Existence of a $\mathbb C$-Banach space isometric to a Hilbert Space but whose norm is not induced by an inner product? As written in the title, does there exists a $\mathbb C$-Banach space isometric to a Hilbert Space but whose norm is not induced by an inner product?
Since an inner product in the Hilbert space has to fulfill the Parallelogram Identity, how could it be that such a $\mathbb C$-Banach space exists?
 A: No such space exists.  Let $H$ be a Hilbert space, and suppose $X$ is a $\mathbb{C}$-Banach space with $T:X\to H$ an isometry.  Then define an inner product on $X$ by $\langle x,y\rangle=\langle T(x),T(y)\rangle$ (where the right-hand side is the inner product of $H$).  Then this inner product induces the norm of $X$, since $$\langle x,x\rangle=\langle T(x),T(x)\rangle=\|T(x)\|^2=\|x\|^2$$ (the last equality being because $T$ is a isometry).
A: Let $f:E \rightarrow F$ an isometry from $E$, a Banach space, to $F$, a Hilbert space with inner product $(x,y) \mapsto L(x,y)$. Then $\forall x \in E, \|f(x)\|_F=\|x\|_E$.
$\forall x,y \in F, \Re L(x,y)=\frac{1}{2}(\|x+y\|_F^2-\|x\|_F^2-\|y\|_F^2)$
because $\|x+y\|_F^2=L(x+y,x+y)=L(x,x)+L(x,y)+L(y,x)+L(y,y)$ and because $L(y,x)= \overline{L(x,y)}$.
And $\forall x,y \in F , \Im L(x,y)=\frac{1}{2}(\|x-iy\|_F^2-\|x\|_F^2-\|y\|_F^2)$
So $L'(x,y)=L(f(x),f(y))=\frac{1}{2}(\|f(x)+f(y)\|_F^2-\|f(x)\|_F^2-\|f(y\|_F^2)+\frac{i}{2}(\|f(x)-if(y)\|_F^2-\|f(x)\|_F^2-\|f(y\|_F^2)$
is an inner product on $E$.
And $\forall x \in E, L'(x,x)=\|f(x)\|_F^2=\|x\|_E^2$.
So $\|.\|_E$ is induced by an inner product.
