# Isometric isomorphism

In the case that $L:B_1 \rightarrow B_2$ is a linear mapping of Banach spaces and $L$ is a isometric isomorphism (bijection and $\|Lx\|_{B_1} = \|x\|_{B_2}$) can I say that $L\overline{L}= 1$ is trivial ? (the bar denotes the complex conjugate);

TIA

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I don't think so, but before I can be sure, I must ask how the complex conjugate of $L$ is defined? – Harald Hanche-Olsen Feb 25 '13 at 21:00
So $\overline{L}$ is defined as $\overline{L}(\alpha x+\beta y)=\overline{\alpha}L(x)+\overline{\beta}L(y)$. Right? – Norbert Feb 25 '13 at 21:00
What is meant by $L\overline{L}$? Is there a multiplication operation on $B_2$? – Zev Chonoles Feb 25 '13 at 21:01
Is there a confusion here with the adjoint of an isometry between Hilbert spaces? For which $LL^*=1$? That particular result has no generalization to Banach spaces that I know of. – Harald Hanche-Olsen Feb 25 '13 at 21:03
@HaraldHanche-Olsen It is worth to mention that this equality makes no sense for Bancach spaces. – Norbert Feb 25 '13 at 21:06

Assume $L\colon H_1\to H_2$ is a linear isometry between Hilbert spaces. Then using the polarization identity $$\langle x,y\rangle=\frac14\sum_{k=0}^3 i^k\lVert x+i^ky\rVert^2$$ we can deduce $\langle Lx,Ly\rangle=\langle x,y\rangle$ for all $x$ and $y$, so that $L^*L=I_1$ (where $I_1$ is the identity on $H_1$). Since $L$ is also assumed to be a bijection, $LL^*=I_2$ follows, where $I_2$ is the identity on $H_2$.