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I want to show that if linear mapping $L:B_1\rightarrow B_2$ is isomorphism of Banach space and $\|L(x)\|_{B_1} =\|x\|_{B_2} $ (surjective and isometry) so it consist that $L$ is isomorphism of Hilbert space, i.e $\langle x,y\rangle =\langle Lx,Ly\rangle $

my way to proove it is by using inner product properties and isometry: $\langle Lx,Ly\rangle =L\langle x,Ly\rangle=L(\langle Ly,x\rangle)^*=LL^*(\langle y,x\rangle)^*=LL^*\langle x,y\rangle=I\langle x,y\rangle=\langle x,y\rangle $

Is it good proof ?

Is legal take out $L$ from the inner product ?

If not what is the right way to do it ?

Thank you.

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I guess, it is meant that $B_1$ and $B_2$ are Hilbert spaces, and $L$ is a surjective isometry $B_1\to B_2$, and you want to show that $L$ also preserves the inner product.

$L$ cannot be taken out from the inner product, it's nonsense.

Else, rather use the polarization identity: $$4\langle x,y\rangle=\|x+y\|^2+i\|x+iy\|^2-\|x-y\|^2-i\|x-iy\|^2\,.$$

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