problem 5.5.63 - Let $\mathcal{H}$ be an infinite-dimensional Hilbert space.

a.) Every orthonormal sequence in $\mathcal{H}$ converges weakly to 0.

b.) The unit sphere $S = \{x:\lVert x\rVert = 1\}$ is weakly dense in the unit ball $B = \{x:\lVert x\rVert \leq 1\}$. (In fact, every $x\in B$ is the weak limit of a sequence in $S$).

Attempted proof a.) Let $\{u_n\}_{1}^{\infty}$ be an orthonormal sequence in $\mathcal{H}$. Then by Bessel's inequality for any $x\in \mathcal{H}$, $$\sum_{n=1}^{\infty}|\langle x,u_n\rangle|^2 \leq \lVert x\rVert^2$$ Suppose $f\in\mathcal{H^*}$ such that $f = \langle \cdot , x\rangle$. Now we know the sum converges by Bessel's Inequality. So, $$\lim_{n\rightarrow \infty}|f(u_n)|^2 = \lim_{n\rightarrow \infty}|\langle x,u_n\rangle|^2 = 0$$ Thus it follows that $$\lim_{n\rightarrow \infty}f(u_n) = 0 = f(0)$$ hence $\{u_n\}_{1}^{\infty}$ converges weakly to $0$

Attempted proof b.) Taking the suggestions from John Dawkins, Let $\{u_n\}_{1}^{\infty}$ be an orthonormal sequence in $\mathcal{H}$. Now define $x_n := x + \beta_n u_n$ to be a sequence of elements of $S$. Pick numbers $\beta_n$ such that $\| x_n \| = 1$.

I am not sure where to go from here, and I am not very sure what it means to be weakly dense.



For a.), forget $x'$ and focus on the left side of Bessel's inequality: What do you know about the terms of a convergent sequence?

For b.), let $\{u_n\}$ be an orthonormal sequence, choose numbers $\beta_n$ such that $x+\beta_nu_n$ has norm $1$, and then show that $\sup_n|\beta_n|<\infty$. Now use this fact in combination with a.).

  • $\begingroup$ I am not completely sure I understand your question in regards to the terms of a convergent sequence, could you clarify? $\endgroup$ – Wolfy Apr 3 '16 at 16:54
  • $\begingroup$ If $\sum_{n=1}^\infty a_n$ converges, then $\lim_n a_n=0$. $\endgroup$ – John Dawkins Apr 3 '16 at 16:57
  • $\begingroup$ I re-edited a.) I think it is correct now. I will use your suggestion for b.), could you provide a bit more insight on your suggestion for b.)? $\endgroup$ – Wolfy Apr 3 '16 at 17:09
  • $\begingroup$ $x_n:=x+\beta_nu_n$ is a sequence of elements of $S$ that will converge weakly to $x\in B$ provided the sequence $\{\beta_n\}$ is bounded. Compute $\beta_n$ and check the boundedness. $\endgroup$ – John Dawkins Apr 3 '16 at 17:31
  • $\begingroup$ I am unsure what it means to be weakly dense. Could you define that for me? Perhaps it would easier to just show your proof of b.) I am not really getting anywhere as of now $\endgroup$ – Wolfy Apr 3 '16 at 17:44

a)It is actually simpler : $\sum_{n=1}^{\infty}|\langle x,u_n\rangle|^2 \leq \lVert x\rVert^2\le +\infty$. So $\lim_{n \to \infty}|\langle x,u_n\rangle|^2=0$, so $\lim_{n \to \infty}\langle x,u_n\rangle = 0$. It is true $\forall x$, so by Riesz representation theorem, $\forall f \in H^*$, $\lim_{n \to \infty}f(u_n)=0$.

b) Let $x_0 \in B-S$. Let $V$ be a weak neighbourhood of $x_0$. $\exists \epsilon>0$ and $\phi_1,...,\phi_n \in H^*$ such as $x_0+V_{\epsilon,\phi_1,...,\phi_n} \subseteq V$. Since $H$ is infinite dimensional, $\exists x\neq 0$ such as $x \in \bigcap_{k=1}^n ker(\phi_k)$. If $f(t)=||x_0+tx||, t\in \mathbb{R}$, $f$ is continuous on $\mathbb{R}$ and we have $f(0)=||x_0||<1$ and $ \lim_{t \to \infty} f(t) =+\infty $. We apply the intermediate value theorem : $\exists t_0>0$ such as $f(t_0)=1$. Then $x_1=x_0+t_0x \in S$ and we have : $|\phi_k(x_1-x_0)|=t_0|\phi_k(x)|=0 \le \epsilon,\space 1\le k \le n$. So $x_1 \in x_0+V_{\epsilon,\phi_1,...,\phi_n}\subseteq V$ and $S \cap V \neq \emptyset$. Finally $S$ is weakly dense in $B$.

  • $\begingroup$ Why did you use the kernal? Your proof is a bit over my head $\endgroup$ – Wolfy Apr 3 '16 at 17:15
  • $\begingroup$ It is related to the neighbourhoods involved in the definition of the weak topology. You have $V_{\epsilon,\phi}(0)=\begin{Bmatrix} x \in H | \space |\phi (x)| \le \epsilon \end{Bmatrix}$ where $\phi \in E^*$. And you have $V_{\epsilon,\phi_1,...,\phi_n}(0)=\bigcap_{k=1}^n V_{\epsilon,\phi_k}(0)$, so in particular $V_{\epsilon,\phi_1,...,\phi_n}(0) \supseteq \bigcap_{k=1}^n ker(\phi_k)$ $\endgroup$ – Bérénice Apr 3 '16 at 19:10
  • $\begingroup$ Interesting, what book have you used for measure theory or functional analysis? $\endgroup$ – Wolfy Apr 5 '16 at 15:59
  • $\begingroup$ Hmm I am french, so I have a pdf with my lecture notes of functional analysis but it is in french :/. $\endgroup$ – Bérénice Apr 5 '16 at 16:10

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