# 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?

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).

• Our teaching assistant told us that such a Banach space exists and I was confused how that is possible. So, his statement was wrong. Thanks for the clarification! Commented Nov 23, 2016 at 9:30

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.

• Thanks for your answer but I was asking for a $\mathbb C$-Banach space which is NOT induced by an inner product. But this one IS induced by it, right? Commented Nov 22, 2016 at 22:28
• @TigerLa He proved that such a Banach space doesn't exist. Commented Nov 22, 2016 at 22:40
• Yes, such a Banach space is always induced by an inner product. Commented Nov 22, 2016 at 22:47
• Ah sorry, was a little too early to comment. Our teaching assistant told us that such a Banach space exists and I was confused how that is possible. So, his statement was wrong. Thanks a lot for the proof! Commented Nov 23, 2016 at 7:31