Let $(\mathcal{H} , (\cdot, \cdot))$ be a Hilbert space over the field $\mathbb{L} = \mathbb{R}$ or $\mathbb{C}$ (so the norm on $\mathcal{H}$ is given by $\|\cdot\| = (\cdot, \cdot)^{\frac{1}{2}}$). If $T: \mathcal{H} \to \mathcal{H}$ is a bounded linear operator, I have seen two different definitions for what it means for $T$ to be an isometry:
$T$ isometric $\iff \|x\| = \|Tx\|$, all $x \in H$ (from page 197 of Reed and Simon's Methods of Modern Mathematical Physics, Vol. 1).
$T$ isometric $\iff (x,y) = (Tx,Ty)$ for all $x,y \in \mathcal{H}$ (from this ProofWiki page on isometry).
It is clear that $2 \implies 1$, and in the case $\mathbb{L} = \mathbb{C}$, we have $1 \implies 2$ by the polarization identity:
$$(x,y) = \frac{1}{4}((\|x+y\|^2 + \|x-y\|^2) - i(\|x + iy\|^2 - \|x - iy\|^2)).$$
My question is, are the definitions equivalent when $\mathbb{L} = \mathbb{R}$? Because in that case, we don't have the polarization identity at our disposal.
Hints or solutions are greatly appreciated!
Following the first comment above: the answer is yes. The definitions are equivalent, even in the case $\mathbb{L} = \mathbb{R}$. This is because there is also a real version of the polarization identity:
$$(x,y) = \frac{1}{4}(||x+y||^2 - ||x-y||^2).$$