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Let $H$ be a real Hilbert space, and let $B : H \times H \to \mathbb{R}$ be bilinear and symmetric. Suppose there is a constant $C$ such that for all $x \in H$, $|B(x,x)| \le C \|x\|^2$. Must $B$ be continuous?

This seems like it should just be a simple polarization argument, but for some reason I can't see it.

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up vote 6 down vote accepted

For $t > 0$, $$4 |B(x,y)| = |B(x/t+ty,x/t+ty) - B(x/t-ty,x/t-ty)| \le C (\|x/t+ty\|^2 + \|x/t-ty\|^2) \le 2 C (\|x\|/t+t\|y\|)^2$$ Take $t = \sqrt{\|x\|/\|y\|}$.

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That, I think, is the argument I was really looking for. – Nate Eldredge Jun 12 '12 at 20:50

I figured this out about 30 seconds after I posted it :-/

Let $B'(x,y) = B(x,y) + C \langle x,y \rangle$. $B'$ is again bilinear and symmetric, and moreover $B'$ is positive semidefinite. So $B'$ satisfies the Cauchy-Schwarz inequality $$|B'(x,y)| \le \sqrt{B'(x,x) B'(y,y)}.$$ But we have $B'(x,x) \le |B(x,x)| + C \|x\|^2 \le 2 C \|x\|^2$, so we now have $$B'(x,y) \le 2 C \|x\| \|y\|$$ hence $B'$ is continuous, and thus so is $B$.

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The constant in the last equation should be $2C$, not $\sqrt{2}C$. – Cocopuffs Jun 12 '12 at 20:43
@Cocopuffs: Fixed. – Nate Eldredge Jun 12 '12 at 20:48

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