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If $V$ is a finite dimensional Hilbert space for any vector $x \in V$ and endomorphism $A$, the function $ Q(x) = \langle Ax, x \rangle $ defines a quadratic form on $V$. Now, I would like to show that $Q$ is continuous. The main thought I have about this is that it is very similar to the (analytic) Riesz representation theorem. If $Q$ were linear, we would be done for Riesz guarantees that all linear forms are continuous. However, $Q$ is actually bilinear so Riesz doesn't apply. Also, I'm aware of the machinery that one can use to solve $n$-linear continuity problems such as this one, i.e., a multilinear map is continuous iff it is bounded iff int is continuous at $0$, etc., and these are the same techniques that one uses with linear maps. This brings me to my question:

Is there a simple way to demonstrate that the function $Q$ as defined above is continuous? Is there perhaps an elementary way to extend Riesz in order to apply it to bilinear functions?

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    Here we say that use powerful tools to solve these problems is to kill an ant with a cannon!2012-03-02

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You prove in the following way let $x_n\rightarrow x$ then let us prove that $Q(x_n)\rightarrow Q(x)$.

Note that $x_n$ is bounded since it is convergent

$|Q(x_n)-Q(x)|=|\langle Ax_n,x_n\rangle-\langle Ax,x\rangle|=|\langle Ax_n,x_n\rangle-\langle Ax,x_n\rangle+\langle Ax,x_n\rangle-\langle Ax,x\rangle|\leq|\langle A(x_n-x),x_n\rangle|+|\langle Ax,x_n-x\rangle|\leq|A||x_n-x||x_n|+|A||x||x_n-x|\leq M|x_n-x|$

Since we have this sequence bounded.