All Questions
28 questions
0
votes
0
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59
views
Decomposition of a contractive representation into an orthogonal sum for the $n$-dimensional case. Has this been done yet?
I know that it has been done for the two-dimensional case. Marek Kosiek showed it in his work "Decomposition of operator representations of the algebra $R(K_1 \times K_2)$" and "...
1
vote
1
answer
151
views
Some operators on spheres
Let $S_2$ be the unit sphere in $\mathbb R^3$ equipped with normalized Haar measure. For a continuous function f and $\delta\in (-1,1)$ define $T_\delta f(x):=\int_{\{y:<x,y>=\delta\}}f(y)d_\...
2
votes
0
answers
94
views
Nemytskij operator for Lebesgue variable UNBOUNDED exponent spaces
Let $f:\Omega\times\mathbb{R}\to\mathbb{R}$ be a Caratheodory function (i.e. $f(x,\cdot)$ is continuous for a.a. $x\in\Omega$ and $f(\cdot,t)$ is measurable for all $t\in\mathbb{R}$), where $\Omega\...
0
votes
0
answers
49
views
Existence of sequence of regular projections
Reading the book :Krasnosel'skii, M.A.; Pustylnik, E.I.; Sobolevskii, P.E.; Zabreiko, P.P. (1976), Integral Operators in Spaces of Summable Functions, Leyden: Noordhoff International Publishing, 520 p....
2
votes
1
answer
149
views
Show that $\|P(f\circ\varphi_{\lambda})-\widetilde{f}(\lambda)\|_p=\|P(f\circ\varphi_{\lambda}-\overline{P(\overline{f}\circ\varphi_{\lambda}}))\|_p.$
Let $\Omega = \mathbb B_n,$ the unit ball in $\mathbb C^n$ and $L^2_a(\Omega)$ be the Bergman space endowed with the normalized volume measure on $\Omega.$ Let $k_{\lambda}$ be the associated Bergman ...
2
votes
1
answer
139
views
Domain of the infinitesimal generator of a composition $C_0$-semigroup
In the paper [1] the following $C_0$-group is presented,
$$
T(t)f(x) = f(e^{-t} x) , \quad x \in (0,\infty) \quad f \in E
$$
where $E$ is an ($L^1,L^\infty$)-interpolation space. In mi case, I'm just ...
1
vote
0
answers
101
views
If we have $\mu_{xy}$, why can we only construct the spectral measure if $\| \mu_{xy} \| \le \| x \| \|y \|$?
Definitions
Representation
Let $X \subset \mathbb{C}^N$ and $\mathcal{A}$ be an algebra in $\mathcal{C}(X)$. Also, we denote $L(H)$ as the set of all linear operators on HIlbert space $H$.
We call $\...
1
vote
1
answer
80
views
Prove that $N_1(A_1,A_2)= N_2(A_1,A_2)$ when $A_1A_2=A_2A_1$ and $A_1$ and $A_2$ are normal operators
Let $\mathcal{L}(E)$ be the algebra of all bounded linear operators on a complex Hilbert space $E$.
On $\mathcal{L}(E)^2$, we have two equivalent norms:
\begin{eqnarray*}
N_1(A_1,A_2)
&=&\sup\...
1
vote
0
answers
135
views
Description of state space of $C(K,M_n)$?
Edit: closed convex hull added.
I am trying to understand the state space of $C(K,M_n)=C(K)\otimes M_n$ for $K$ a compact space.
My guess would be that these are the closed convex hull of states on $C(...
4
votes
1
answer
330
views
Uniform boundedness principle for almost surely converging sequence of operators
I'd like to do the following: I consider a separable Banach space $X$ with a probability measure $\mu$ on the Borel $\sigma$-algebra $\mathcal B(X)$. Additionally, I have a sequence of measurable, ...
3
votes
0
answers
101
views
Pettis vs. Dunford integrability of operator valued functions
Given a Banach space $X$ and a measure space $(\Omega ,\mu )$, one says that a function
$$
f:\Omega \to X
$$
is Dunford integrable, or scalarly integrable if, for every $\varphi $ in the ...
2
votes
1
answer
135
views
A non-condensing operator with a power condensing
Let $\alpha$ to be the Kuratowski measure of non-compactness, in a Banach space $E$.
It's very easy to prove that $\alpha (D_1\times D_2)\leq \alpha (D_1)+\alpha (D_2)$, where $D_1$ and $D_2$ are ...
2
votes
0
answers
40
views
Invariance of simple functions
Let $(A(s),D(A(s)))_{s\in\mathbb{R}}$ be a family of unbounded operators on a Banach space $X$ and $g:\mathbb{R}\rightarrow X$ be a simple function, i.e.,
\begin{align*}
g=\sum_{i=1}^n{x_i\textbf{1}_{...
