Show that the set $C := \left\{\left.\sum_{j=1}^\infty\lambda_j e_j\right|\, \lambda_j \in \mathbb{K},\,|\lambda_j| \leq a_j\right\}$ is compact in H.

Question

I am trying to work out question 8.17 from the book 'Functional Analysis: an elementary introduction' by Haase. The question is formulated as follows. Let $(e_j)_{j\geq 1}$ be an orthonormal system in a Hilbert space $H$, and let $a_j \geq 0$ be scalars with $\sum_{j=1}^\infty a_j^2 < \infty$. Show that the set \begin{align} C := \left\{\left.\sum_{j=1}^\infty\lambda_j e_j\right|\, \lambda_j \in \mathbb{K},\,|\lambda_j| \leq a_j\right\} \end{align} is compact in $H$.

This is my attempted solution: I would go with the definition of sequential compactness here. My plan was either to construct a pairwise orthogonal sequence. Then, showing that $ \sum_{j=1}^\infty \|f_n\|^2 < \infty$ would imply that $\sum_j f_n$ converges in H. I thought about using Bessel's inequality, which turns out here to be an equality, since for some $f\in C$ we have \begin{align} \langle f, e_k\rangle = \langle\sum_{j=1}^\infty \lambda_j e_j, e_k\rangle = \sum_{j=1}^\infty \lambda_j \langle e_j, e_k\rangle = \lambda_k. \end{align} Thus, we can write $f$ as follows \begin{align} f = \sum_{j=1}^\infty\lambda_j e_j = \sum_{j=1}^\infty\langle f, e_j\rangle e_j = Pf, \end{align} the standard abstract Fourier series. From this, we can deduce that \begin{align} \|f\|^2 = \sum_{j=1}^\infty |\langle f,e_j\rangle |^2 = \sum_{j=1}^\infty |\lambda_j|^2 \left(\leq \sum_{j=1}^\infty a_j^2 < \infty\right), \end{align} using Pythagoras.
Now, To show that $C$ is compact we have to show that every sequence in $C$ has a converging subsequence. Let $(f_n)_{n\geq 1}$ be a sequence in $C$. Let $(f_{n_k})_{k\geq 1}$ be a subsequence, we need to show that $(f_{n_k})_{k\geq 1}$ converges to some $f$ in $C$.
Each $f_n$ is of the form \begin{align} f_n = \sum_{j=1}^\infty \lambda_{n,j} e_{n, j}, \end{align} the coefficients, $\lambda_{n, \cdot}$, depend on $n$, and so does the 'positions', $e_{n, \cdot}$, where we place these coefficients. The sequence $(f_n)_{n\geq 1}$ is not pairwise orthogonal. \begin{align} \langle f_n, f_m\rangle = \left\langle \sum_{j=1}^\infty \lambda_{n, j}e_{n,j}, \sum_{k=1}^\infty \lambda_{m, k}e_{m,k}\right\rangle = \sum_{j=1}^\infty\sum_{k=1}^\infty \lambda_{n,j}\lambda_{m, k} \langle e_{n,j}, e_{m,k}\rangle \end{align} we can say nothing about the term $\langle e_{n,j}, e_{m,k}\rangle$. We could also try to show that this set $C$ is closed in some other compact set $D$. But then we would have to construct some set $D$, and I do not really get far with this approach either.
Any help is much appreciated!

Answer 1

Proving compactness by showing that every sequence has a convergent subsequence is a nice strategy, and I think you can make this work. Can I suggest going about it using a diagonal argument?

Suppose $(f_1, f_2, f_3, \dots)$ is a sequence in $C$, and write each $f_n$ in the form $f_n = \sum_{j = 1}^\infty \lambda_{n, j} e_j$.

1. Using the compactness of the closed ball $\overline{B(0, a_1)} \subset \mathbb K$, one can show that there exist positive integers $n^{(1)}_1 < n^{(1)}_2 < n^{(1)}_3 < \dots$ such that $(\lambda_{n^{(1)}_1, 1}, \lambda_{n^{(1)}_2, 1}, \lambda_{n^{(1)}_3, 1}, \dots)$ is a convergent sequence.
2. Hence, using the compactness of the closed ball $\overline{B(0, a_2)} \subset \mathbb K$, one can show that there exist positive integers $n^{(2)}_1 < n^{(2)}_2 < n^{(2)}_3 < \dots$, chosen from the set $\{ n^{(1)}_1 , n^{(1)}_2 , n^{(1)}_3 , \dots \}$, such that $(\lambda_{n^{(2)}_1, 1}, \lambda_{n^{(2)}_2, 1}, \lambda_{n^{(2)}_3, 1}, \dots)$ and $(\lambda_{n^{(2)}_1, 2}, \lambda_{n^{(2)}_2, 2}, \lambda_{n^{(2)}_3, 2}, \dots)$ are both convergent sequences.

And so on.

Having done all of this, one can conclude that $(\lambda_{n^{(1)}_1, j}, \lambda_{n^{(2)}_2, j}, \lambda_{n^{(3)}_3, j}, \dots)$ is a convergent sequence, for any $j$.

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To summarise, the above argument shows that:

There exists a subsequence $(f_{n_k})_{k \in \mathbb N}$ of $(f_n)_{n\in \mathbb N}$ such that the sequence $(\lambda_{n_k, j})_{k\in\mathbb N}$ converges, for any $j$.

(Take $n_k := n^{(k)}_k$.)

The final step is to show that $(f_{n_k})_{k \in \mathbb N}$ is itself convergent (as a sequence in the Hilbert space $H$).

This is where your ideas about Pythagoras' theorem come in. Here's the rough idea:

For each $j$, define $\lambda_j := \lim_{k \to \infty} \lambda_{n_{k}, j}$. Also define $f := \sum_{j = 1}^\infty \lambda_j e_j$. (Here, you should convince yourself that this infinite series is well defined, and that this $f$ is in fact ab element of $C$.) The goal is to prove that $(f_{n_k})_{k \in \mathbb N}$ converges to $f$ as a sequence in $H$.

So pick an $\epsilon > 0$. Since $\sum_{j = 1}^\infty |a_j|^2 < \infty$, there exists a $J$ such that $\sum_{j = J + 1}^\infty |a_j|^2 < \epsilon$.

Also, for each $j$, there exists a $K_j$ such that $k \geq K_j$ implies $(\lambda_{n_k, j} - \lambda_j)^2 < \epsilon$.

Thus one can show that if $k \geq \max \{ K_1, \dots K_J \}$, then $$ \left\| f_{n_k} - f \right\|^2 = \sum_{j = 1}^\infty (\lambda_{n_k, j} - \lambda_{j})^2 < J \epsilon + 4\epsilon.$$ (Here, Pythagoras' theorem is used.)

Feel free to leave a comment if you have questions about filling in the details.