0

Let $\big(X,\langle\ \rangle\big)$ be a Hilbert space over $K$. I want to prove the following

If $(x_n)_{n\in\Bbb N}$ is an orthonormal sequence in $X$ $\Rightarrow\; x_n\to0$ weakly

My attempt:

Let $f\in X^*$ so, by Riesz's theorem, there $\exists\;y_f\in X$ s.t. $\;f(x)=\langle x,y_f\rangle\;\forall x\in X$ and $\|y_f\|=\|f\|$. In particular, we get that $f(x_n)=\langle x_n,y_f\rangle\;\forall n\in\Bbb N$. So, fixing $f$ and by C-B-S inequality we get that:

$$ \big|\;f(x_n)\;\big|=\big|\; \langle x_n,y_f \rangle \;\big|\le\|x_n\|\|y_f\|=\|f\|\;\;\forall n\in\Bbb N $$

and since we can do this for every $f$ we take we can make it as small as wanted so that: $$ f(x_n)\to0\;\;\forall f\in X^* $$ whichs means that $x_n\to0$ weakly.
Am I doing something wrong? Would appreciate your comments and help if it's not complete.

Arnulf
  • 1,231

1 Answers1

2

Weak convergence means: for every $f$ the sequence $\{f(x_n)\}$ tends to $0$. When you are proving a statement that begins with "for every $f$", you don't get to choose $f$.

So, there is no reason for $\|f\|$ to be small. Your argument doesn't work.

Instead, use Bessel's inequality $\sum |\langle y_f, x_n\rangle|^2 \le \|y_f\|^2$, and the fact that the terms of a convergent series must tend to zero.