[proofplan]
The implication from [inner product](/page/Inner%20Product) preservation to isometry follows by evaluating the identity at $y=x$ and using the norm induced by the Hilbert inner product. Conversely, an isometry preserves the squared norm of every vector; applying this to the vectors $x+y$ and $x-y$ in the real case recovers the real polarization identity. In the complex case, applying norm preservation to $x+i^k y$ for $k=0,1,2,3$ recovers the complex polarization identity, with the convention that the inner product is linear in the first variable.
[/proofplan]
[step:Recover norm preservation from inner product preservation]
Assume that
\begin{align*}
(Vx,Vy)_K = (x,y)_H
\end{align*}
for every $x,y \in H$. Taking $y=x$ gives
\begin{align*}
\|Vx\|_K^2 = (Vx,Vx)_K = (x,x)_H = \|x\|_H^2
\end{align*}
for every $x \in H$. Since both norms are non-negative, taking square roots gives
\begin{align*}
\|Vx\|_K = \|x\|_H
\end{align*}
for every $x \in H$. Hence $V$ is a [Hilbert space](/page/Hilbert%20Space) isometry.
[/step]
[step:Use real polarization when the scalar field is $\mathbb{R}$]
Assume that $\mathbb{F}=\mathbb{R}$ and that $V$ is a Hilbert space isometry. Thus
\begin{align*}
\|Vz\|_K = \|z\|_H
\end{align*}
for every $z \in H$. Since $V$ is linear, for all $x,y \in H$ we have
\begin{align*}
V(x+y) = Vx+Vy
\end{align*}
and
\begin{align*}
V(x-y) = Vx-Vy
\end{align*}
The real polarization identity in a real Hilbert space gives
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\left(\|Vx+Vy\|_K^2-\|Vx-Vy\|_K^2\right)
\end{align*}
Using linearity of $V$ and norm preservation, this becomes
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\left(\|V(x+y)\|_K^2-\|V(x-y)\|_K^2\right)
\end{align*}
and hence
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\left(\|x+y\|_H^2-\|x-y\|_H^2\right)
\end{align*}
Applying the real polarization identity in $H$ gives
\begin{align*}
(Vx,Vy)_K = (x,y)_H
\end{align*}
[guided]
Assume that $\mathbb{F}=\mathbb{R}$ and that $V$ is a Hilbert space isometry. By definition, this means that
\begin{align*}
\|Vz\|_K = \|z\|_H
\end{align*}
for every $z \in H$.
We want to prove an identity involving $(Vx,Vy)_K$, so we need a way to express an inner product using only norms. In a real Hilbert space, the relevant formula is the real polarization identity:
\begin{align*}
(a,b) = \frac{1}{4}\left(\|a+b\|^2-\|a-b\|^2\right).
\end{align*}
We apply this identity in the Hilbert space $K$ with $a=Vx$ and $b=Vy$. This gives
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\left(\|Vx+Vy\|_K^2-\|Vx-Vy\|_K^2\right).
\end{align*}
Now we use the linearity of the map $V:H\to K$. Since $V \in \mathcal{L}(H,K)$, it is linear, so
\begin{align*}
Vx+Vy = V(x+y)
\end{align*}
and
\begin{align*}
Vx-Vy = V(x-y)
\end{align*}
Substituting these equalities into the preceding display gives
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\left(\|V(x+y)\|_K^2-\|V(x-y)\|_K^2\right).
\end{align*}
The isometry hypothesis now applies to the two vectors $x+y \in H$ and $x-y \in H$. Therefore
\begin{align*}
\|V(x+y)\|_K = \|x+y\|_H
\end{align*}
and
\begin{align*}
\|V(x-y)\|_K = \|x-y\|_H
\end{align*}
Squaring both equalities and substituting gives
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\left(\|x+y\|_H^2-\|x-y\|_H^2\right).
\end{align*}
Finally, applying the real polarization identity in the Hilbert space $H$ gives
\begin{align*}
\frac{1}{4}\left(\|x+y\|_H^2-\|x-y\|_H^2\right) = (x,y)_H.
\end{align*}
Hence
\begin{align*}
(Vx,Vy)_K = (x,y)_H
\end{align*}
for all $x,y \in H$.
[/guided]
[/step]
[step:Use complex polarization when the scalar field is $\mathbb{C}$]
Assume that $\mathbb{F}=\mathbb{C}$ and that $V$ is a Hilbert space isometry. Let $i \in \mathbb{C}$ denote the imaginary unit. For all $x,y \in H$, the complex polarization identity with the convention that the inner product is linear in the first variable gives
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\sum_{k=0}^{3} i^k \|Vx+i^k Vy\|_K^2.
\end{align*}
Since $V$ is complex-linear,
\begin{align*}
Vx+i^k Vy = V(x+i^k y)
\end{align*}
for each $k \in \{0,1,2,3\}$. Therefore
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\sum_{k=0}^{3} i^k \|V(x+i^k y)\|_K^2.
\end{align*}
By norm preservation applied to each vector $x+i^k y \in H$,
\begin{align*}
(Vx,Vy)_K = \frac{1}{4}\sum_{k=0}^{3} i^k \|x+i^k y\|_H^2.
\end{align*}
Applying the complex polarization identity in $H$ gives
\begin{align*}
(Vx,Vy)_K = (x,y)_H.
\end{align*}
[/step]
[step:Combine the two scalar-field cases]
If the scalar field is $\mathbb{R}$, the real polarization argument proves that every Hilbert space isometry $V \in \mathcal{L}(H,K)$ preserves inner products. If the scalar field is $\mathbb{C}$, the complex polarization argument proves the same conclusion. Together with the first step, this proves the equivalence for Hilbert spaces over either $\mathbb{R}$ or $\mathbb{C}$.
[/step]
