1
$\begingroup$

Let $\mathbb{H}$ be a Hilbert space. Let $\{x_n\}$ be a sequence in $\mathbb{H}$ with the property that $\langle x,x_n \rangle\to 0$ as $n\to\infty$ for $x\in\mathbb{H}$. Show that $\sup\{\|x_n\|:n=1,2,3,...\}<\infty$.

So this is what I have:

As $\langle x,x_n \rangle\to 0$, then for $\epsilon > 0$ there exists $N\in\mathbb{N}$ such that $\|\langle x,x_n \rangle-0\|<\epsilon, \forall n\geq N$. Also, as $\mathbb{H}$ is a Hilbert space, then we know that Then either $x=0$ or $x_n\to 0$. If $x_n\to 0$ then $x_n$ is bounded by some value $b$, and therefore $\sup\{\|x_n\|:n=1,2,3,...\}\leq b\lt\infty$. Now if $x=0$ and $x_n\to y$ where $y\neq\infty$, then once again $x_n$ is bounded by some $b$ and $\sup\{\|x_n\|:n=1,2,3,...\}\leq b\lt \infty$. Finally, let $\langle x,x_n \rangle \to 0$, $x=0$ and $x_n\to\infty$, then $\lim_{n\to\infty}\langle x,x_n\rangle=\langle \lim_{n\to\infty} x,\lim_{n\to\infty}x_n\rangle=\langle x,\infty\rangle$ which is undefined (so this cannot be, based on the asumptions). Therefore for all valid cases, $\sup\{\|x_n\|:n=1,2,3,...\}\lt\infty$.

Any comments would be appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ You cannot conclude directly that either $x=0$ or $x_n\to 0$. You need to apply Uniform Boundedness principle here. $\endgroup$ Oct 20 '14 at 6:13
  • $\begingroup$ The Baire category theorem should feature in your proof... $\endgroup$
    – copper.hat
    Oct 20 '14 at 6:23
1
$\begingroup$

Let $T_nx = \langle x_n , x \rangle$. We have $\sup_n |T_n x| < \infty$, hence by the Banach Steinhaus theorem we have $\sup_n \|T_n\| < \infty$. Since $\|T_n\| = \|x_n\|$, we have the desired result.

$\endgroup$
7
  • $\begingroup$ So the answer you have provided makes perfect sense (I was looking up Banach Steinhaus, and it seems to work perfectly). I am just curious as to your comment above about the need to use the Baire category theorem - why would that be needed in this proof? @copper.hat $\endgroup$ Oct 20 '14 at 6:46
  • $\begingroup$ The Banach Steinhaus is proved using the Baire category theorem... $\endgroup$
    – copper.hat
    Oct 20 '14 at 6:51
  • $\begingroup$ Hahaha, okay. But we have already proven the Banach Steinhaus theorem, so the actual proof wouldn't reference Baire category - it would just be inferred through the use of Banach Steinhaus, correct? @copper.hat $\endgroup$ Oct 20 '14 at 6:57
  • $\begingroup$ Correct, I really meant to say that since the Baire category theorem was missing from your proof that something must not be right... $\endgroup$
    – copper.hat
    Oct 20 '14 at 6:58
  • 1
    $\begingroup$ Cauchy Schwartz gives $\|T_n x\| \le \|x_n\| \|x\|$, so $\|T_n \| \le \|x_n \|$. Then $T_n x_n = \|x_n\|^2$, which gives equality. $\endgroup$
    – copper.hat
    Oct 20 '14 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.