Questions tagged [compact-operators]
A compact operator is an operator from normed space $X$ to a normed space $Y$, such that image of every bounded subset of $X$ is relatively compact in $Y$. It's used with (functional-analysis) and (operator-theory) tags.
1,394 questions
0
votes
0
answers
22
views
Compact operator condition
In John B. Conway's "A Course in Functional Analysis" on pg 46. #6 says "Show that if $T: H \rightarrow K$ is a compact operator and $\{e_n\}$ is any orthonormal sequence in H, then $||...
- 377
2
votes
1
answer
104
views
Compact operators in Hilbert space
I have tried to solve this exercise:
Let $H$ be a Hilbert space and let $T = T^* \in B_{\infty}(H)$ a compact operator. Given $\psi_0 \in H$ consider the equations:
\begin{align*}
(1)\quad &T\psi =...
7
votes
1
answer
224
views
Equivalent formulations of the approximation property for Banach spaces
We say that a Banach space $Y$ satisfies approximation property, if for any Banach space $X$, any compact operator $K\colon X\to Y$ and any $\varepsilon>0$, there exists an operator of finite rank $...
- 71
5
votes
1
answer
369
views
Exercise 1.9 in Topics of Banach Space Theory (Albiac)
I was able to solve part a using his hint, I need help with part b. I will write my solution for part a and an idea for part b, which is probably gibberish: I tried fixing it for a long time to no ...
- 3,801
8
votes
1
answer
152
views
Does the norm decreases, along the limit on compactly embedded Banach subspace?
My question is the following:
Let $X, Y$ be Banach spaces such that $X$ is compactly embedded in $Y$. Let $\{f_n\}_{n \ge 1}$ be a sequence in $X$, and let $f \in X$, such that $f_n \to f$ in $Y$.
...
- 762
10
votes
1
answer
409
views
Norm estimate for an injective compact operator
Let $(X,\| \cdot \|_{X}),(Y,\| \cdot\|_{Y})$ be Banach spaces, $X$ not reflexive, and let $T:X \rightarrow Y$ be a compact operator that is injective. Further let $|\cdot|$ be another norm on $X$ that ...
- 187
6
votes
0
answers
140
views
How to find the spectrum of $T+T^*$ for this compact operator $T$?
The question is related to this stackexchange question
We fix a parameter $0<q<1$. Let $l^2=l^2_{\geq 1}$. We define an operator $T$ on $l^2$ by
$$
T(e_n)=q^n\sqrt{1-q^{2n}}e_{n+1}.
$$
Its ...
- 1,214
0
votes
2
answers
39
views
Operators in $K(H)^*$
Im working through Sunders An Invitation to Von Neumann Algebras and exercise 0.2.2 states that
If $\omega\in K(H)^*,$ then $\omega$ admits a decomposition $$\omega=\sum_{n=1}^\infty\alpha_n\omega_{\...
- 513
8
votes
2
answers
604
views
Proving an operator is compact.
I'm studying functional analysis and I ask this question mainly because I want a feedback on my proof and suggestions to formalize better my argument (if not already well formalized); any other hint ...
5
votes
1
answer
78
views
T/F: $X, Y$ Banach spaces. $T\in L(X, Y)$ is a compact norm-attaining operator $\iff T^*\in L(Y^*, X^*)$ is also a compact norm-attaining operator.
Let $X$ and $Y$ be Banach spaces. Denote $L(X, Y)$ as the space of bounded linear operators from $X$ to $Y$. Then is the following result true:
$$ T\in L(X, Y) ~\text{is compact and norm-attaining} \...
- 836
4
votes
1
answer
138
views
Characterization of compactness for $M_g(f) = gf$ on Hardy space
Let $1 \leq p \leq +\infty$ and $g \in \mathcal{H}(\mathbb{D})$. When is the operator $ M_g : H^p(\mathbb{D}) \to H^p(\mathbb{D})$, $M_g(f) := gf$ compact in terms of $g$? I would like to give a ...
- 53
5
votes
0
answers
145
views
About the Daugavet property
A Banach $X$ space has the Daugavet if for each rank-one operator $T$ on $X$ satisfies the so-called Daugavet equation
$$
\|I+T\|=1+\|T\| , \label{1} \tag{1}
$$
where $I$ estands for the identity ...
- 176
2
votes
1
answer
99
views
Volterra operator being compact when $1<p<\infty$ using weak convergence
Let $X=L^p([0,1];\mathbb{C})$ and $Y=C([0,1];\mathbb{C})$. Consider the Volterra operator $V:X \to Y$ defined by
$$
V(f)(x)=\int_{[0,x]}f(t)dt
$$
then show that $V$ is a compact operator.
I can ...
- 677
2
votes
1
answer
33
views
Is countable eigenvalues converging to 0 a sufficient condition for compactness? [closed]
Compact operators are known to have countable eigenvalues with $0$ as the unique possible accumulation point. Is that condition also sufficient to make an operator compact? If not, I would like to see ...
- 29
2
votes
1
answer
121
views
The integral equation $f(t) = \int_0^t K(t,s) f(s) ds$ has only the trivial solution over $C[0,1]$.
I have just been reading a little bit about compact operators and the Fredholm alternative, and I'm trying to solve the following problem:
Let $K \in C[0,1]^2$. Show that for every $g \in C[0,1]$ the ...
- 951