In a Hilbert space $v \in l^2, A = \{x \in l^2: |x(n) |\le |v(n)|$ for all$ n\}$, prove $A$ is compact

Question

Say $[x_n]$ is an sequence in A, then $\lVert x_n - x_m\rVert$ is bounded because $|x_n(k)| \le |v(n)|$ is bounded. How do I prove this mathematically? and How do I even say $\lVert x_n - x_m\rVert \to 0 $

Answer 1

Let $x_n$ be a sequence in $A$ and use notation $x_n(k)$ to denote the $k$-th component of $x_n$. You have $|x_n(k)|≤|v(k)|$, so $x_n(k)$ is always a bounded sequence in $\Bbb C$ for all $k$.

If we look at the $x_n(k)$ seperately for each $k$, we are in the situation that we have a countable number of bounded sequences. The diagonalisation trick allows you to find a subsequence $x_{n_j}$ so that $x_{n_j}(k)$ converges for any $k$.

So now we have a sequence $x_{n_j}$ in $A$ whose components converge, if we can show that this implies $x_{n_j}$ is convergent we will have shown that $A$ is sequentially compact, as then our arbitrary sequence $x_n$ has a convergent subsequence.

First, let the limits of the components $x_{n_j}(k)$ be called $x(k)$. Second, from $|x_{n_j}(k)|≤|v(k)|$ it follows that $|x(k)|≤|v(k)|$ and then from $\sum_k |v(k)|^2<\infty$ that $\sum_k|x(k)|^2<\infty$. So the components $x(k)$ actually do determine an element $x$ of $\ell^2$ (that also lies in $A$).

Now from $\sum_k |v(k)|^2<\infty$ you've got for any $\epsilon$ a $K$ so that $\sum_{k=K+1}^\infty |v(k)|^2<\epsilon$. On the other hand $\sum_{k=0}^K|x_{n_j}(k)-x(k)|^2\to0$ as $j\to\infty$, so you've got a $J$ where if $j>J$ this term is smaller than $\epsilon$. It follows

$$\|x_{n_j}-x\|^2=\sum_{k=0}^K|x_{n_j}(k)-x(k)|^2+\sum_{k=K+1}^\infty|x_{n_j}(k)-x(k)|^2≤\sum_{k=0}^K|x_{n_j}(k)-x(k)|^2+4\sum_{k=K+1}^\infty |v(k)|^2≤5\epsilon $$ whenever $j>J$. Since $\epsilon$ was arbitrary this implies convergence of $x_{n_j}$ to $x$.