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Let $T:X \times X \to \mathbb{R}$ be a continuous bilinear operator defined on a normed linear space $X$ s.t.

$T(\alpha x + \beta y,z) = \alpha T(x,z) + \beta T(y,z)$) and $T(x,y) = T(y,x)$.

Does there exist a constant $C$ s.t. $||T(x,y)|| \leq C$ $||x||$ $||y|| \forall x,y$?

I know that the result is true if $X$ and $Y$ are complete spaces, by using the uniform boundedness principle on $T$ as a continuous function of x for fixed y (and/or the other way around).

However, I'm not sure if completeness is necessary, since it is true that a continuous linear operator $T: X \to \mathbb{R}$ has the property $||T(x)|| \leq C ||x|| \forall x$ on any normed linear space $X$ (although linear and bilinear operators are not exactly the same).

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up vote 3 down vote accepted

I think you can show this using the same argument as in the continuous linear operators. Since T is continuous, then $U=T^{-1}( (-1,1) )$ is open and contains $(0,0)$. find a c>0 small enough such that if $|x|,|y|\leq c$ then $(x,y)\in U$ and then for a general point (x,y) you have

$|T(x,y)| = |T(\frac{c x}{|x|} \frac {|x|}{c}, \frac{c y}{|y|} \frac {|y|}{c} )| = \frac {|x||y|}{c^2} |T(\frac{c x}{|x|} , \frac{c y}{|y|})| \leq \frac {|x||y|}{c^2}$

if x=0 or y=0 then T(x,y)=0 so you can use the argument above for $x,y \neq 0$.

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Thanks! That looks good to me. –  user1736 Oct 14 '10 at 6:36

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