Compact operator

Question

Please, help with the problem. Is the Hilbert operator, which defined by matrix $ \{ a_{i,j} = \frac{1}{i+j} \}^{\infty}_{i,j=1}$ , compact on space $l_2$ ?

I will add my thoughts about the problem as soon as I formulate and issue them

I think that it can be proved that Hilbert operator is bounded on $L_2 (\mathbb{N}, 2^{ \mathbb{N} }, \#)$, where $\#$ is the counting measure, and that the embedding operator $Jx=x$ from $L_2 (\mathbb{N}, 2^{ \mathbb{N} }, \#)$ to $l_{2}$ is compact. It will mean that the Hilbert operator is compact as the composition of compact and bounded operators

Answer 1

Try to prove on your own the continuity of the operator $T\phi(j)=\sum^\infty_{k=1}\frac{1}{j+k}\phi(k)$ on $\ell_2(\mathbb{N})$. As for compactness, this is from a paper by Barria and Halmos. Consider the sequence $\phi_n(j)=\frac{1}{n}\mathbb{1}_{\{1,\ldots,n^2\}}(j)$. This is a unit-norm sequence in $\ell_2$.

• Check that for any $\phi\in\ell_2$, $\langle\phi_n,\phi\rangle=\sum^\infty_{j=1}\phi_n(j)\phi(j)\xrightarrow{n\rightarrow\infty}0$. This means that $\phi_n$ converges weakly to $0$ in $\ell_2$.
• Check that $\langle\phi_n,T\phi_n\rangle\geq\frac12$.
• From this, deduce that $T$ cannot be compact, for if $T\phi_{n_k}\rightarrow\phi\in\ell_2$, then \begin{align} \frac12\leq\langle\phi_{n_k},T\phi_{n_k}\rangle &= \langle\phi_{n_k},T\phi_{n_k}-\phi\rangle +\langle\phi_{n_k},\phi\rangle\\ &\leq \|T\phi_{n_k}-\phi\|_2+\langle\phi_{n_k},\phi\rangle\xrightarrow{k\rightarrow\infty}0 \end{align} which is absurd.