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Suppose that $q$ is a quadratic form on $\mathbb{R}^n$, $q(x)=(x,Ax)$ say (or $q(x)=x^TAx$ if you prefer that notation). Then one could consider the quantity $$ \sup\{ \left|q(x)\right| : \left\| x \right\| \leq 1 \}. $$ Is this an interesting quantity? In particular when the norm is the $p$-norm for $p \neq 2$?

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Note that $\sup \{ |q(x)|: \| x \| \le 1 \} = \sup \{ |q(x): \| x \| =1 \}$ by applying linearity: $q(\lambda x)=\lambda^2 q(X)$, so $q(x) = \|x\|^2 q(x/\|x\|)$. –  Johannes Kloos May 3 '12 at 22:07

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It is, indeed. You can state the problem as that of maximizing the quadratic form $x^TAx$ subject to the constraint $x^Tx=1$. It is well known that the maximum you seek is exactly the maximum eigenvalue of $A$ and the x attaining this maximum is the eigenvector associated to this eigenvalue. Thus, the quantity you are asking about is just the maximum eigenvalue's norm.

More generally, given $A$ symmetric (no loss of generality here...) and $B$ positive definite, you can prove that $$max \{x^TAx\;\;\colon\;\; x^TBx=1\}$$ is the maximum eigenvalue of $B^{-1}A$ at $x$ the corresponding eigenvector.

Best, Nel

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As I read the question, the OP was more interested in the case of the $p$-norm with $p\neq 2$. Can you say anything in this case? –  Fabian Feb 22 '13 at 4:02

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