convex subset of topological dual

Question

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

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

Let $t\in [0,1]$ and $L,L'\in F(u)$ and $L''=tL+(1-t)L'$.

It's clear that $\left<L'', u\right> = ||u||^2_E$ but I am stuck in proving that $||L''||_{E^∗} = ||u||_E$.

Using triangular inequality we have $||L''||_{E^∗} \le ||u||_E$ but how to get equality?

Thank you for your help.

Answer 1

For simplicity I will write $\|u\|$ for $\|u\|_E$ and $L(v)$ for $ \langle L, v \rangle$. $\|tL+(1-t)L'\|\geq (tL+(1-t)L')(v)$ where $v=\frac u {\|u\|}$. Hence $\|tL+(1-t)L'\|\geq (t\|u\|^{2}+(1-t)\|u\|^{2}) /\|u\|=\|u\|$.

Answer 2

$||L''||_{E^{*}} = ||tL+(1-t)L'||_{E^{*}} = \sup_{u \in E} \frac{\langle tL+(1-t)L', u\rangle}{||u||_{E}} = \sup_{u \in E} \frac{t\langle L,u\rangle +(1-t)\langle L',u\rangle}{||u||_{E}} = \sup_{u \in E} \frac{t||u||_{E}^2 +(1-t)||u||_{E}^2}{||u||_{E}} \geq ||u||_E$