Let $X$ be an infinite dimensional normed space over $\mathbb R$ or $\mathbb C$, $X^*$ be the dual space, and $\mathbb{B}$ be the closed unit ball. Part of a proof I’m reading asserts the following identity:
$$\mathbb{B}=\bigcap_{f\in X^*,\|f\|\leq 1}\{x\in X: |f(x)|\leq 1\}.$$
I’m trying to understand why this is true. The forward inclusion is straightforward — if $\|x\|\leq 1,$ the for each $f\in X^*$ with $\|f\|\leq 1$ we have $|f(x)|\leq \|f\|\|x\|\leq 1$. The reverse inclusion feels less straightforward. My professor said it has to do with the Hahn-Banach theorem but I don’t see the connection. Any hints for why this holds?
Suppose $\|x\| > 1$, then define $f(tx) = \|x\|t$ on the linear space $\operatorname{span} \{ x\}$ and extend using Hahn Banach to the entire space. Then $\|f\| = 1$ but $|f(x)|>1$.