I just need a sanity check. Can someone verify this proof for me?

Theorem: Suppose $X$ is a normed linear space, and $x_n$ is an unbounded sequence. Then, there exists some $f \in X^*$, the topological dual, such that $f(x_n)$ is unbounded.

Proof. Suppose that $f(x_n)$ is bounded for all $f \in X^*$. Then, the set $\lbrace \widehat{x}_n \rbrace \in X^{**}$ has the property that $\widehat{x}_n(f)$ is bounded for all $f \in X^*$. By the Uniform Boundedness Principle on the complete space $X^*$, $\lbrace \widehat{x}_n \rbrace$ must be bounded. Since the canonical embedding of $X$ into $X^{**}$ is an isometry, $\lbrace x_n \rbrace$ is also bounded.

In particular, I just want to be certain that the completeness of $X$ is not required here.


1 Answer 1


Yes, it's fine. In particular, there is no need to suppose that $X$ is a Banach space.


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