If $\{x^*(x_\lambda); \lambda\in\Lambda\}$ is bounded for each $x^*\in X^*$, then the set $\{x_\lambda; \lambda\in\Lambda\}$ is bounded

Question

Problem

$X$ is a Banach space. $\{x_\lambda\}_{\lambda\in\Lambda} \in X.$

$X^*$ is the dual space.

Show the following.

If $\forall x^*\in X^*, |x^*(x_\lambda)|\le M \Rightarrow \{\|x_\lambda\|\}_{\lambda\in \Lambda}$ is bounded.

($M$ is a constant)

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My trial (by contradiction method)

If $\{\|x_\lambda\|\}_{\lambda\in \Lambda}$ is not bounded, $\forall M\in\mathbb R$, $\exists\lambda_1\in\Lambda, \|x_{\lambda_1}\|\gt M$.

Thus, we can think $x^*\in X^*$ such that $x^*(x_{\lambda_1})=\|x_{\lambda_1}\|$.

$\therefore x^*(x_{\lambda_1})\gt M$.

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I am not confident about my proof. Does $x^*$ described in my trial exist? I think I can make a linear operator $x^*$ such that $x_{\lambda_1}\mapsto \|x_{\lambda_1}\|$, and others (which are not related with $x_{\lambda_1}$) $\mapsto 0$.

Thanks for reading my question.

Answer 1

A problem with your solution: $x^*(x_{\lambda}) = \| x_\lambda\|$ is not a linear functional and I'm unconvinced that you can find a linear functional $x^*$ such that $x^*(x_{\lambda_M}) = \|x_{\lambda_M}\|$ for each $\lambda_M$ (note, in your solution $\lambda_1$ should actually depend on $M$, hence I used $\lambda_M$ instead).

What you want to use here is the Banach-Steinhaus theorem (also called the principle of uniform boundedness) which says the following: let $B$ be a Banach space, $C$ be a normed vector space and $\{T_\alpha\}_{\alpha \in \mathcal A}$ be a collection of bounded linear maps $T_\alpha:B \to C$. If the set $$\{\|T_\alpha(x)\|_C\}_{\alpha \in \mathcal A}$$ is bounded for each $x \in B$, then $$\{\|T_\alpha\|_{B\to C}\}_{\alpha \in \mathcal A}$$ is also bounded. Here $\|T_\alpha\|_{B \to C}$ is the operator norm of $T_\alpha$ as a map from $B$ to $C$.

How to apply this to our problem? Since $X$ is a Banach space, so is $X^*$. Now each $x \in X$ can be considered as a linear map $x:X^* \to \mathbb C$ by the rule $x(x^*) := x^*(x)$ for each $x^* \in X^*$. With this association, the operator norm $\|x\|_{X^*\to \mathbb C}$ is equal to the original norm $\| x\|_X$. Considering $\{x_\lambda\}$ as a collection of bounded linear maps, we have then assumed that for any $\lambda$ the set $$\{\|x_\lambda(x^*)\|_{X^*}\}_{x^* \in X^*}$$ is a bounded set for each $x^* \in X^*$. Then by the Banach Steinhaus theorem, the set $$\{\|x_\lambda\|_{X^* \to \mathbb C}\}_{x \in X}$$ is also bounded. But then from $\|x_\lambda\|_{X^*\to\mathbb C} = \|x_\lambda\|_{X}$, we conclude that $\{x_\lambda\}$ is bounded.