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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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    Unless I'm missing something (always a possibility) you don't need anything fancy: Is the map $\langle\,,\,\rangle\colon V \times V \to \mathbf C$ continuous? Is the map $V \to V \times V$ given by $x \mapsto (Ax, x)$ continuous?2012-03-02
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    I don't know why you even speak about Riesz. $V$ is finite dimensional, so you can fix a basis $\{ e_1, \dots, e_n \}$. If you write $x = \sum x_i e_i$, $Q(x)$ becomes a quadratic polynomial in the coefficients. That quadratic is clearly continuous, so we're done. Is there a problem in that? (I'm wondering why you didn't think of this.2012-03-02
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    @PatrickDaSilva You know, it seems lately that every time I ask a question around here somebody rags on me for asking something "obvious" or "simple" or something I "should have thought of". I ask questions on here to get assistance on questions that, in fact, I don't know the answer to. If I knew the answer, or if the answer was something I could "think of" I wouldn't ask the question in the first place. The reason I bring up Riesz is because this problem reminds me of the problem that theorem addresses. So, after this, I think I will quit asking questions around here for awhile since....2012-03-02
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    ...I'm obviously dull and can't think up the answers on my own.2012-03-02
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    I'm sorry if I made you feel offended, but when you want to show that something "quite elementary" is true and referring to something as powerful as "Riesz's representation theorem", I'm quite confused. Don't feel insulted. My comment was not about saying "Why didn't you think about this?" but more like "How did you get to think about Riesz's representation theorem without thinking of this?"... I was really confused about how could that happen. For me it feels like someone trying to compute $\sum_{i=1}^n i$ by using integrals ; it's just weird. Can you put some light on your confusion?2012-03-02
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    Perhaps that with context it makes more sense.2012-03-02
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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.