Brezis' exercise 6.5: $V$ is a Hilbert space and $V \subset \ell^2$ with compact injection

Question

I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,

Let $(\lambda_n)_{n\ge 1}$ be a sequence of positive numbers with $\lim_n \lambda_n = \infty$. Let $V$ be the space of sequences $(u_n)_{n\ge 1}$ such that $$ \sum_{n=1}^\infty \lambda_n |u_n|^2 < \infty. $$ The space $V$ is equipped with the inner product $$ [u, v] := \sum_{n=1}^\infty \lambda_n u_nv_n \quad \forall u,v \in V. $$ Prove that $V$ is a Hilbert space and $V \subset \ell^2$ with compact injection.

There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?

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We define a $\sigma$-finite measure $\mu$ on $\mathbb N^*$ by $$ \mu (n) := \lambda_n \quad \forall n \in \mathbb N^*. $$

Then $V=L^2(\mathbb N^*, \mu, \mathbb R)$. Then $V$ together with $[\cdot, \cdot]$ is a Hilbert space. Because $\lim_n \lambda_n = \infty$, there is $N \in \mathbb N^*$ such that $\lambda_n \ge 1$ for $n \ge N$. Let $\alpha := \sup_{1 \le n \le N} \frac{1}{\lambda_n} < \infty$. Then for $u \in V$, $$ \begin{align} \sum_{n=1}^\infty |u_n|^2 &= \sum_{n=1}^N |u_n|^2 + \sum_{n=1}^{N+1} |u_n|^2 \\ &\le \alpha \sum_{n=1}^N \lambda_n |u_n|^2 + \sum_{n=1}^{N+1} \lambda_n |u_n|^2 \\ &\le (\alpha+1) \sum_{n=1}^\infty \lambda_n |u_n|^2 . \end{align} $$

Then the linear injection $J:V \to \ell^2, (u_n) \mapsto (u_n)$ is continuous. Let $B$ be the closed unit ball of $V$. Let's prove that $J(B)$ has compact closure in $\ell^2$. Because $\ell^2$ is complete, it remains to prove that $J(B)$ is totally bounded in $\ell^2$. Let $|\cdot|_V$ and $|\cdot|_{\ell^2}$ be the corresponding norms of $V$ and $\ell^2$. Let $u = (u_n) \in B$ and $v = (v_n) \in B$. Then $$ \begin{align} |u|_V^2 &= \sum_{n=1}^\infty \lambda_n |u_n|^2, \\ |Ju|_{\ell^2}^2 &= \sum_{n=1}^\infty |u_n|^2. \end{align} $$

Fix $\varepsilon \in (0, 1)$. Because $\lim_n \lambda_n = \infty$, there is $N \in \mathbb N^*$ such that $\lambda_n \ge \frac{1}{\varepsilon}$ for $n \ge N$. Then $$ \frac{1}{\varepsilon} \sum_{n=N}^\infty |u_n|^2 \le \sum_{n=N}^\infty \lambda_n |u_n|^2 \le 1. $$

Then $\sum_{n=N}^\infty |u_n|^2 \le \varepsilon$ and thus $\sum_{n=N}^\infty |u_n -v_n|^2 \le 4\varepsilon$. It remains to control the set $C:=\{(x_n) \in B : x_n=0 \quad \forall n \ge N\}$. Clearly, $C$ is bounded in $\ell^2$ and contained in a finite-dimensional subspace, so $C$ is totally bounded $\ell^2$. This completes the proof.

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Update I have added more explanation below.

Because $C$ is totally bounded, it has an $\varepsilon$-covering $\{x^1, \ldots, x^M\}$. We claim that $\{x^1, \ldots, x^M\}$ is a $5\varepsilon$-covering of $B$. Let $u = (u_n) \in B$. There is $x^k$ for some $k \in \{1, 2, \ldots, M\}$ such that $$ \sum_{n=1}^{N-1} |x^k_n-u_n|^2 \le \varepsilon^2 \le \varepsilon. $$

Because $x^k, u \in B$, we get $\sum_{n=N}^{\infty} |x^k_n-u_n|^2 \le 4\varepsilon$. Then $$ \begin{align} |x^k-u|_V^2 &= \sum_{n=1}^{N-1} |x^k_n-u_n|^2 + \sum_{n=N}^{\infty} |x^k_n-u_n|^2 \\ &\le \varepsilon + 4\varepsilon. \end{align} $$

Clearly, $5\varepsilon$ can be arbitrarily small.

Answer 1

I don't understand the last part of the proof in the original post. Here's another way of showing the compactness

Let $H=\ell^2(\Bbb{N}^+,\Bbb{C})$, $\{e_n\}_{n=1}^\infty$ the standard basis of $H$ and $T:H\to H$ be defined by $T(v) = \sum_{n=1}^\infty \frac{1}{\sqrt{\lambda_n}} \langle v, e_n\rangle e_n$. $T$ is a multiplication operator. It is bounded with $\|T\|=\max_{n=1}^\infty \frac{1}{\sqrt{\lambda_n}}$.

The range of $T$ is $V$ as defined in the original post. $T$ by the formula defining it is a compact operator (see here), so the closure of $T(B)$ is compact in $H$. If you want an explicit proof, Let $T_N(v) = \sum_{n=1}^N \frac{1}{\sqrt{\lambda_n}} \langle v, e_n\rangle e_n$. Then $T_N$ are finite rank and $\|T-T_N\| = \max_{n=N+1}^\infty \frac{1}{\sqrt{\lambda_n}}\rightarrow 0$ as $N\rightarrow \infty$. So $T$ is a norm limit of finite rank operators, and as shown here it is compact.