Given a symmetric bilinear form $f:V\times V \to K$ , where $V$ is a vector space and $K$ is an appropriate field, define the quadratic form $q:V \to K$ as $q(v):= f(v,v)$.
The Polarisation Formula states that $f(x,y) = 1/2\big( q(x+y) – q(x) – q(y)\big)$, which is easily proven.
This means that any symmetric bilinear form $f:V\times V \to K$ is fully determined by the values $f(v,v)$ for all $v \in V$.
I now want to prove the following theorem:
Prove that any symmetric $k$-linear form $M:V\times\cdots \times V \to K$ is determined by the values $M[v]^k := M[v,…,v]$ for all $v\in V$.
How does that work?
Best Answer
You can be completely explicit in this matter. For $T_j$ in a commutative algebra $$ T_1T_2\dots T_k=\frac{1}{2^k k!}\sum_{\epsilon_j=\pm 1} \epsilon_1\dots\epsilon_k(\epsilon_1T_1 +\dots+\varepsilon_{k}T_{k})^k. $$ The following lemma in available in the Euclidean case.
Lemma. Let $V$ be an Euclidean finite-dimensional vector space, and $A$ a symmetric $k$-multilinear form. We have $ \sup_{\Vert T\Vert=1} \vert{A T^k}\vert =\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert. $
This lemma is a consequence of the 1928 paper by O.D. Kellogg [MR1544896]. This is not true in the non-Euclidean case where the inequality $$ \sup_{\Vert T\Vert=1} \vert{A T^k}\vert \le \sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert\le \kappa_k \sup_{\Vert T\Vert=1} \vert{A T^k}\vert, $$ holds true in general with the best constant $ \kappa_{k}= k^k/k!. $