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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?

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

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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.

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  • $\begingroup$ I think I am supposed to use the uniform boundedness principle and I am not sure if I can just assume that it is bilinear when it say linear. $\endgroup$ – Matriz Nov 8 '18 at 12:58

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