3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ – TigerLa Nov 22 '16 at 22:28
  • 2
    $\begingroup$ @TigerLa He proved that such a Banach space doesn't exist. $\endgroup$ – Dan Saattrup Nielsen Nov 22 '16 at 22:40
  • $\begingroup$ Yes, such a Banach space is always induced by an inner product. $\endgroup$ – marco2013 Nov 22 '16 at 22:47
  • $\begingroup$ 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! $\endgroup$ – TigerLa Nov 23 '16 at 7:31
4
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$ – TigerLa Nov 23 '16 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.