0
$\begingroup$

Let X be a normed space and Y a Banach space, and let $\{T_n\}_{n \in \mathbb{N}} \subset B(X,Y)$, show that the following are equivalent:

i) $sup_{n \in \mathbb{N}} \|T_{N}\| < +\infty$;

ii) for any $x \in X$, $sup_{n \in \mathbb{N}} \|T_n(x)\| <+\infty$;

iii) for any $\phi \in Y^*$, $sup_{n \in \mathbb{N}}|\phi(T_n(x))| < +\infty$.

Here $Y^*$ denotes the topological dual of the space

My attempt so far:

For i -> ii I argued that for any $x \in X \; \|T_n(x)\| < \|T_n\|\|x\|$

As for ii -> iii I used a similar argument as before by saying that the norm of $\phi$ is finite

As for the rest, my teacher advised we prove that iii->ii and ii->i. Now ii->i is pretty simple, it's the well known Banach-Steinhaus theorem, but I don't see a way to prove iii -> ii.

Thanks in advance.

$\endgroup$
1
  • $\begingroup$ iii) says "for any $\phi$" but it doesn't say "for any $x$". $\endgroup$
    – GEdgar
    Jul 2, 2021 at 0:44

1 Answer 1

0
$\begingroup$

The proof of $iii\rightarrow ii$ is also an application of Banach Steinhaus. Let $sup_{n \in \mathbb{N}}|\phi(T_n(x))| < +\infty\Rightarrow sup_{n \in \mathbb{N}} |(T_n(x))^{**}(\phi)|<\infty$ where $(T_n(x))^{**}:X^*\rightarrow \mathbb{C}$ s.t. $(T_n(x))(\phi)=\phi(T_n(x))$ and the map $X\rightarrow X^{**}$ where $x\mapsto x^{**}$ is an isometric embedding i.e. $\|T_n(x)\|=\|T_n(x)^{**}\|$. Now applying Banach Steinhaus on the space $X^*$, you get that $sup_{n \in \mathbb{N}} \|{(T_n(x))^{**}}\|<\infty\Rightarrow sup_{n \in \mathbb{N}} \|(T_n(x))\|<\infty$.

Little more details: Define $X^{**}$ to be the dual space of $X^*$ i.e. $\{\psi:X^*\rightarrow \mathbb{C} \ni \psi$ is a bounded linear functional$\}$. Now there is an embedding of $X\rightarrow X^{**}$ where $x\mapsto x^{**}$ such that $x^{**}(\phi)=\phi(x)$. You can easily show that $x^{**}$ is a bounded linear functional by a Corollary of the Hahn Banach theorem. The corollary states that given $x\in X$ $\exists \phi_x\in X^* \ni \phi_x(x)=\|x\|$ and $\|\phi_x\|=1$.

First oberve that $\|x^{**}\|=sup_{\phi\in X^*(\neq 0)}\frac{|x^{**}(\phi)|}{\|\phi\|}=sup_{\phi\in X^*(\neq 0)}\frac{|\phi(x)|}{\|\phi\|}\leq sup_{\phi\in X^*(\neq 0)}\frac{\|\phi\|\|x\|}{\|\phi\|} =\|x\|$. Thus $ \|x^{**}\|\leq \|x\|$. Again by the corollary stated above $\exists \phi_x\in X^* \ni \phi_x(x)=\|x\|$ and $\|\phi_x\|=1$. This implies $x^{**}(\phi_x)=\phi_x(x)=\|x\|$, and hence $\|x^{**}\|= \|x\|$. Thus $x^{**}$ is a bounded linear functional.

$\endgroup$
2
  • $\begingroup$ Thank you for the answer, but the solution is really confusing, you didn't define the functions and embeddings properly, making it hard for anyone to read and understand $\endgroup$ Jul 1, 2021 at 14:01
  • $\begingroup$ @GuilhermeTakata Look at the details now, tell me if you require any more. $\endgroup$ Jul 1, 2021 at 23:54

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.