It can be shown that when X, Y, and Z are all Banach spaces (or at least when X or Y are Banach spaces) over the number fields R or C, and when B : X×Y →Z is a bilinear function, the continuity of B for each component implies its overall continuity.
However, I'm curious about the scenario when X, Y, and Z are merely normed spaces, not necessarily complete. Is it possible for a bilinear mapping to be continuous per component, but not continuous as a whole?
The abovementioned proposition can be proven using the Uniform Boundedness Principle. Therefore, it seems natural to search for an example that fails to satisfy the UBP as a typical counterexample to this statement. Despite this, I'm finding it challenging to conceive a meaningful counterexample. Could anyone provide an instance that illustrates this, or give some guidance on how to think about this problem?
Take $X=Y=c_{00}$ be the space of sequences with finitely many non-zero entries supplied with the $l^2$-norm.
Define $$ B(x,y):=\sum_{k=1}^\infty k^3 x_k y_k. $$ For fixed $x$ the map $y\mapsto B(x,y)$ is continuous. Same for $x\mapsto B(x,y)$ for $y$ fixed.
However, $B$ is not continuous, as $B(k^{-1}e_k,k^{-1}e_k)=k\not\to0=B(0,0)$, where $e_k=(0,\dot,0,1,0,\dots)$ is the standard unit vector.