Trying to prove that operator is compact

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.