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3 votes
1 answer
75 views

What is wrong with this proof that a linear, bounded, time invariant operator on $L_p$ must be a convolution?

I'm trying to understand if this is true and how to prove it, "If $T$ is a bounded, time invariant operator on $L_p(\mathbb{R})$, then $T$ is a convolution operator.'' Here's an attempt at a ...
travelingbones's user avatar
4 votes
1 answer
150 views

Proving that operator in $L^2[0,1]$ is compact

I need help with some functional analysis: Let $A$ be a continuous linear operator on $L^2[0,1]$ and for any $f \in L^2[0,1]$ the function $Af$ is Lipschitz continuous. Show that $A$ is compact. It is ...
Plumbum's user avatar
  • 51
1 vote
0 answers
31 views

Boundedness of an Integral Operator on $L^p(\mu)$

I realize this question has been asked way too many times, for instance, it's this exact problem which I will put below for convivence: Let $(X,\Omega, \mu)$ be a $\sigma$-finite measure space with $...
Isochron's user avatar
  • 1,399
1 vote
1 answer
34 views

Finding the conjugate operator of the following operator

Let $A$ be linear operator from $l^2$ to $l^2$ such that $Ax =y^0 \cdot \sum_{1}^{\infty}{x_k} $ , where $y^0 \in l^2$ — fixed element. Show, that conjugate operator $A^*$ exists and find it. Show, ...
Metal Sonic's user avatar
-1 votes
1 answer
35 views

Some intersection of Schatten spaces gives a weak Schatten space?

Assume $1\leq p <\infty$. It is known that for all $q>p$ $$ S^{p} \subset S^{p,\infty} \subset S^{q}. $$ Out by curioisity, it is true that $\cap_{q>p} S^{q}=S^{p,\infty}$ ? Recall that $$ ...
Liam's user avatar
  • 323
2 votes
3 answers
319 views

How to compute the numerical radius of the right shift operator?

Let $T$ be the right shift operator on $\ell^2$ defined by $T(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$. The numerical radius of $T$ is defined by $w(T)=\sup\{|\langle Th,h\rangle|:\, \|h\|=1\}$. It is well ...
DeltaEpsilon's user avatar
  • 1,130
2 votes
2 answers
114 views

Estimate norm of convolution operator

I'm trying to find the operator norm for $T: L^2([0,1])\to L^2([0,1])$, defined as $Tf(x)=\int_{[0,1]}|\sin(x-y)|^{-\alpha}f(y)dy$, where $0<\alpha<1$. Using an upper bound on $|\sin(x)|\geq |x|/...
mtcicero's user avatar
  • 529
1 vote
1 answer
58 views

Showing that the integral operator $Ku(x):=\int_{A}k(x,y)u(y)dy$ is a Hilbert-Schmidt operator if $k\in L^2(A^2)$ for $A\subset\mathbb{R}$

Let $A\subset\mathbb{R}$ and $k\in L^2(A^2;\mathbb{C})$. I am trying to understand how we can conclude that the integral operator $K$ defined to act on $L^2(A;\mathbb{C})$ by $Ku(x) := \int_Ak(x,y)u(y)...
Cartesian Bear's user avatar
2 votes
1 answer
74 views

For which $p\in [1,+\infty]$ is the following operator continuous?

For each $n\in \Bbb N$, $n\ge1$ and for each $x\in \Bbb R$, consider the following operator: $T_n: L^2(\Bbb R) \to L^p(\Bbb R) $ defined as $$T_nf(x)=n^{3/4} \int_x^{x+1/n} f(t)\mathrm dt $$ The ...
Sine of the Time's user avatar
1 vote
0 answers
35 views

If the multiplication operator has full domain, the multiplier is essentially bounded.

From: Norm of multiplication operator in $L_p$ [NOTE: Unlike this question, here, $p$ can be $\infty$.] Let $Ω$ be an open set in $R^n$, and let $a$ be a measurable complex-valued function on $Ω$. The ...
isz's user avatar
  • 31
2 votes
1 answer
90 views

Calculating operator norm of operator $(Ax)(k) = \int_{0}^{1} x(t)\operatorname{arcctg}\left((-1)^kt\right) \,\mathrm{d}t$

I have no idea how to solve this: $$A: CL_1[0,1] \rightarrow \ell_\infty$$ $$(Ax)(k) = \int_{0}^{1} x(t)\operatorname{arcctg}\left((-1)^kt\right) \,\mathrm{d}t,\quad\forall x \in CL_1[0,1],\quad\...
worst functional analytic's user avatar
0 votes
3 answers
115 views

Brezis' exercise 8.28.4: how to prove that $\langle Tf, f \rangle \ge 0$?

Let $I$ be the open interval $(0, 1)$ and $H := L^2 (I)$ equipped with the usual inner product $\langle \cdot, \cdot \rangle$. Consider the linear map $T: H \to H$ defined by $$ (Tf) (x) = \int_0^x t ...
Akira's user avatar
  • 17.6k
2 votes
1 answer
56 views

Must the domain of a Hilbert space operator be limited to square-summable elements? If so, what type of space does not restrict, thus?

Background: The operation of bandlimiting is a projection operator, $P$, into a separable Hilbert space. For example, with inner product, $\int_{-1/2}^{1/2}g(f)\overline{h(f)}df$, there is an ...
EEatWork's user avatar
1 vote
2 answers
113 views

The norm, resolvent set, and adjoint of the operator $T(x_1, x_2, \ldots) = (x_2, x_3, \ldots)$ on $\ell_1$

Let $T$ be the operator on $\ell_1$ such that $$T(x_1, x_2, \ldots) = (x_2, x_3, \ldots).$$ I am going through an example in Reed & Simon's book on functional analysis which makes the following ...
CBBAM's user avatar
  • 6,275
2 votes
1 answer
43 views

For which values of the parameter $r$ the operator is compact?

I have an operator $$A : L_{2}[0,1] \to L_{2}[0,1]$$ which is $$(Ax)(t)={t^{r-1}}\int_{0}^{t}\frac{x(s)}{s^r}ds$$ I already proved that operator A is bounded in $L_{p}[0,1]$ space when $p\in(1,\infty)$...
Alisa Libon's user avatar

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