Let $E$ a vector space and $f\in E^*$. Why $\|f\|:=\sup_{x\in E, \|x\|\leq 1}|f(x)|=\sup_{x\in E, \|x\|\leq 1}f(x)$?

Question

Let $E$ a vector space and $f\in E^*$. By definition $$\|f\|_{E^*}:=\sup_{x\in E, \|x\|\leq 1}|f(x)|.$$ But why do we have $$\|f\|_{E^*}=\sup_{x\in E, \|x\|\leq 1}f(x) \ \ ?$$

Because linear functional are not positive, do they ?

Answer 1

It is because $f(-x) = -f(x)$.

Answer 2

Umberto's answer is direct and efficient. I thought it might help to fill in the details.

Let $S = \left\{x \in E \mid \Vert x \leq 1 \Vert \right\}$, $m = \sup_{x\in S} f(x)$, and $M = \sup_{x\in S} |f(x)|$ for brevity. The claim is that $m = M$.

Since $f(x) \leq |f(x)|$ for all $x$, it's clear that $m \leq M$. Choose any $x \in S$. Then either $f(x) \geq 0$ or $f(x) < 0$. In the former case, $$ |f(x)| = f(x) \leq m $$ and in the latter case: $$ |f(x)| =-f(x) = f(-x) \leq m $$ The second equality comes from the fact that $f$ is linear, and the inequality from the face that $x \in S \implies -x \in S$. So either way, $|f(x)| \leq m$.

Thus $m$ is an upper bound to $|f|$ on $S$. Since $M$ is the least upper bound of $|f|$ on $S$, $M \leq m$.