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Let $B$ be a symmetric bilinear form over a Euclidean space $E$. Say that $|B(v,v)|\le c\|v\|^2$ for every $v\in E$, for some $c\ge0$. Then $$4B(v,w)=B(v+w)+B(v-w)$$ yields $2|B(v,w)|\le c(\|v\|^2+\|w\|^2)$. Replacing $v$ by $tv$, $w$ by $t^{-1}w$ and minimizing over $t$, we get $|B(v,w)|\le c\|v\|\cdot\|w\|$. Notice that we can't use Cauchy-Schwarz if $B$ is indefinite.

I am interested in symmetric $n$-linear forms $B$ over $E$.

If $B(v,\ldots,v)|\le c\|v\|^n$ for every $v\in E$, is it true that $|B(v_1,\ldots,v_n)|\le c\|v_1\|\cdots\|v_n\|$ ?

At least, there exists a constant $\gamma_n$ such that $B(v,\ldots,v)|\le c\|v\|^n$ implies $|B(v_1,\ldots,v_n)|\le \gamma_nc\|v_1\|\cdots\|v_n\|$. This is because the map which, to a homogeneous polynomial of degree $n$ over $E$, associates its polar form (a symmetric $n$-linear form) is well-defined and linear. What is the best constant $\gamma_n$ ?

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2 Answers 2

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After a bit of searching, I found that $\gamma_n=1$ for all $n$.

This is known as "van der Corput-Schaake inequality" (1935), discovered before by Szegö (1928), and mentioned to have been known to Banach in the Scottish Book.

The proof proceeds by showing that if $P$ is homogeneous of degree $n$ on a euclidean space, one has, for $B$ the unit ball, $$\sup_{B\times B} |\nabla P(x)\cdot y|\leq n \sup_B |P|,$$ reducing first to the 2-dimensional case and reasoning about trigonometric polynomials.

This implies by differentiating successively that $$\sup_{x_i\in B}|D^nP(x_1,\dots,x_n)| \leq n! \sup_B |P|,$$ hence the result since $D^nP/n!$ is the polarization of $P$.

I must admit that I don't understand the proof yet.

Here is a link to van der Corput-Schaake's article

http://archive.numdam.org/article/CM_1935__2__321_0.pdf

EDIT: I found more recent references with simpler proofs, for instance in Chapter 4 of DeVore and Lorentz "Constructive approximation" (Springer 1993), cf the Szegö inequality, and in an expository text by Lawrence A. Harris. Hope this helps.

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  • $\begingroup$ The result you mentioned requires the n-linear form be continuous. It is fine in this case because (perhaps) $E$ is finite dimensional in OP's assumption. $\endgroup$
    – booksee
    Jun 7, 2015 at 5:27
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The question is developed here :

http://www.math.tsukuba.ac.jp/~wkbysh/note3.pdf

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