Proving that $\exists f \in X^*$ : $f(x) = \|x\|^2$ and $\|f\| = \|x\|$

Question

Exercise :

Let $X$ be a normed space. Prove that for all $x \in X$ there exists $f \in X^*$, such that $f(x) = \|x\|^2$ and $ \|f\| = \|x \|$.

Thoughts :

I apologise for not providing a proper attempt but this is one of the first such exercises I'm handling, so I seem at loss.

Initially, I thought about the Riesz Representation Theorem, which could yield the result straightforward, but the space we are working over must be a Hilbert Space, which we do not know in the given exercise.

The second possible solution could (and probably should) be based on the Hahn-Banach Theorem (or one of its results/applications) but I cannot see a way out.

Any hints or elaborations will be greatly appreciated.

Answer 1

If $x=0$, we are done. Now let $x \in X$ and $x \ne 0$. A consequence of the Hahn-Banach theorem is the existence of some $g \in X^*$ with

$g(x)=||x||$ and $||g||=1.$

Now it is easy to see that $f:=||x||g$ has the desired properties.

Answer 2

Hint:

Fix an $x\in X$. If $x=0$ then the answer is rather easy, so we can assume $x\neq 0$.

Next, define a suitable functional on the linear hull of $x$, which is the linear subspace $V:=\{ \alpha \cdot x | \alpha \in \mathbb R\}$. Then extend this functional onto the whole space $X$ using the Hahn-Banach theorem.