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Suppose we have a separable Hilbert space (thus with a countable basis) and that we to represent an operator in matrix form, i.e:

$$A: H \rightarrow H \\ \; \; \; \; \; \;x \;\rightarrow \sum_{j \in \mathbb{N}}\left(\sum_{k \in \mathbb{N}} a(j,k)\cdot<x,e_k> \right)e_j$$

Given that the series of complex numbers $\sum_{k \in \mathbb{N}} a(j,k)\cdot<x,e_k>$ converges and that $\sum_{j \in \mathbb{N}}\left(\sum_{k \in \mathbb{N}} a(j,k)\cdot<x,e_k> \right)e_j$ converges in $H$ how do I prove without further assumptions that $A$ is a bounded operator?

I can prove that for each $j \in \mathbb{N}$ the sucession $(a(j,k))_k \in l^2 (\mathbb{N},\mathbb{C})$ and that for each $k \in \mathbb{N}$ the sucession $(a(j,k))_j \in l^2 (\mathbb{N},\mathbb{C})$.

However I always get to a point where I require that $\sum_{j \in \mathbb{N}}\sum_{k \in \mathbb{N}} \left |a(j,k) \right|^2 < \infty$ which is a condition that I can show to be necessary and sufficient for $A$ to be a Hilbert-Schmidt operator. Any help would be welcome.

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  • $\begingroup$ Yes, I meant a bounded operator. I edited it, thanks. $\endgroup$ – grizzlyjoker Apr 11 '13 at 19:52
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As pointed out by Elias, the Uniform Boundedness Principle (UBP) is the key. But the assumption does not imply that $A$ be Hilbert-Schmidt, as it is satified by every bounded linear operator $A$ on $H$. Indeed, denoting by $P_n$ the finite rank projection onto $\mbox{span}(e_0,\ldots,e_n)$, we have that $P_n$ tends to the identity in the strong operator topology (i.e. pointwise). So for every $n$, $P_nAP_mx$ tends to $P_nAx$ as $m$ tends to $+\infty$, which proves that the coordinates of $Ax$ in the basis are given by the converging series $\sum_{k=0}^{+\infty} A_{jk}\langle x,e_k\rangle$ where $A_{jk}=\langle Ae_k,e_j\rangle$ are the matrix coordinates of $A$ in the basis. And then $P_nAx$ tends to $Ax$ as $n$ tends to $+\infty$, which proves the second part, namely that $\sum_{j=0}^{+\infty}\left(\sum_{k=0}^{+\infty}A_{jk}\langle x,e_k\rangle\right)e_j$ converges to $Ax$ in $H$.

Conversely, which is what you are asking about, assume that $(A_{jk})$ is a doubly infinite sequence such that $\sum_{j=0}^{+\infty}\left(\sum_{k=0}^{+\infty}A_{jk}\langle x,e_k\rangle\right)e_j$ has converging coordinates, and converges in $H$ for every $x\in H$. We will show altogether that this defines a bounded linear operator on $H$. For every $n,m$, consider the bounded linear operator $A^{(n,m)}$ defined by $A^{(n,m)}x:=\sum_{j=0}^n\left(\sum_{k=0}^mA_{jk}\langle x,e_k\rangle\right)e_j$. By the first part of the assumption, we see that $A^{(n,m)}$ converges to $A^{(n)}$ pointwise as $m$ tends to $+\infty$, where $A^{(n)}x:=\sum_{j=0}^n\left(\sum_{k=0}^{+\infty}A_{jk}\langle x,e_k\rangle\right)e_j$. By the UBP, it follows that $A^{(n)}$ is a bounded linear operator on $H$. Finally, note that the second part of the assumption yields that $A^{(n)}$ converges pointwise to $A$ given by $Ax:=\sum_{j=0}^{+\infty}\left(\sum_{k=0}^{+\infty}A_{jk}\langle x,e_k\rangle\right)e_j$ as $n$ tends to $+\infty$. By the UBP again, we can conclude that $A$ is a bounded linear operator on $H$.

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Hint Use the Uniform boundedness principle or better yet just do $x=\sum_{v=1}^{\infty}\sum_{u=1}^{\infty}a(u,v)e_u$ in $\sum_{j \in \mathbb{N}}\left(\sum_{k \in \mathbb{N}} a(j,k)\cdot<x,e_k> \right)e_j<\infty$ for proving $\sum_{j \in \mathbb{N}}\sum_{k \in \mathbb{N}} \left |a(j,k) \right|^2 < \infty$.

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    $\begingroup$ The series $\sum_{u=1}^{+\infty}e_u$ does not define a vector of $H$. $\endgroup$ – Julien Apr 11 '13 at 20:25
  • $\begingroup$ Hmm, we didn't learn the uniform boundedness principle so far. About the second aproach: how to we know that this particular x converges? If it does that would seem to prove that every bounded operator is Hilbert-Schmidt operator $\endgroup$ – grizzlyjoker Apr 11 '13 at 20:28
  • $\begingroup$ Hmm, i got it by another means: using the Jensen inequality and using some results about $l^2$ sequences. Thanks for the help! $\endgroup$ – grizzlyjoker Apr 12 '13 at 9:03
  • $\begingroup$ @julien I corrected my little mistake in my answer. $\endgroup$ – MathOverview Apr 12 '13 at 11:29
  • $\begingroup$ Hmm nvm, it doesn't work: i can prove that $\sum_{k \in \mathbb{N}} \left |a(j,k) \right |$ converges but i also need (in order for my argument to work) that $\sum_{j \in \mathbb{N}} \left |a(j,k) \right |$ which I cannot prove $\endgroup$ – grizzlyjoker Apr 12 '13 at 21:38

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