Question about separable, infinite dimensional Hilbert spaces

Question

Here's the problem:

Let $H$ be a separable, infinite dimensional Hilbert space. Prove the following two statements:

1. If $\{e_n\}_{n \in \mathbf{N}}$ is an orthonormal basis of $H$, and $\lambda \in H^\ast$, then $\lambda(e_n) \to 0$.
2. Let $x \in H$ such that $\|x\| \leq 1$. Prove that there is a sequence $\{x_n\} \subset H$ such that $\|x_n\| = 1$, for each $n$ and $\lim_n \lambda(x_n) = \lambda(x)$.

For the first part, I think I can do it as follows: by Riesz Lemma, $\lambda(x) = \langle x,y \rangle$ for a fixed $y \in H$, and any $x \in H$. By orthonormality, $\|y\|^2 = \sum_1^{\infty} |\langle e_n, y \rangle |^2$, so thus $\langle e_n, y \rangle \to 0$, so $\lambda(e_n) \to 0$. I'm a little concerned since I didn't use the separable hypothesis. Is what I've done thus far correct? How do I do part 2 of the question above? I expected to also have to use that we know $H$ is isometrically isomorphic to $\ell^2$.

Answer 1

For 2) one can use (1). If $\|x\|=1$ then the constant sequence $x_n=x$ is enough.

Assume $\|x\|<1$. To construct $x_n$ we add to $x$ a multiple of $e_n$: $$ x_n := x + a_n e_n $$ with $a_n\in\mathbb R$. Then $\|x_n\|=1$ if and only if $$ 1 = \|x\|^2 + a_n^2 + 2a_nRe \langle x,e_n\rangle. $$ or $$ (a_n + Re \langle x,e_n\rangle)^2 + \|x\|^2 - 1 - (Re \langle x,e_n\rangle)^2 =0 $$ or $$ a_n = -Re \langle x,e_n\rangle \pm \sqrt{ 1+ (Re \langle x,e_n\rangle)^2-\|x\|^2} . $$ It follows that $(a_n)$ is a bounded sequence. This implies by (1) that $$ \lambda(x_n-x) = \lambda(a_ne_n)\to 0 $$ for all $\lambda\in H^*$

Answer 2

Separability is contained in the fact that the space has a countable Hilbert basis. Non-separable Hilbert spaces have Hilbert bases in the form $\{e_i\}_{i\in \kappa}$ for some cardinal $\kappa>\aleph_0$. Actually, (1) can be extended to the non-separable case in this way:

Let $H$ be an infinite-dimensional Hilbert space, let $\{e_i\}_{i\in \kappa}$ be a Hilbert basis and $\lambda\in H^*$. Then, for all injective sequences $\left\{e_{i_n}\right\}_{n\in\Bbb N}$, $\lim\limits_{n\to\infty} \lambda\left(e_{i_n}\right)=0$

The proof is basically what you did, because Parseval and Riesz hold in the non-separable case too. Only thing to observe in Parseval is that the set of $i\in \kappa$ such that $\langle e_i,y\rangle\ne 0$ must be at most countable - and so you are only left with the two cases when $i_n$ is frequently in that set or eventually outside.

On to (2). This topological proof works in complex and real spaces, but the idea is inherently real.

Call $S^H=\{x\in H\,:\,\lVert x\rVert=1\}$. We only need to prove that there is a sequence $x_n\in S^H$ such that $\lim_n\lvert \lambda(x_n)\rvert=\lvert\lambda(x)\rvert$, because we can then consider (unless $\lambda(x)= 0$) the sequence $x'_n=\frac{\lambda(x)\overline{\lambda(x_n)}}{\lvert\lambda(x)\rvert\lvert \lambda(x_n)\rvert}x_n$. Recall that $S^H$ is connected unless $H=\Bbb R$. Therefore, $\lvert \lambda(S^H)\rvert$, the image of $\lambda(S^H)$ under the modulus function, is a connected subset of $[0,\infty)$, namely, an interval $J$ with extremal points $a,b$ such that $0\le a\le b\le\infty$. We want to prove that $a\le\lvert\lambda(x)\rvert\le b$: if that's the case, there is a (perhaps constant) sequence $y_n\in J$ such that $y_n\to \lvert\lambda(x)\rvert$; by the definition of $J$, $y_n=\lvert \lambda(x_n)\rvert$ for some $x_n\in S^H$. Notice that, since point (1) says that any Hilbert basis is a sequence such that $\lambda(e_n)\to 0$, we know that $a=0$. Moreover, by the definition of $\lVert \lambda\rVert$, we know that $b=\lVert\lambda\rVert$. But, by definition of $\lVert \lambda\rVert$, we know that $0\le\lvert\lambda(x)\rvert\le\lVert \lambda\lVert$ holds for all $\lVert x\rVert\le 1$. QED

Added: It must be noted, though, that all this machinery is unnecessary, because a much stronger version of point (2) not only holds, but it can be proved with nothing but elementary linear algebra.

Lemma: Let $V$ be a normed vector space over the field $\Bbb K$ ($\Bbb R$ or $\Bbb C$), and such that $\dim V\ge2$. Let $\lambda:V\to\Bbb K$ be continuous and linear. Then, for all $\alpha$ such that $\lvert \alpha\rvert\in\{0\}\cup(0,\lVert\lambda\rVert)$ there is $x$ such that $\lVert x\rVert=1$ and $\lambda(x)=\alpha$. As a corollary, for all $x$ such that $\lVert x\rVert\le 1$ there is $y$ such that $\lVert y\rVert=1$ and $\lambda(y)=\lambda(x)$.

Proof: As before, we only need to prove that there is $y\in S^V$ such that $\lvert \lambda(y)\rvert =\lvert\alpha\rvert$, so we may assume $\alpha\in[0,\lVert\lambda\rVert)\cup\{0\}$. If $\alpha=0$, we only need to find a element of $S^V\cap \ker\lambda$, which exists because $\dim\ker\lambda\ge1$. If $0<\alpha<\lVert\lambda\rVert$, consider $y_1\in S^V\cap\ker\lambda$ and $y_2$ such that $\alpha<\lambda(y_2)\le \lVert\lambda\rVert$. Notice that, necessarily, $y_2$ and $y_1$ aren't linearly dependent. The function $$g:[0,1]\to\Bbb R\\ t\mapsto\left\lvert\lambda\left(\frac{ty_1+(1-t)y_2}{\lVert ty_1+(1-t)y_2\rVert}\right)\right\rvert$$ is therefore well defined and continuous. Since $g(1)\le\alpha\le g(0)$, there must be $t_\alpha$ such that $g(t_\alpha)=\alpha$. In other words, $y=\frac{t_\alpha y_1+(1-t_\alpha)y_2}{\lVert t_\alpha y_1+(1-t_\alpha)y_2\rVert}$ is what we were looking for.