If the supremum of absolute value of bounded linear functional is finite in a normed linear space $X$ then supremum of $\|x_\alpha\|$ is also finite.

Question

I was supposed to use the Principle of Uniform bounded to prove the following assertion:

If $\{x_\alpha\}$ be the set of elements in a normed linear space $X$ and

$\sup\limits_\alpha |f(x_α)| < \infty$ for any $f \in X^*$ (Topological Dual) then $\sup\limits_\alpha \|x_α\| < \infty$

I don't have any clue to start this. Any hint is appreciated.

Answer 1

I think you can use a corallary of the Hanh Banach Theorem which says for every $x_\alpha\neq 0$, there exists a $f \in X^*$ such that $||f||=1$ and $f(x_\alpha)=||x_\alpha ||$

Answer 2

Hint: consider the family of functionals $\Lambda_\alpha\in X^{**}$ defined by $\Lambda_\alpha(f) := f(x_\alpha)$ for every $f\in X^*$.