$\Vert x\Vert$ in terms of bounded linear functional $f$.

Question

Let $X$ be a normed space, $X'$ be its dual and $f \in X'$. Then we have $$\Vert x\Vert=\sup_{f \neq0}{|f(x)| \over \Vert f\Vert}$$ Then, is the following also true $$\Vert x\Vert=\sup_{\Vert f\Vert=1} |f(x)|$$?

Answer 1

Let's rename $x$ to $x_0$. For $x_0 = 0$, the claim is easily derived using linearity. For $x_0 \neq 0$ one can apply a corollary of Hahn-Banach ensuring the existence of a functional $x_0' \in X'$ with $x_0'(x_0) = \Vert x_0 \Vert$ and $\Vert x_0' \Vert = 1$. Therefore we have

$$\Vert x_0 \Vert \le \sup_{\Vert f \Vert = 1} |f(x_0)|$$

Since $f \in X'$, we also have $\sup_{\Vert f \Vert = 1} |f(x_0)| \le \sup_{\Vert f \Vert = 1} \Vert f \Vert \Vert x_0 \Vert = \Vert x_0\Vert$. Combining both results yields the claim.