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So I wrote this down a week ago and cannot figure out what I was thinking. Not sure if this is correct.

Context we have a bilinear operator $B:X\times Y\to \mathbb{K}$.

Is it true that $$\sup_{x\in X, y\in Y} \|B(x,y)\|< \infty \implies |B(x,y)|\leq K \|x\|\|y\|$$

So the absolute value is a norm on the reals, complex in one dimension so that's not a problem because the norms are equivalent.

So the implication works for the $x,y$ supremum case. But does it immediately follow for all other $(x,y)$? I cannot remember what I was thinking. Can anyone help?

Matthew
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    You may want to take that supremum for (at most) unit vectors. – Berci Mar 19 '17 at 20:33
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    You may want to take the $Sup_{|x|=1|y|=1}|B(x,y)|$. – Tsemo Aristide Mar 19 '17 at 20:33
  • @Berci I start with $\sup_{x\in X} |B_x | < \infty$ and I want to end up with $|B(x,y)| \leq K |x| |y|$ for all $x,y$ in $X\times Y$. If I use the supremum over unit vectors wouldn't that ruin my goal? I'm trying to prove that the bilinear operator is bounded. $X,Y$ are banach spaces. I used the uniform boundedness principle to arrive to $\sup_{x\in X} |B_x | < \infty$ where $B_x$ is the map where $x$ is kept constant. – Matthew Mar 19 '17 at 21:09

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$$\sup_{x\in X, y\in Y} \|B(x,y)\|< \infty \implies B=0$$

Emanuele Paolini
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