Surjection that Preserves Inner Product is Linear
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Theorem
Let $H, K$ be Hilbert spaces, and denote by ${\innerprod \cdot \cdot}_H$ and ${\innerprod \cdot \cdot}_K$ their respective inner products.
Let $U: H \to K$ be a surjection such that:
- $\forall g, h \in H: {\innerprod g h}_H = {\innerprod {Ug} {Uh} }_K$
Then $U$ is a linear map, and hence an isomorphism.
Proof
Let $x, y \in H$.
Let $\alpha \in \GF$.
By surjectivity of $U$, choose $z \in H$ such that $Uz = \map U {\alpha x + y} - \paren { \alpha Ux + Uy }$.
Then:
\(\ds {\innerprod {Uz} {Uz} }_K\) | \(=\) | \(\ds {\innerprod {\map U {\alpha x + y} - \paren{\alpha Ux + Uy } } {Uz} }_K\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {\innerprod {\map U {\alpha x + y} } {Uz} }_K - \paren{ \alpha {\innerprod {Ux} {Uz} }_K + {\innerprod {Uy} {Uz} }_K}\) | by linearity in the first coordinate | |||||||||||
\(\ds \) | \(=\) | \(\ds {\innerprod {\alpha x + y} z }_H - \paren{ \alpha {\innerprod x z }_H + {\innerprod y z }_H}\) | as $U$ preserves the inner product | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | by linearity in the first coordinate |
By positivity, $Uz = {\bf 0}_K$.
Hence:
- $\map U {\alpha x + y} = \alpha U x + U y$
Thus, $U$ is linear.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text I.5$: Exercise $9$