# Weak convergence as convergence of matrix elements

Let $H$ be a Hilbert space with orthonormal basis $(e_h)_{h \in \mathbb{N}}$ and let $(A_n)_{n \in \mathbb{N}}$ and $A$ be bounded linear operators. We say that $A_n$ converges weakly to $A$ if

$$\forall \xi, \eta \in H, \quad (\eta, A_n\xi)\to (\eta, A\xi).$$

Question Is it true that $A_n$ converges weakly to $A$ if and only if

$$\forall h, k \in \mathbb{N},\quad (e_h, A_n e_k) \to (e_h, Ae_k)?$$

Indeed, I'm wondering if it is correct to think at weak convergence as convergence of the matrix entries associated to the operators $A_n$ and $A$.

Thank you.

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I think it's not the case. Put $A_n(e_k)=0$ if $k\neq n$ and $A_n(e_n)=ne_n$. Then $A_n$ is a bounded linear operator with norm equal to $n$, and for all $h,k\in\mathbb N$, $A_ne_k=0$ if $n\geq k+1$, so the property $(e_h, A_ne_k)\to (e_h,Ae_k)$ is satisfied for $A=0$. But we can't have weak convergence, since the principle of uniform boundedness implies that $\sup_n\lVert A_n\rVert$ is finite.

But we can show that the property you mentioned and $\sup_n\lVert A_n\rVert<\infty$ is equivalent to have weak convergence.

Indeed, if we have weak convergence then for all $v$, the sequence $\{A_n v\}$ converges weakly to $Av$, so it's bounded and $\sup_n\lVert A_n\rVert<\infty$.

Conversely, if $\sup_n\lVert A_n\rVert<\infty$ and $(e_h,A_ne_k)\to(e_h,Ae_k)$ for all $h,k\in\mathbb N$, we fix $u,v\in H$, and $\delta>0$. Let $u',v'\in \operatorname{Span}\{e_h,h\in\mathbb N\}$ such that $\lVert u-u'\rVert\leq \delta$ and $\lVert v-v'\rVert\leq \delta$. We have, writing $M:=\sup_{n\in\mathbb N}\lVert A_n\rVert$: \begin{align*} |(u,A_nv)-(u,Av)|&\leq |(u,A_nv)-(u',A_nv)|+|(u',A_nv)-(u',A_nv')|\\ &+|(u',A_nv')-(u',Av')|+(u',Av')-(u,Av')|+|(u,Av')-(u,Av)|\\ &\leq \lVert u-u'\Vert M\lVert v\rVert +\lVert u'\rVert M\lVert v-v'\rVert + |(u',A_nv')-(u',Av')|\\ &+\lVert u'-u\rVert \lVert A v'\rVert+ \lVert u\rVert\lVert A\rVert\lVert v-v'\rVert\\ &\leq \left(M\lVert v\rVert+M\lVert u'\rVert+\lVert Av'\rVert +\lVert u\rVert\lVert A\rVert\right)\delta+|(u',A_nv')-(u',Av')|, \end{align*} so for all $\delta >0$, $$\limsup_n|(u,A_nv)-(u,Av)|\leq \left(M(\lVert v\rVert+\lVert u\rVert+\delta)+(\lVert v\rVert+\lVert u\rVert+\delta)\lVert A\rVert\right)\delta,$$ hence $\lim_{n\to\infty}|(u,A_nv)-(u,Av)=0$ for all $u,v\in H$.

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+1 of course. A linguistic quibble: I would drop the "the" in the last two sentences. I would write "We can't have weak convergence since the uniform boundedness principle implies that $\sup_{n}{\ \|A_n\|}$ is finite." – t.b. Dec 11 '11 at 12:56
A style quibble: Couldn't you verify directly that you don't have weak convergence by taking $\xi=\eta=\sum{1\over n} e_n$ (no need to appeal to UBP)? – David Mitra Dec 11 '11 at 13:41
@DavidMitra Yes, but I wrote that in order to make a link with what we have to add to get the weak convergence. – Davide Giraudo Dec 11 '11 at 13:52
Nice counterexample! In the follow up, how can you be sure that $A_n v$ converges weakly for all $v$? (You say: "$A_n v$ converges weakly ... so it's bounded"). Maybe you mean that $v \in \text{span}(e_n)$? In the following, as far as I can see, you only need boundedness of operator $A$, which we have by hypothesis. – Giuseppe Negro Dec 11 '11 at 23:31
Ok, now I understand perfectly. Thank you very much! – Giuseppe Negro Dec 12 '11 at 8:54

If it is possible to write $\zeta = \sum_{i=0}^{\infty}a_{k}e_{k}, \eta=\sum_{i=0}^{\infty}b_{k}e_{k}$ with the $a_{k}, b_{k} = const. \forall k \in\mathbb{N}$, the the weak convergence as defined above follows directly from the fact that the direct product on the Hilbert space is linear, that the operator we're dealing with is linear, and that we can write the $\zeta, \eta$ can be written as linear combinations of the o.n.b.. The "if part" doesn't follow, however, since you only regard the diagonal entries of the operator. This goes in the same direction as the person answering the question before me already said. You should easily find plenty of examples that disprove the if part.

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Thank you for your answer. However, I find it somewhat unclear. Expanding $\zeta, \eta$ as you write, the "sandwich" $(\zeta, A_n \eta)$ becomes $\sum_{k, h}a_kb_h(e_k, A_n e_h)$, and it does not contain only diagonal entries of $A_n$: indeed, $k$ and $h$ are allowed to differ. – Giuseppe Negro Dec 12 '11 at 8:57