Surjection that Preserves Inner Product is Linear

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.

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.

$\blacksquare$