prove a set is non empty using Hahn-Banach theorem

Question

Let $(E, ||.||_E )$ be a banach space, and its $E^*$ topological dual.

For $u ∈ E, F(u) =\{L\in E^*, ||L||_{E^∗} = ||u||_E, \left<L, u\right> = ||u||^2_E \}$

Prove $F(u)$ is non empty and closed.

What I did:

I proved that $F(u)$ is closed as $F(u)=f^{-1}(||u||_E) \cap g^{-1}(||u||_E^2)$ with $f$ and $g$ continuous functions.

For non emptiness, I tried to use Hahn-Banach theorem but I couldn't identify the vector spaces to use.

Thank you for your help.

Answer 1

One of the consequences of the Hahn-Banach theorem is the following one.

Proposition Let $E$ be a normed space, and $u\in E$. Then there exists a bounded linear functional $F$ on $E$ such that $\|F\|=1$ and $Fu=\|u\|$.

Proof. If $u=0$, the assertion is obvious. Assume $u\neq 0$, and let $M$ be the one-dimensional subspace of $E$ spanned by the vector $u$. Define a linear functional $f$ on $M$ by $f(tu)=t\|u\|$. The norm of $f$ is: $$\|f\|=\sup\{|f(tu)|:\|tu\|\leq 1\}=\sup\{|t|\|u\|: |t|\leq\frac{1}{\|u\|}\}=1$$ and $f(u)=\|u\|$. By the Hahn-Banach theorem, there exists a linear functional $F$ defined on $E$ such that $F$ agrees with $f$ on the subspace $M$, and such that $\|F\|=\|f\|$. Thus, $Fu=fu=\|u\|$ and $\|F\|=1$. Q.E.D

In your problem, Given $u\in E$ the proposition guarantees the existence of $F\in E^*$ such that $\|F\|=1$ and $Fu=\|u\|$. Take $L=\|u\|F$. Then $Lu=\|u\|Fu=\|u\|^2$, and $\|L\|=\|u\|$.