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).
Best Answer
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 $x,y \neq 0$.