I was asked to prove the following:
Let $X,Y$ Banach spaces and $a : X\times Y \to \mathbb{R}$ a bilinear function.
Assume that for each $x$ fixed the application $T_x(y) := a(x,y)$ is continuous, the same holding for each $y$ fixed. I.e. the application $T_y(x):= a(x,y)$ is continuous. Prove that there exists $C > 0$ such that for all $(x,y) \in X\times Y$ holds $|a(x,y)| \le C\|x\|_X\|y\|_Y.$
My attempt:
Our hypothesis implies that $\sup_{x\in X} |T_x(y)| <C_y$ and $\sup_{y\in Y} |T_y(x)| < C_x,$ where the subscripts indicate that such constants my chance with $x$ and $y$. By Banach-Steinhaus we get that there exists $C_1,C_2 > 0$ such that $$|T_x(y)| \le C_1\ \|y\|, ~ \forall (x,y) \in X\times Y$$ $$|T_y(x)| \le C_2 \|x\|, ~ \forall (x,y) \in X\times Y.$$ But since $|T_x(y)| = |T_y(x)| = |a(x,y)|$ we get that $$|a(x,y)| \le C_1\|y\|, ~\forall (x,y) \in X\times Y$$ $$|a(x,y)| \le C_2\|x\|, ~ \forall (x,y) \in X\times Y.$$
How can I find $C >$, that supposed to relate with $C_1,C_2$ such that $$ |a(x,y)| \le C\|x\|\|y\|, ~ \forall (x,y) \in X\times Y$$ ?
Restrict your attention to the unit ball. Define $$\mathcal{T} = \left\{ T_x : \|x\| \le 1 \right\}.$$
The family is pointwise bounded: for each $y \in Y$ there exists a constant $C_y$ with the property that $$\sup_{T_x \in \cal T} \|T_x y\| \le C_y$$ so by Banach-Steinhaus there is a constant $C$ with the property that $$\sup_{T_x \in \cal T} \|T_x y\| \le C.$$ Thus if $\|x\|,\|y\| \le 1$ then $|a(x,y)| \le C$, which is equivalent to what you need.