# Convergence weakly in Hilbert space. Why $$\lim_k \langle f_{n_k},e_j\rangle =\langle f,e_j\rangle$$ with $$f=\sum_{k=1}^{\infty} {a_k}^{k} e_k$$? I dont understand the diagonal argument...

Maybe it's easier to strip this down to its essentials. You have infinite sets $$(e_i)_i$$ and $$(f_i)_i$$. I will write the inner product suggestively, for convenience, $$\langle f,e\rangle=f(e).$$

$$(f_1(e_1),f_2(e_1),f_3(e_1),\cdots)$$ is a bounded sequence of real numbers, so it has a convergent subsequence:

$$\tag 1(f_{n_1}(e_1),f_{n_2}(e_1), f_{n_3}(e_1),\cdots)\to \ell_1.$$

Now consider $$(f_{n_1}(e_2),f_{n_2}(e_2), f_{n_3}(e_2),\cdots)$$. This is also a bounded sequence of real numbers, so it has a convergent subsequence:

$$\tag2 (f_{n_{k_1}}(e_2),f_{n_{k_2}}(e_2),f_{n_{k_3}}(e_2),\cdots )\to \ell_2.$$

Notice that the $$(f_{n_{k_{i}}})_i$$ is a subsequence of $$(f_{n_i})_i$$, so $$(f_{n_{k_1}}(e_1),f_{n_{k_2}}(e_1),f_{n_{k_3}}(e_1),\cdots )\to \ell_1,$$ too.

We continue, for $$e_3:\ (f_{n_{k_1}}(e_3),f_{n_{k_2}}(e_3),f_{n_{k_3}}(e_3),\cdots )$$ is a bounded sequence of real numbers, so it has a convergent subsequence:

$$\tag3 (f_{n_{k_{j_1}}}(e_3),f_{n_{k_{j_2}}}(e_3),f_{n_{k_{j_3}}}(e_3),\cdots )\to \ell_3.$$

Observe that $$(f_{n_{k_{j_1}}}(e_2),f_{n_{k_{j_2}}}(e_2),f_{n_{k_{j_3}}}(e_2),\cdots )\to \ell_2$$ because it is a subsequence of $$(2)$$. And

$$(f_{n_{k_{j_1}}}(e_1),f_{n_{k_{j_2}}}(e_1),f_{n_{k_{j_3}}}(e_1),\cdots )\to \ell_1$$ because it is a subsequence of $$(1)$$.

Can you see what the next step would be? Take $$(3)$$ and substitute in $$e_4$$ for $$e_3$$ and repeat the analysis.

I hope now it is clear what is happening. Think of this process as an infinite matrix, in which the $$k$$th row consists of $$(f_{k,i})^{\infty}_{i=1}$$, and is a subsequence of each of the rows above it. When we substitute $$e_1$$ into the matrix, $$every$$ row converges to $$\ell_1$$. When we substitute $$e_2$$ into the matrix, every row from the $$second$$ row on, converges to $$\ell_2$$. Etc. By the time we get to the $$k$$th row, every row, from the $$k$$th row on, converges to $$\ell_k$$, when we substitute $$e_k$$ into the matrix.

So now the question is: how do we get a $$single$$ sequence $$(f_i)$$ out of this matrix, so that when we substitute $$any$$ vector $$e_k$$ in, we get $$(f_i(e_k))\to \ell_k?$$.

Easy: we go down the diagonal. We take the sequence $$(f_{kk})_k$$, which we may as well just call $$(f_k)$$. This gives us what we want because now if we take an arbitrary $$e_k$$, as soon as $$j\ge k,$$ the sequence $$(f_j(e_k),f_{j+1}(e_k),\cdots )$$ is a subsequence of a tail of the $$k$$th row of the matrix, and to must converge to $$\ell_k.$$