(x) \to 0$ uniformly on $\tilde{E}$. It follows ...

4

This is another example. Consider the Hardy operator in $L^2(0,1)$, $$Hf(x)=\frac{1}{x}\int_0^x f(t)\, dt,$$ which is bounded by Hardy inequality. If $f_n=\sqrt{n} \chi_{[0,1/n]}$, then $\|f_n\|_2=1$ and $f_n\to 0$ weakly but $\|Hf_n\|_2 \geq 1$ so that $H$ is not compact. If $g_n(x)=\sin (n\pi x)$, then $(g_n)$ is an orthonormal basis but $Hg_n(x)=\frac{2 \... 13$T$is not necessarily compact. Let me produce a counterexample. Let$H$be any infinite dimensional real or complex separable Hilbert space. Let$(f_{j,k})_{1\leq k\leq j},(e_{j})_{j=1}^{\infty}$be orthonormal bases for$H$. Then let$T:H\rightarrow H$be the bounded linear operator defined by letting$T(f_{j,k})=\frac{1}{\sqrt{j}}\cdot e_{j}$whenever$1\...

3

You need to think about what Theorem 3.5 tells you! Firsly, Defintion 2.13 in that chapter tells us what a normal functional is. If $M \subseteq B(H)$ is a von Neumann algebra, then $\omega$ is a normal functional exactly (by definition) when $\omega$ is $\sigma$-weakly continuous. Now Theorem 3.5 tells us that if $M$ is a $W^*$-algebra with (as you say, ...

1

In Hilbert spaces the situation is crystal clear by singular value decomposition: If $K$ is compact between two Hilbert spaces then one can write $K$ as $$Kx = \sum \lambda_n\langle v_n,x\rangle u_n$$ where $\lambda_n>0$, $(u_n)$ is an orthonormal basis for the range of $K$, and $(v_n)$ is an orthonormal basis for orthogonal complement of the null space ...

4

The following ramblings are just night thoughts kicked off by your question (and so should really be a comment, but I am not entitled) which I am posting in the hope that they may be of interest to you. Firstly, your operator $K$ has more properties than you mention. It is not just a right inverse—it is, as near as can be, also a left inverse, just missing ...

10

Well, if $A$ is bounded and $B$ is compact then $AB$ is compact, so $AB$ cannot be the identity unless the Banach space is finite dimensional. Thus the left (or right) inverse of a compact operator on an infinite dimensional Banach space, if it exists, must be unbounded. There's no implication in the other direction --- the inverse of an unbounded operator ...

0

Answer I am interested in forming a factorization for $f(z) = ({\bar{z} -a})^{-1} \cdot ({z -b}^{-1})$ of the following form $$f(z) = \sum_{k=0}^\infty g_k(a,b) h_k(z, \overline{z}),$$ where $a, b \in \mathbb{C}$, $|{a}/{z}|, |{b}/{z}| <1$, $g_k$ depends only on $a$ and $b$, and $h_k$ depends only on $z$ and $\overline{z}$. We ...

7

Historically, bases in functional analysis arose from reflection on a collection of concrete examples, such as trigonometric functions, orthogonal polynomials, spherical harmonics, eigenfunctions of the Laplacian, etc. What's in common in these examples is that each appeared as a tool to solve a concrete problem by diagonalizing an operator. E. g. you know ...

2

I understand that you are happy with the case $y\geq 0$, so let's assume that for simplicity (I expect that small modifications should deal with arbitrary self-adjoint $y$). In that case, and if $1<p<\infty$, it is true that $\| u_n^* y - y\|_p \to 0$. I am not sure about the extreme cases $p=1,\infty$, where the uniform convexity argument breaks down. ...

4

This is an addition to Yemon Choi's answer. In fact, a real-valued function $f$ defined on a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$, for instance, extends uniquely to ad additive function, not to a linear function (it only needs to be linear over $\mathbb{Q}$). This can be seen readily defining $$F(x)=\sum_{i=1}^n q_i f(b_i),$$ where $b_i\in H$ ...

1

You haven't really defined $(d/dy)^{-1}$, but let's pick, for instance, $(d/dy)^{-1} f(y) = \int_{0}^{y} dy' f(y')$. Then, your two expressions for $H(y)$ in general are not equal. A simple counterexample is the case $T(y)=y$. In this case, \begin{eqnarray} \frac{1}{y^{-n} (d/dy)^{-n} e^{-(d/dy)y(d/dy)} T(y)} &=& \frac{1}{y^{-n} (d/dy)^{-n} (y-1)} \\...

11

The following perhaps could be a comment, but I thought that maybe I should state it as an answer so it can get positive or negative feedback. To my mind, whenever I have taught linear algebra in the past, I emphasises that the great power of having a basis B for a vector space V, to study a linear map T from V to some other vector space W it is enough to ...

6

So, we are dealing with locally convex topological vector spaces. I think, in general, given a family of seminorms, you would need to consider the finite intersections of the open balls they form, see wikipedia article. So a basic open set about $0$ is of the form $$\{ x : p(x)<r \ (p\in F) \}$$ where $F$ is a finite subset of the set $\mathcal P$ of ...

10

If $f$ is bounded on the imaginary line, (and has exponential type) then $f$ has completely regular growth in the sense of Levin-Pfluger, with indicator $c|\cos\theta|$. This implies that density of zeros on the positive ray must be zero. Moreover, density of zeros in any angle $|\arg z|<\pi/2-\epsilon$ and in the vertical angle is zero. Boundedness on ...

2

If $(d/dy)g(y)=g’(y)$, so no operator commutator, then $$e^{d/dy}g(y)=g(y+1),$$ so $$f(y)=y^{-n}/g(y+1).$$ If instead you choose the operator identity $(d/dy)g(y)=g’(y)+g(y)d/dy$, then $$f(y)=(1/g(y))e^{-d/dy}y^{-n}.$$

2

Actually the above three conditions yielding a counterexample are satisfied by any counterexample and this is easy to be checked. However such a space has extremely peculiar structure. In particular it does not contain any subspace of the form Y + Z with Y Hilbertian and Z indecomposable. This follows from the previous positive partial answer. That means ...

2

I think that the following provides a partial positive answer to Question I. Fact If X is separable, Hereditarily Prime and Decomposable then it is isomorphic to a Hilbert space. The proof goes as follows. We write X as V + W with both of infinite dimension.Also both are isomorphic to the space X. Let Z be a subspace of V and Y = Z + W. Then W is a ...

2

In addition to @PietroMajer's good answer, I'd want to make a point about "vector-valued integrals" (and related), that the Bochner integral gives a construction (of something we want, with certain obvious/natural properties), but/and we have to prove that this construction succeeds. Oppositely, we can go the Gelfand-Pettis route, and "define&...

4

I understand you assume that $T:[0,c]\to\mathcal{B(H)}$ is Bochner integrable in order to write $\int_0^c T(t)dt$ as Bochner integral. Then for any $x\in\mathcal H$ the map $[0,c]\ni t\mapsto \mathcal H$ is also Bochner integrable and the identity you wrote holds. More generally: for a measure situation $(X,\mathcal S,\mu)$, a couple of B-spaces $\mathbb E$...

1

The purpose of this note is not to answer my own question, but to share a sufficient condition with everyone: If $A$ is amenable, then Q1 & Q2 are answered affirmatively. Notation: $A\hat{\otimes}_{\pi} A$ is the projective tensor product of $A$ with itself. For an $A$-bimodule $M$, the bimodule center is defined by \$\mathcal{Z}(A,M) =\{\beta\in M: a\...

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