## 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

Let us set $L_1=\sum_{1\le j\le m}X_j^2$, where the $X_j$ are smooth real vector fields satisfying Hörmander's condition so that $L_1$ is subelliptic with an estimate
$$
\Vert L_1 u\Vert_{H^s}\ge C\Vert u\Vert_{H^{s+\frac{2}{k+1}}},
$$
for any $u\in C^\infty_c$ with a diameter of the support small enough included in a fixed compact set.
We define $L_p=(L_1)^...

3

Let us discretise the problem by setting $a_n=2^{-n}f(2^n)$, $b_n=2^{-n-1}\Delta_f(2^n)$. Then your relation becomes,
$$b_n=a_n-a_{n+1}.$$
since $a_n,b_n$ are non-negative, we conclude that
$$\sum_{n=1}^\infty b_n<\infty.$$
This is a necessary and sufficient condition. Indeed, take any summable sequence $b_n$ of positive numbers, then we can define
$$a_n=\...

answered 22 hours ago

Alexandre Eremenko

72.4k77 gold badges210210 silver badges346346 bronze badges

2

Generalising your example slightly to $f(x) := \frac{x}{g(x)}$ where $g$ is increasing (but sublinear) we get $$\Delta_f(x) := 2 \frac{x}{g(x)} - \frac{2x}{g(2x)} = 2x \left( \frac{1}{g(x)} - \frac{1}{g(2x)} \right)$$ so by picking suitable $g$ it should be possible to get arbitrarily close to linear growth. Consider, for example, composing the inverse ...

5

There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. We use the notation of the post, and assume that $T$ is ergodic. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part ...

1

I asume $M$ is a surface and $SM$ is the unit circle bundle. The bundle $SM\to M$ is also a principal $S^1$-bundle and the obvious $S^1$ action on $SM$ that preserves the natural metric on $SM$. $\newcommand{\ii}{\boldsymbol{i}}$ For $p\in SM$ and $e^{\ii t}\in S^1$ denote by $e^{\ii t}p$ the action of $e^{\ii t}$ on $p$. For $\newcommand{\bR}{\...

2

This is just fleshing out my earlier comment. Here's the proof that if the maximal ergodic theorem holds, then the set of $L^1$ functions for which almost everywhere convergence holds is closed.
Let $S=\{f\in L^1\colon P_nf(x)\text{ converges $\mu$-a.e.}\}$.
Define $Mf(x)=\sup_n |P_nf(x)|$ and assume that a maximal inequality holds in the form $\mu(\{x\colon ...

7

Nik Weaver's answer gives a nice counter-example. Let me just say a few words of context. Kernels $K$ for which $KT$ is trace-class for all trace-class $T$ are called Schur multipliers. (Not to be confused with an unrelated, and more common, term from group theory). I believe this is because of Schur's 1911 paper (Crelle's journl, in German).
A more ...

8

It's a little more complicated than I thought! Frederik Ravn Klausen pointed out an error. Still, I maintain that the product needn't even be bounded.
As the answer to this question shows, in $M_n$ you can find a unitary $U$ and an matrix $A$ whose entries all have modulus 1, whose Hadamard product $A\bullet U$ has operator norm $\sqrt{n}$. Using the duality ...

4

The point $x_\mu$ is the barycentre of the measure $\mu$. Its existence and uniqueness is Proposition 1.1 in "Lectures on Choquet's theorem" by Phelps.

3

Yes. This follows from the uniform convexity of the Schatten $p$ classes.
Indeed, for every unitary diagonal operator $u$, we have $\| (x + u x u^*)/2\|_p \geq \| E( x+uxu^*)/2\|_p = \|E(x)\|p \geq 1-\delta$ ($E$ is a contraction). Therefore, we have $\|x - uxu^*\|_p\leq\varepsilon$ (where $\varepsilon = \varepsilon(p,\delta)$ is given by uniform convexity, ...

12

The answer is no. Let $\ell^\infty$ be the space of bounded sequences equipped with the $\sup$ norm, and $c_0$ its closed subspace of sequences convergent to zero. The quotient $\ell^\infty/c_0$ is then a Banach space when equipped with the quotient norm, and it is consistent that it does not have any nonzero continuous functionals - see a discussion in this ...

13

No, this is equivalent to the Hahn–Banach theorem already in the case of Banach spaces.
See in my write up: https://arxiv.org/abs/2010.15632

0

This is not true in the case of $H^{-1}(\Omega) = H_0^1(\Omega)^*$ (real spaces), where $\Omega \subset \mathbb R^d$ is bounded:
In https://math.stackexchange.com/questions/336834/decomposition-of-functionals-on-sobolev-spaces, we see that $|\psi|$ cannot be defined for all $\psi \in H^{-1}(\Omega)$.
Even if $|\psi|$ can be defined (as a measure), it might ...

2

The definition makes no sense due to the mixing up of "relatively" (weakly) compact and (not relatively) compact.
I guess that what you mean is:
$\psi$ is strongly-weakly proper on closed balls (that is, the intersection of preimages of weakly compact sets with closed balls are compact). For strongly-weakly continuous maps and strong/weak measures ...

