Showing that inf$_{x \in M}||x||_{E} = \frac{1}{||f||}$ for $E$ a Banach space, $f \in E'$ and $M = \{x \in E: f(x) =1\}$ Sorry if the question title is a bit cramped -- but I thought, in the end, it was short enough to fit. (If it's not standard notation, by $E'$ I mean the dual of $E$.) I think the question is quite clear. My progress so far: I can show that the infimum is at least $\frac{1}{||f||}$ like this: for any $x \in M$, $|f(x)| \leq ||f||\cdot ||x||_{E}$, so by the definition of $f$, $1 \leq ||f||\cdot ||x||_{E}$, and so $||x||_{E} \geq \frac{1}{||f||}$ for any $x\in M$. Thus inf$_{x\in M}||x||_{E} \geq \frac{1}{||f||}$. But I'm not sure where to begin on proving the reverse inequality. I imagine you need to pick a specific element of $M$ which has norm $\frac{1}{||f||}$, which you should be able to do since $M$ is closed and convex, but I don't know how to go about finding that element.

For all this assume $f\
eq0$.

First, if $x\in M$ then $f(x)=1 ≤ \|f\|\,\|x\|$ and thus $\|x\|≥\frac1{\|f\|}$. The inequality holds when you pass to the infimum.

Next note that $\|f\|=\sup_{x\in E,\, \|x\|=1}|f(x)|$. For that reason for any $\epsilon>0$ there exists a norm one vector $x_\epsilon\in E$ so that $$\|f\|≥|f(x_\epsilon)|≥\|f\|-\epsilon.$$ By making $\epsilon$ small we may assume that $f(x_\epsilon)\
eq0$, then $\frac{x_\epsilon}{f(x_\epsilon)}\in M$ and you get: $$\frac1{\|f\|-\epsilon}≥ \left\|\frac{x_\epsilon}{f(x_\epsilon)}\right\|≥\inf_{x\in M}\|x\|.$$ This can be done for any $\epsilon>0$ small enough, so take $\epsilon\to0$ to achieve the remaining inequality.