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Let X be a normed space, Z⊆X, and for all $f \in X^*$ set $\{ x: x \in \mathbb{R} ~,~ \exists z \in Z ~~with~~ x=f(z) \}$ is bounded, prove that Z is bounded. I was thinking of using the Theorem (Bounded linear functionals) on page 223 of the book kreyszig introductory functional analysis with applications, but I could do the demonstration, if anyone has any ideas I would appreciate it.

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For fixed $z$, let $x_{z}^{\ast\ast}\in X^{\ast\ast}$ be such that $x_{z}^{\ast\ast}(f)=f(z)$, so for fixed $f\in X^{\ast}$, $|x_{z}^{\ast\ast}(f)|\leq M_{f}$ for all $z$ and hence by Uniform Boundedness Principle we have $\|x_{z}^{\ast\ast}\|\leq M$ for all $z$, then for each $z$, choose an $f$ such that $\|z\|=|f(z)|$ and $\|f\|=1$, then $\|z\|\leq\|x_{z}^{\ast\ast}\|\|f\|\leq M$.

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