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.

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

1 Answer 1


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.

  • $\begingroup$ 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

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