Brezis- Functional Analysis, Corollary 1.4

Question

I'm having trouble understanding the proof to this corollary. It goes

For every $x \in E$ we have \begin{align*} ||x|| = \sup_{f \in E^*,||f|| \leq 1} |\langle f,x \rangle| = \max_{f \in E^*, ||f|| \leq 1} |\langle f,x \rangle| \end{align*}

where $E^*$ is the dual space of $E$, they are using the notation $|\langle f,x \rangle| = f(x)$, and $||f|| = \sup_{x \in E, ||x|| \leq 1} |f(x)|$.

The first line of the proof goes,

It is clear that \begin{align*} \sup_{f \in E^*, ||f|| \leq 1} |\langle f,x \rangle| \leq ||x|| \end{align*}

Why is this obvious? For any $f \in E^*$ such that $||f|| \leq 1$, we just know that for any $x \in E$ such that $||x|| \leq 1$, we have $|\langle f,x \rangle| \leq 1$. I tried reasoning by first supposing that $\sup_{f \in E^*, ||f|| \leq 1} |\langle f,x \rangle| > ||x||$ so that there would exist an $f \in E^*$ with $||f|| \leq 1$ and $|\langle f, x \rangle| > ||x||$ but I don't see how this leads to a contradiction.

Any hints? Maybe I'm fundamentally misunderstanding something.

Answer 1

We have the highlighted inequality because for $f\in E^*$ and $x\in E$ we have $$|f(x)|\leq\|f\|\|x\|,$$ and you're taking the supremum over $\|f\|\leq1$. This can be seen by observing that for $x\in E$ nonzero, $x/\|x\|$ has norm $1$, so $$\left|f\left(\frac{x}{\|x\|}\right)\right|\leq \sup_{x \in E, ||x|| \leq 1} |f(x)|=\|f\|.$$ Multiplying by $\|x\|$ yields $|f(x)|\leq\|f\|\|x\|$.

If you have any further questions, let me know and I will edit.