$a: E\times F\to G$ bilinear separately continuous implies continuous? Let $E$, $F$ and $G$ be Banach spaces and let $a$: $E \times F \to G$ be a bilinear map which is separately continuous, that is
$$\forall x \in E \textrm{ the map } y \mapsto a(x,y) \textrm{ is continuous}$$
and
$$\forall y \in F \textrm{ the map } x \mapsto a(x,y) \textrm{ is continuous}$$
Show that $a$ is bounded, i.e., there exists a constant $M\geq 0$ such that
$$\|a(x,y)\|_G \leq M \|x\|_E \|y\|_F,\quad\forall(x,y)\in E \times F.$$
 A: For each $x\in E$ the operator $a_x: F\to G:y\mapsto a(x,y)$ is continuous. Similarly, for each $y\in Y$ the operator $a_y:E\to G:x\mapsto a(x, y)$ is continous. Since $a_x$ is continuous for any $x\in E$, there is a constant $c_x>0$ such that $\Vert a_x(y)\Vert\leq c_x\Vert y\Vert$ for all $y\in F$.
For any $y\in S_F:=\{y\in F:\Vert y\Vert_F=1\}$ we have
$$
\Vert a_y(x)\Vert_G=\Vert a_x(y)\Vert_G\leq c_x\Vert y\Vert_F=c_x.
$$
Consider family of operators $\mathcal{A}=\{a_y:y\in S_F\}$ between Banach space $E$ and $G$. By uniform boundedness principle the family $\mathcal{A}$ has a global constant $M>0$ such that $\Vert a_y\Vert\leq M$ for all $y\in S_F$. Now for any $x\in E$, $y\in F\setminus\{0\}$ we have
\begin{align}
\Vert a(x,y)\Vert_G
&=\left\Vert \Vert y\Vert_F a\left(x,\frac{y}{\Vert y\Vert_F}\right)\right\Vert_G \\
&=\Vert y\Vert_F \Vert a_{\frac{y}{\Vert y\Vert_F}}(x)\Vert_G \\
&\leq\Vert y\Vert_F \Vert a_{\frac{y}{\Vert y\Vert_F}}\Vert\Vert x\Vert_E \\
&\leq M\Vert y\Vert_F \Vert\ x\Vert_E \\
\end{align}
For $y=0$ the inequality is obvious.
