Let $X$ be a Banach space. Show that:

(1) A set $A^{\prime} \subset X^{\prime}$ is bounded in $X^{\prime}$ if and only if $M_{x}=\left\{f(x): f \in A^{\prime}\right\}$ is bounded for every $x \in X$.

(2) A set $A \subset X$ is bounded in $X$ if and only if $M_{f}=\{f(x): x \in A\}$ is bounded for every $f \in X^{\prime}$.

Here $X'$ denotes the set of bounded linear functionals on $X$. For the 2nd part of the question, here is my attempt:

$\rightarrow:$ Assume that $M_{f}=\{f(x): x \in A\}$ is bounded.

Thus, $ \forall f \in X^{\prime} \exists c>0: \sup _{a \in A} f(a) \leq c.$

Then for each $a \in A$, consider the linear functional $F_x$ on $X'$ defined as: $F_x(g) := g(x)$.

By definition, it is easy to see that $F_x$ is linear and bounded. Moreover,

$||F_x||_{X''}=sup_{g \neq 0} \cfrac{|F_x(g)|}{||g||} \leq ||x||$.

For any $x \in X$, there exists $g^* \in X'$ such that $||g^*{||}_{X}=1$, thus $g^*(x)=||x||$. Thus, $F_x(g^*)=g^*(x)=||x||$ implying that $||x{||}_X = F_x(g^*) \leq |F_x(g)|=||F_x{||}_{X''}$. So, we have $||F_x{||}_{X''} = ||x||$.

Now, by assumption $\{F_x(g) : x \in A \}$ is bounded, since $F_x(g)=g(x)$, where $g\in X'$ and $sup _{x \in A} \leq c$ for some $c>0$. By the Uniform Boundedness Principle, $||F_x{||}_{X''} = ||x||$ is bounded for $x \in A$. So, the set $A$ is bounded.

Is my attempt correct? If it is possible, could you help me on the reverse direction for 2nd part? and the both sides in 1st? Thanks in advance.

  • 1
    $\begingroup$ The directions $\Rightarrow$ follow directly from $$ \vert f(x) \vert \leq \Vert f \Vert_{X'} \Vert x \Vert_X.$$ The $\Leftarrow $ for $1.)$ is directly the uniform boundedness principle. And yes, your proof is fine, maybe a bit long, but correct. $\endgroup$ May 14 at 20:31


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