Continuous Quadratic Form $\implies$ Continuous Sesquilinear Form Given a Hilbert space $\mathcal{H}$.
Consider a quadratic form $q:\mathcal{H}\to\mathbb{C}$.
Define its inducing sesquilinear form:
$$s:\mathcal{H}\times\mathcal{H}\to\mathbb{C}: s(x,y):=\frac{1}{4}\sum_{k=1}^4 i^k q(x+i^ky)$$
How to prove that:
$$|q(x)|\leq\|q\|\cdot\|x\|^2\text{ for all }x\in\mathcal{H}\implies|s(x,y)|\leq\|s\|\cdot\|x\|\cdot\|y\|\text{ for all }x,y\in\mathcal{H}$$
and especially: $\|s\|=\|q\|$
 A: Using the Parallelogram law $\|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2}$:
$$
   |s(x,y)| \le \frac{\|q\|}{4}\sum_{n=0}^{3}\|x+i^{n}y\|^{2}
      = \|q\|\{\|x\|^{2}+\|y\|^{2}\}.
$$
I'm going to assume that you know somehow that your definition of $s$ is sesquilinear; this is a separate issue. Then, if $x \ne 0$, $y \ne 0$, the above implies
$$
       \frac{|s(x,y)|}{\|x\|\|y\|}
               = \left|s\left(\frac{1}{\|x\|}x,\frac{1}{\|y\|}y\right)\right|
               \le 2\|q\|,
$$
which gives $|s(x,y)| \le 2\|q\|\|x\|\|y\|$ for $x\ne 0$, $y\ne 0$. The cases where $x=0$ and $y=0$ are handled by dividing by $(\|x\|+\epsilon)$ and/or $(\|y\|+\epsilon)$ instead, and letting $\epsilon \downarrow 0$ after clearing denominators.
It's not true in general that $|s(x,y)| \le \|q\|\|x\|\|y\|$. However, if $q$ is real, then you get this improved estimate because you can choose $\theta \in [0,2\pi)$ such that $e^{i\theta}s(x,y)$ is real, which forces
$$
         e^{i\theta}s(x,y) = s(e^{i\theta}x,y)= \frac{1}{4}\sum_{n=0}^{3}i^{n}q(e^{i\theta}x+iy) \\
              = \frac{1}{4}\{ q(e^{i\theta}x+y)-q(e^{i\theta}x-y)\},
$$
which cuts the estimated bound in half.
