compact projection

Question

Show that the operator $T: l^2\rightarrow l^2$ defined by $T({x_n })={( x_n)/2^n } $is compact. How one can show that the sequenc ${ََ T(x_n)}$ is contain convergent subsequence if ${x_n }$ is bounded? I know that $T({x_n })={( x_n)/2^n }$ is bounded.

Answer 1

Suppose that $x^{(n)}$ is a bounded sequence in $\ell^2$. Then, in particular $x^{(n)}_1$ is bounded. Therefore there is a subsequence $x^{(n_k^1)}$ such that $x^{n_k^1}_1$ converges to some $x_1$. We can even take it such that $|x^{n_k^1}_1-x_1|<\frac{1}{k}$ for all $k=1,2,...$

Since $x^{(n_k^1)}_2$ is also bounded, there is a subsequence $x^{(n_k^2)}$ of $x^{n_k^1}$ such that $x^{(n_k^2)}_2$ converges to some $x_2$, and we can take it such that $|x^{n_k^2}_2-x_2|<\frac{1}{k}$

Continue in this way to get (nested) subsequences $x^{(n_k^t)}$, $t=1,2,3,...$, such that $|x^{n_k^t}_t-x_t|<\frac{1}{k}$

Then define $y^{(k)}=x^{(n_k^k)}$. We have that $y^{(k)}_n\to x_n$ for each $n$. Moreover, if we put $x=(x_k)$ then $$\|T(y^{(k)})-T(x)\|_2^2=\sum_n\frac{|y^{(k)}_n-x_n|^2}{2^{2n}}=\sum_n\frac{|x^{n_k^k}_n-x_n|^2}{2^{2n}}\leq\frac{1}{k}\sum_n\frac{1}{2^{2n}}<\frac{1}{k}$$