Better reduce your theorem..
Theorem (Isometry)
Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.
Let R be a bounded operator from $\mathcal{H}$ into $\mathcal{K}$. TFAE:
- $R$ has a left inverse by its adjoint, i.e. $R^*R=1$
- $R$ is an isometry, i.e. $\|R\varphi\|=\|\varphi\|$ for all $\varphi\in\mathcal{H}$.
- $R$ preserves scalar products, i.e. $\langle R\varphi,R\psi\rangle=\langle\varphi,\psi\rangle$ for all $\varphi,\psi\in\mathcal{H}$.
Proof (Polarization)
Proving by circular chain:
(1=>2): Scalar product induces norm:
$$\|R\varphi\|^2=\langle R\varphi,R\varphi\rangle=\langle R^*R\varphi,\varphi\rangle=\langle\varphi,\varphi\rangle=\|\varphi\|^2$$
(2=>3): Norm generates scalar product:
$$\langle R\varphi,R\psi\rangle=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|(R\varphi)+i^\alpha (R\psi)\|^2=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|(\varphi)+i^\alpha (\psi)\|^2=\langle\varphi,\psi\rangle$$
(3=>1): Scalar product defines adjoint:
$$\langle R^*R\varphi,\chi\rangle=\langle R\varphi,R\psi\rangle=\langle\varphi,\psi\rangle=\langle 1\varphi,\psi\rangle$$
(3=>1): Scalar product is non-degenerate:
$$\langle R^*R\varphi,\psi\rangle\equiv\langle 1\varphi,\psi\rangle\implies R^*R=1$$
Concluding the first lemma.
Lemma (Inverses)
Given plain spaces $X$ and $Y$.
Let $F$ be a function from $X$ into $Y$. TFAE:
- $F$ is injective resp. surjective.
- $F$ has a left resp. right inverse, i.e. $LF=1_X$ resp. $FR=1_Y$.
Remark (Uniqueness)
Left resp. right inverses are not necessarily unique!
Lemma (Left vs. Right)
Given plain spaces $X$ and $Y$.
Let $F$ be a function from $X$ into $Y$. Then:
$$LF=1_X\quad FR=1_Y\implies L=R$$
That is left and right inverse become unique and agree.
Proof (Identity)
Identity function is a unit:
$$L=L1_Y=L(FR)=(LF)R=1_XR=R$$
Concluding the third lemma.