I have already tried but I failed. I can't show it is. I used this way:
$| (Kf)_n |^2 \leq c_n \|f\|^2$, and therefore $\|Kf\|^2 \leq \|c\| \|f\|^2$, so that $\|K\| \leq \sqrt{\|c\|} $ is bounded, but I cant take result.
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Do you know that you can think of the Hilbert-Schmidt norm as the square root of the sum of the squares of the entires of the infinite matrix representing the operator? – Potato May 11 '14 at 20:26
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No i dont know is it useful – user149385 May 11 '14 at 20:53
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From wiki article we know that $$ \Vert K\Vert_{HS}^2=\sum_{i\in\mathbb{N}}\Vert K(e_i)\Vert^2 $$ Now we compute $$ \Vert K(e_i)\Vert^2 =\sum_{n\in\mathbb{N}}|K(e_i)_n|^2 =\sum_{n\in\mathbb{N}}\left|\sum_{j\in\mathbb{N}} k_{j+n}(e_i)_j\right|^2 =\sum_{n\in\mathbb{N}}\left|\sum_{j\in\mathbb{N}} k_{j+n}\delta_{i,j}\right|^2 =\sum_{n\in\mathbb{N}}|k_{i+n}|^2 $$ $$ \Vert K\Vert_{HS}^2 =\sum_{i\in\mathbb{N}}\Vert K(e_i)\Vert^2 =\sum_{i\in\mathbb{N}}\sum_{n\in\mathbb{N}}|k_{i+n}|^2 =\sum_{i\in\mathbb{N}}\sum_{m=i+1}^\infty|k_m|^2 =\sum_{m=1}^\infty(m-1)|k_m|^2 \leq\sum_{m=1}^\infty m c_m^2 \leq\sum_{m=1}^\infty m c_m\cdot c_m \leq\sum_{m=1}^\infty c_m =\Vert c\Vert_{\ell_1(\mathbb{N})}<+\infty $$ Thus $\Vert K\Vert_{HS}\leq \Vert c\Vert_{\ell_1(\mathbb{N})}^{1/2}$, then $K$ is a Hilbert-Schmidt operator.
Norbert
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