9

Corollary 5.4 in my book (Lipschitz Algebras, 2nd edition): Let $V$ be a Banach space. Then there is a norm 1 linear projection from ${\rm Lip}_0(V)$ onto $V^*$. If $V$ is separable then there is a weak* continuous norm 1 linear projection.
This follows from Theorem 2 of "On nonlinear projections in Banach spaces" by Lindenstrauss (Michigan Math J ...

2

This does not answer the question asked (see the other answer for a good counterexample) and I don't know if it's relevant to the paper. However, if $u$ is sufficiently far from $\varphi_1$, the first eigenfucntion, you do get a positive answer. Below I assume $\Omega$ is connected (you'd have to work out what happens more carefully if it's not).
First, we ...

2

The first is not true, and probably also the others.
Take $L^2(0, \pi)$ and $u_1=\sin x$, $u_2=\sin (2x)$, so that $E_2=\{u=a\sin x+b \sin (2x)\}$ and $u \geq 0$ iff $a \geq 0$ and $2|b| \leq a$. If $v=\alpha sin x+\beta \sin (2x)$, then $\|u-v\|_2^2=\frac{\pi}{2} \left((a-\alpha)^2+(b-\beta)^2\right)$ and, if $v_\epsilon=\sin x-\frac{1+\epsilon}{2}\sin (2x)$...

3

Set $c_n:=\frac{1}{\alpha_n}$. Clearly, for any $N>0$ one must have $\DeclareMathOperator{\spa}{span}$ $\DeclareMathOperator{\conv}{conv}$ $\newcommand{\bsB}{\boldsymbol{B}}$
$$
\bsB_1\cap\spa\{e_1,\dotsc, e_N\} \subset\conv\{ \pm\alpha_1 e_1,\dotsc,\pm\alpha_N e_n\}.
$$
This happens if and only if
$$
f_N(\alpha):=\min\{ x_1^2+\cdots +x_N^2;\;\;c_1x_1+\...

3

$\newcommand\tk{\tilde k}\newcommand\ip[2]{\langle #1,#2\rangle}$Let $k$ be a reproducing kernel of a reproducing kernel Hilbert space (RKHS) $H:=\mathcal H$ of real-valued functions on a set $X$. Then
$$\ip f{k_x}=f(x)\tag{1}$$ and
$$k(x,y)=\ip{k_x}{k_y}=k_x(y)\tag{2}$$
for all $f\in H$ and all $x$ and $y$ in $X$, where $\ip\cdot\cdot$ is the inner product ...

5

The spaces $[0,1]$, $[0,1]^2$, and $S^1$ are all isomorphic as measurable spaces, including their sets of measure 0, as required by the Gelfand-type duality for measurable spaces.
For instance, the isomorphism $[0,1]→S^1$ is given by identifying $S^1=[0,1]/(0∼1)$. Since the points $0$ and $1$ have measure 0,
and maps that differ on a set of measure 0 are ...

6

For a category $\mathcal{C}$, let $\mathcal{C}'$ denote the full subcategory of $\mathcal{C}$ whose objects are the non-terminal objects of $\mathcal{C}$.
In a category, say that an object $Y$ is final if for every object $X$ there exists an epimorphism $X\to Y$.
In turn, say that an object of $\mathcal{C}$ is pre-final if it is a final object of $\mathcal{C}...

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.

reference-request fa.functional-analysis ap.analysis-of-pdes ca.classical-analysis-and-odes elliptic-pde

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, ...

fa.functional-analysis oa.operator-algebras operator-theory von-neumann-algebras topological-vector-spaces

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
\begin{equation}
f(z) = \sum_{k=0}^\infty g_k(a,b) h_k(z, \overline{z}),
\end{equation}
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 ...

fa.functional-analysis oa.operator-algebras operator-theory von-neumann-algebras topological-vector-spaces

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 ...

answered Sep 29 at 17:20

Alexandre Eremenko

72.4k77 gold badges210210 silver badges346346 bronze badges

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
$$f(y)=(1/g(y))e^{-d/dy}y^{-n}.$$

answered Sep 29 at 13:31

Carlo Beenakker

134k1212 gold badges325325 silver badges464464 bronze badges

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\...

Top 50 recent answers are included

#### Related Tags

fa.functional-analysis × 7590real-analysis × 1053

ap.analysis-of-pdes × 1032

banach-spaces × 954

reference-request × 814

operator-theory × 753

oa.operator-algebras × 632

measure-theory × 514

sobolev-spaces × 484

pr.probability × 466

ca.classical-analysis-and-odes × 410

hilbert-spaces × 358

gn.general-topology × 343

harmonic-analysis × 327

fourier-analysis × 300

sp.spectral-theory × 297

dg.differential-geometry × 265

c-star-algebras × 256

linear-algebra × 244

cv.complex-variables × 209

banach-algebras × 206

mp.mathematical-physics × 176

elliptic-pde × 169

topological-vector-spaces × 159

mg.metric-geometry × 151