Convergence weakly in Hilbert space.

Question

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...

Answer 1

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 $\ell_k.$