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Suppose $B: K\times K \to \mathbb{R}$ is a bilinear functional. Take $\|(x,y)\|_{K\times K} := \sqrt{\|x\|_{K}^{2}+\|y\|_{K}^{2}}$ for all $(x,y) \in K\times K$. Show $B$ is continuous iff $B$ is bounded.

This is easy to show when the mapping is from two different spaces, but with the product space $K\times K$ and this $\|(a,b)\|_{K\times K}$ norm, I am stuck. What am I missing?

edit: I have gotten stuck at this point for showing continuity:

For all $\epsilon > 0$, choose $\delta$ = ?, so we then have: \begin{align} |B(x,y) - B(a,b)| &= |B(x-a,b) + B(x-a,y-b) + B(x,y-b)| \\ &\leqslant |B(x-a,b)| + |B(x-a,y-b)| + |B(x,y-b)|\\ &\stackrel{\text{bounded}}{\leqslant} c\|(x-a,b)\|_{K\times K}+c\|(x-a,y-b)\|_{K\times K}+c\|(x,y-b)\|_{K\times K}\\ &=c\bigg[\sqrt{\|x-a\|_{K}^{2}+\|b\|_{K}^2} + \sqrt{\|x-a\|_{K}^{2}+\|y-b\|_{K}^2} + \sqrt{\|x\|_{K}^{2}+\|y-b\|_{K}^2}\bigg]\\ &\leqslant \ldots \\ &< \epsilon \end{align}

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    The same argument as when the two spaces are different works. Nobody forces you to dwell on the fact that both factors here are equal. – 2017-02-05
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    @DanielFischer I made an edit, do you have any comments? – 2017-02-06
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    Since a bilinear map is, unless trivial, not uniformly continuous, you cannot choose $\delta$ independent of the point. And the boundedness of a bilinear functional doesn't mean $\lvert B(x,y)\rvert \leqslant c\lVert (x,y)\rVert_{K\times K}$ - the only bilinear functional that has such a bound is $0$ - it means $\lvert B(x,y)\rvert \leqslant c\cdot \lVert x\rVert_K \cdot \lVert y\rVert_K$. Pick $x,y$ with $B(x,y) \neq 0$ and consider $f(t) = \lvert B(tx,ty)\rvert - c\lVert(tx,ty)\rVert_{K\times K}$ to see that. – 2017-02-06

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