Showing $x \mapsto B(x,y), y\mapsto B(x,y)$ continuous $\implies B(x,y)\le c\|x\|\|y\|$ for all $x,y\in H$, where $B:H\to H$ sesquilinear, $H$ Hilbert

Question

Let $\mathscr{H}$ be a Hilbert space and $B: \mathscr{H} \times \mathscr{H} \to \mathbb{K}$ be a sesquilinear form, and $\mathbb{K}$ denote $\mathbb{R}$ or $\mathbb{C}$. Show that 1. implies 2., where

1. The mapping $y \mapsto B(x,y)$ is continuous for each $x \in \mathscr{H}$ and analogously for $x \mapsto B(x,y)$.

2. $B$ is bounded, i.e. there exists a $c > 0$ such that $| B(x,y) | \le c \| x \| \| y \|$ for all $x,y \in \mathscr{H}$.

This answer on a similar post suggests to use the uniform boundedness principle but I am not sure how to do that and would appreciate a hint. What I don't unterstand in particular is how the boundedness of the family of linear operators helps me with this proof.

Instead I started out like this:

Since $B$ is linear in it second argument (sesquilinear in its first), we know that every map $B_x: \mathscr{H} \to \mathbb{K}$ defined by $y \mapsto B(x,y)$ is linear. Since it is continuous by assumption we know that it is bounded, i.e. $\| B_x y\| = \| B(x,y) \| \le c \| y \|$ for all $y \in \mathscr{H}$ and some $c > 0$.

Let $x \in \mathscr{H} \setminus \{0\}$ (we can w.l.o.g. assume that $x \ne 0$ since then $B_x \equiv 0$) and $x' := x_0 + \frac{\delta x}{\| x \|}$ for some $x_0 \in \mathscr{H}$. Then we have $\| x - x' \| \le \delta$ and \begin{equation*} \frac{\delta}{\| x \|} | B(x,y) | \overset{\textrm{(L)}}{=} \left| B\left(\delta \cdot \frac{x}{\| x \|}, y \right)\right| \overset{\textrm{(L)}}{=} | B(x' - x_0, y) | = | B(x',y) - B(x_0,y) | \le 1, \end{equation*} where the last inequality follows from the continuity of $B_y$ in $x_0$. This implies $ | B(x,y) | \le \frac{1}{\delta} \| x \|$.

Is this correct and if yes, how can I proceed?

Answer 1

I'd like to see a proof that does $not$ use the uniform boundedness principle. Here is the standard proof, or at least the one I know:

$\mathscr A=\{B(-,y):y\in \mathcal H;\ \|y\|\le 1\}$ is pointwise bounded by hypothesis, so by the uniform boundedness principle, $\mathscr A$ is uniformly bounded, which means that

$\sup_{\|y\|\le1}\|B(-,y)\|\le c<\infty.$ Thus,

$\|B(x,y)\|\le c$ for all $\|x\|,\|y\|\le 1$ and now, using linearity of $B$, this implies that

$\|B(x,y)\|\le c\|x\|\|y\|$ for all $x,y\in H.$