# 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.

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### How to prove that an operator is compact?

Consider $T\colon\ell^2\to\ell^2$ an operator such that $Te_k=\lambda_k e_k$ with $\lambda_k\to 0$ as $k \to \infty$ how to prove that it is compact?
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### Compact operator with closed range has finite dimensional range

Let $X,Y$ be Banach Spaces, and let $T\in K(X,Y)$ be a compact operator from $X$ to $Y$. I have to prove that $T(X)$ is closed in Y if, and only if, $\dim(T(X))<\infty$. Can anybody help me with ...
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### What makes compact operators special?

I would like to understand why compact operators are considered so special to consider them as an extra class of operators. Over Hilbert spaces these (as far as I know) these are the ones with ...
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### Proof that operator is compact

Prove that the operator $T:\ell^1\rightarrow\ell^1$ which maps $x=(x_1,x_2,\dots)$ to $\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\dots\right)$ is compact. For an arbitrary sequence $x^{(N)}\in\ell^1$ one ...
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### Set of all compact operators $K(H)$ is the unique ideal in $B(H)$?

I want to show that the set of all compact operators $K(H)$ is the unique ideal in $B(H)$. Is there any relation between invertibility and compactness of an operator?
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### Is the Neumann series a compact operator?

Let $X$ be an infinite dimensional Banach space and $A:X\to X$ be a compact operator with the operator norm $\|A\|<1$. Then $I-A$ is invertible and the Neumann series $$S_N = \sum_{k=0}^N A^k$$ ...
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### Possible flaw in “proof” that a sum of two compact operators is compact

If X and Y are Banach spaces, and $A: X \to Y$, $B: X \to Y$ are both compact operators, then $A + B$ is compact. A + B is compact if and only if for every bounded sequence $\lbrace x_n \rbrace$ in X,...
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### Is convolution operator compact?

I know convolution is not a Hilbert–Schmidt integral operator, but it needs more to tell if convolution is compact or not.
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### If $T^2$ is a compact operator on a Banach space then is $T$ also compact?
If $T^2$ is a compact operator on a Banach space $X$ then is it true that $T$ is also compact operator on X? Here $T:X\to X.$ First I was trying to prove(assuming it is true). Since $X$ is Banach ...