3
votes
1
answer
274
views
Function square-integrable
Let $f$ be an arbitrary function in $L^2(0,\infty)$ and consider the function
$$(g_f)(y) = \frac{1}{y-x_0} \int_{0}^{\infty} f(x) \left(\frac{xy}{(x^2+y^2+1)}\right)^2 \ dx$$
where $x_0$ is an ...
2
votes
2
answers
431
views
References for the Spectral Theorem ( Multiplication Operator Form)
Let $A_1$ and $A_2$ be two commuting self-adjoint (or normal) operators on an infinite-dimensional complex Hilbert space $E$, then there exists a measure space $(X,\mathcal{E},\mu)$,
two functions $\...
1
vote
1
answer
165
views
Integral function $z(x):=\int_{Y} f(x,y)d\mu(y)$ continuous?
Let $z(x):=\int_{Y} f(x,y)d\mu(y)$ for $x \in \mathbb R$ be an integral function where $\mu$ is a finite(!) Borel measure on $Y$ and $x \mapsto f(x,y)$ is continuous for every $y.$
Moreover, we know ...
2
votes
1
answer
140
views
Measurability of $T \to \Pi_{ker T}$ w.r.t. SOT
I came across the following technical question, to which I could not - after some time of thinking - find an answer:
Let $\mathcal{U},\mathcal{H}$ be two real (in general infinite dimensional) ...
2
votes
0
answers
58
views
Absolute continuity of DOS measure for Schrödinger operators
Kotani theory gives roughly that for ergodic operators there is a certain equivalence between absolutely continuous spectrum and an absolutely continuous density of states measure.
I would like to ...
1
vote
1
answer
287
views
Interpolation between $L^1$ and $L^2$ spaces
I was wondering whether the following interpolation between $L^1$ and $L^2$ spaces is true:
Let $f \in \mathbb{R}^n$ be such that
$$ \alpha_1:= \int_{\mathbb{R}} \left\lVert f(x_1,\cdot,....\cdot) \...
11
votes
3
answers
3k
views
Dual space of $L^2(\mathbb{R},L^1(0,1))$?
I was wondering what the dual space of $L^2(\mathbb{R},L^1(0,1))$ is? (equipped with Lebesgue measures)
Formally, one would suspect that it is just $L^2(\mathbb{R},L^{\infty}(0,1))$. But this may be a ...
3
votes
0
answers
225
views
Defining a trace-class operator with a Bochner integral
I had asked this question previously on Math.StacheExchange but did not get an answer there in several months. This isn't strictly speaking research level mathematics but I hope it is sufficiently ...
4
votes
1
answer
2k
views
Operator topologies on $L^{\infty}(X,\mu )$
Let $(X,\mu )$ be a measure space. Then, $L^2(X):=L^2(X,\mu )$ is a Hilbert space in the usual way and we may view $L^{\infty}(X):=L^{\infty}(X,\mu )$ as a subalgebra of bounded operators on $L^2(X)$ ...
4
votes
1
answer
283
views
Absolutely continuity in variation of constant formula
We are talking here about the initial value problem on some Hilbert space $H$
$$y'(t)=Ay(t)+f(t), \\ y(0)=y_0 \in D(A).$$(Problem 1.13 in the reference)
Then $y(t)=e^{At}y_0 + \int_0^t e^{A(t-s)}f(s) ...
4
votes
1
answer
313
views
Invariant subspaces are reducing subspaces in $L^2(\mu)$; where $\mu$ is a singular measure w.r.t Lebesgue measure
I have already posted this question on math.stackexchange but didn't get any answer. I hope this is the right place to ask this question.
Recently I was reading a book "Operator Function and system" ...
7
votes
1
answer
306
views
An indicator of a planar subset as an element of a tensor product
Denote $I=(0, 1)$, and let $\mu$ be the Lebesgue measure on $I$. Does there exist a function $f$ on $I\times I$ viewed as an element of the space $L^\infty(\mu\times\mu)$ such that
$$
f^2=f
$$
(that ...
3
votes
1
answer
216
views
Extension of a bilinear functional
Does any one know an example of a bilinear functional $B:C(X)\times C(Y)\to {\bf R}$ ($X$ and $Y$ are open subsets of Euclid spaces) which cannot be extended continuously to a measure $\mu:C(X\times Y)...
2
votes
1
answer
577
views
When is the bound in Riesz-Thorin Interpolation Theorem attained?
Let me recall the statement of Riesz-Thorin theorem (see also http://en.wikipedia.org/wiki/Riesz%E2%80%93Thorin_theorem).
Theorem (Riesz-Thorin): Let $(X,\mu)$ and $(Y,\nu)$ be $\sigma$-finite ...
5
votes
1
answer
2k
views
definition of operator valued integral with spectral measure
I am trying to make sense of some operators that come up on Buchholz and Summers' work on warped convolutions (two works on arxiv: 2008 and 2011).
There, they work on a Hilbert space $H$ and on the ...