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Consider $T\colon\ell^2\to\ell^2$ an operator such that $$T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 $$ Does anyone know how to prove that it is compact?

I understand that I have to find a converging semisequence in $\ell^2$.

Thank you for your help.

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I understand that a linear operator $T:E_1\to E_2$ is considered compact if for every bounded sequence $(a_n)$ in $E_1$; $(Ta_n)$ has a converging semisequence in $E_2$. So I understand that I have to find a converging semisequence in $\ell^2$ and show that $(X_n)$ is bounded. I got some hints but Im having trouble in understanding the whole picture here. – jeremy Mar 20 at 18:29

3 Answers

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Try the sequence of operators $T_n:\ell^2\rightarrow\ell^2$ defined by $$T_n(x)=(2^{-1}x_1,...,2^{-n}x_n,0,0,...)$$

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I understand that a linear operator $T:E_1\to E_2$ is considered compact if for every bounded sequence $(a_n)$ in $E_1$; $(Ta_n)$ has a converging semisequence in $E_2$. So I understand that I have to find a converging semisequence in $\ell^2$ and show that $(X_n)$ is bounded. I got some hints but Im having trouble in understanding the whole picture here. – jeremy Mar 20 at 18:22
Show that $T_n$ converges to $T$. Remember that if $T_n$ is a sequence of compact operators converging to $T$ in the dual norm, then $T$ is compact. – Tomás Mar 20 at 19:00
Thanks, but isn't $T_n(x)$ you proposed in your answer supposed to be a semisequence of $T((X_n))$? – jeremy Mar 21 at 9:18
I guess Im missing something here. Tomas, Some more detail would be really helpfull. anyways thanks for all the help :) – jeremy Mar 21 at 10:56
Dear @Jeremy, you want to show that $T$ is compact. One way to do this, is to show that $T$ is the limit of a sequence of compact operators. That' what I showed, in fact, i have showed that $T$ is a sequence of finite rank operators and every finite rank operator is compact. – Tomás Mar 21 at 12:57
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To show that this operator is compact you need to show that it sends every bounded region to pre-compact. Pre-compact sets are locally-bounded sets. You can try to prove (it's very well-known) that locally boundness condition in $l^2$ is equivalent to both:

a) all elements from set are uniformly bounded, and

b) for every $\epsilon > 0$ there is a number $N$, such that the following is true : norm of a tail (i.e. for element $x = (x_i)_{i=0}$ the tail is $x^N = (x_{i + N})_{i=0}$) of any element from set is bounded by $\epsilon$.

Using this criteria the problem becomes obvious.

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I understand that a linear operator $T:E_1\to E_2$ is considered compact if for every bounded sequence $(a_n)$ in $E_1$; $(Ta_n)$ has a converging semisequence in $E_2$. So I understand that I have to find a converging semisequence in $\ell^2$ and show that $(X_n)$ is bounded. I got some hints but Im having trouble in understanding the whole picture here. – jeremy Mar 20 at 18:23
Do you have any problems in understanding locally bounding criteria? – Sergey Finsky Mar 21 at 5:39
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There is a useful characterization of self-adjoint compact operators on a separable Hilbert space.

These are the operators which can be diagonalized in an orthonormal basis with diagonal $(\lambda_n)$ such that $\lim \lambda_n =0$.

Your operator is of this form, so it is compact.

The direction $\Rightarrow$ is the difficult part. A proof is given in the link above. For the converse (which is what you need here), one easily proves that such operators are norm limits of finite rank operators. And limits of finite rank operators are compact.

share|improve this answer
I understand that a linear operator $T:E_1\to E_2$ is considered compact if for every bounded sequence $(a_n)$ in $E_1$; $(Ta_n)$ has a converging semisequence in $E_2$. So I understand that I have to find a converging semisequence in $\ell^2$ and show that $(X_n)$ is bounded. I got some hints but Im having trouble in understanding the whole picture here. – jeremy Mar 20 at 18:22
If $T$ has finite rank, it is compact because your sequence $Ta_n$ is bounded in a finite-dimensional space (where compact=bounded+close), so there exists a converging subsequence. Now try to prove that if $T_k$ is a sequence of compact operators which converges in norm to $T$, then $T$ is compact. You can do this by CAntor diagonal extraction. If $a_n$ is bounded,there is a subsequence such that $T_1(a_{\phi_1(n)})$ converges. Then extract a subsequence from the latter which also converges for $T_2$, etc... – julien Mar 20 at 21:49

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