For each functional $f$ there exist an $x$ such that some properties hold

Question

I'm trying to prove or disprove the following statement:

Let $(X,\|\cdot\|)$ be a normed $\mathbb{R}$-vector space.

1. For each $f\in X^*$ there exists $x\in X$ with $\|x\|\leq 1$ such that $f(x)=\|f\|$

Now intuitively I would say this is true since the operatornorm is defined as $$\|f\|=\sup\limits_{\|x\|\leq 1}|f(x)|$$ So we could choose $$f(x):=\sup\limits_{\|x\|\leq 1}|f(x)|$$

But I'm not sure if the following still holds: $$\sup\limits_{\|x\|\leq 1}|f(x)|=\left\|\sup\limits_{\|x\|\leq 1}|f(x)|\right\|$$

What also keeps me doubting my solution is the fact that I did not use the Hahn-Banach theorem. I had to prove another statement that was very similar but about the existance of a linear functional for each $x\in X$ with $f(x)=\|f\|$, where I used the Hahn-Banach theorem.

Answer 1

When we choose $x$ so that $f(x)=\sup_{\|x\|\leq 1}|f(x)|$, we're assuming the result. That is, we're assuming that there exists $x\in X$ with $\|x\|\leq 1$ so that $f(x)=\|f\|$.

Otherwise, I'll point out that the result is true when $X$ is finite-dimensional. In finite-dimensional normed vector spaces the closed unit ball $\{x\in X:\|x\|\leq 1\}$ is compact, meaning that $f$ achieves a maximum value on this ball, and this maximum value must be $\sup_{\|x\|\leq 1} |f(x)|$. The result can fail, however, when $X$ is infinite-dimensional (because the closed unit ball is no longer necessarily compact). For examples of this failure, see this question.

Answer 2

As Austin has pointed out, the statement is not necessarily true in the infinite-dimensional case.

I would like to add that (in the Banach space case) there actually exists a precise characterization of when your statement is true: it is true if and only if $X$ is reflexive.

This is the content of James' theorem.

Edit: I'm of course talking about Banach spaces here. I initially overlooked that you are concerned with general normed spaces. Thanks to Aweygan for pointing this out.