# Applying the uniform boundedness principle

I try to solve the following:

Let $$X,Y$$ be Banach spaces and $$B:X\times Y\to \mathbb{C}$$ a linear map such that $$x_n\to 0\implies B(x_n,y)\to 0\ \forall y\in Y$$ $$y_n\to 0\implies B(x,y_n)\to 0\ \forall x\in X$$ Prove that $$x_n\to 0$$ and $$y_n\to 0\implies B(x_n,y_n)\to 0$$.

I try to apply the uniform boundedness principle so my idea so far: Define the operators $$B_n:=B(x_n,\cdot)$$ and $$B^n:=B(\cdot,y_n)$$. We have pointwise convergence $$\lim\limits_{n\to\infty}B_ny\to0 \quad \lim\limits_{n\to\infty}B^nx\to0$$ hence by applying the uniform boundedness principle we have also uniform boundedness for $$B_n$$ and $$B^n$$. Now I somehow try to argue that this implies that $$B_n^n:=B(x_n,y_n)$$ also has uniform boundedness since it converges on every point pointwise.

Is this correct?

• Do you mean that $B$ is bilinear, and not linear? – supinf Nov 8 '18 at 10:49
• @supinf it just says linear. – Matriz Nov 8 '18 at 12:55

If $$B$$ is indeed bilinear, see the lemma on this page. Note that $$x_n\to 0\implies B(x_n,y)\to 0\ \forall y\in Y$$ is equivalent to the linear map $$X\to \mathbb{C},x\mapsto B(x,y)$$ is bounded for all $$y\in Y$$.
If $$B$$ is indeed linear, then there is an easy proof: $$B(x_n,y_n)=B(x_n,0)+B(0,y_n) \to 0.$$ The convergence holds because we can apply the conditions to $$0\in X$$ and $$0\in Y$$, respectively.