# Boundedness of a subset of a Banach Space

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.

• 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. May 14 at 20:31