a Hilbert space is strictly convex Prove that a Hilbert space is strictly convex in the following sense: for all $u, v \in E \times E,$ with $u \ne v$, $\| u \|_E =
\| v \|_E = 1$ and for all $t \in ]0, 1[$ we have
$$|tu + (1 − t)v| < 1.$$
I tried prove that a norm is strictly convex in a Hilbert space but couldn't go far.
Thank you for your help or any reference to a book on this subject.
 A: My argument is for a real space. Let $t\in(0,1).$ We have $$\|tu+(1-t)v\|^2=t^2\|u\|^2+2t(1-t)\langle u,v\rangle+(1-t)^2\|v\|^2$$ (here we need a scalar field to be $\Bbb R$, because to use a symmetry of the inner product). Since $\|u\|=\|v\|=1$, and $\langle u,v\rangle\le\|u\|\|v\|=1$ (the Schwarz inequality), then $$\|tu+(1-t)v\|\le 1.$$ Suppose that $\|tu+(1-t)v\|=1.$ Then $t^2+(1-t)^2+2t(1-t)\langle u,v\rangle=1,$ whence $$2t(1-t)\bigl(1-\langle u,v\rangle\bigr)=0.$$ As a consequence (remember $t\in(0,1)$), $$\langle u,v\rangle=1=\|u\|\|v\|.$$ Then $u,v$ are linearly dependent, $u=\alpha v$ and $$1=\|u\|=|\alpha|\|v\|$$ and $\alpha=\pm 1.$ $\alpha=1$ is impossible by $u\ne v$. Then $u=-v$. This means that $$\|tu+(1-t)v\|=|2t-1|<1$$ because $t\in(0,1)$. Of course, this contradicts our conjecture $\|tu+(1-t)v\|=1.$
In a complex space the argument should be more careful.
At the beginning we will have $$\|tu+(1-t)v\|^2=t^2+(1-t)^2+2t(1-t)\text{Re}(\langle u,v\rangle).$$ Further continue also with Schwarz inequality.
