# How can I prove that the norm of the following operator is $\frac{1}{m!}$?

I'm trying to prove that the norm of the multilinear symmetric operator $A$ is $\frac{1}{m!}$ where $A$ is defined as:

$$A(x_1,\dots, x_m) = \frac{1}{m!} \sum_{\sigma \in S_m} \xi_1(x_{\sigma(1)} ) \dots \xi_m(x_{\sigma(m)} )$$

$\xi_i$ are the coordinate functionals ($\xi_i(e_j)=\delta_{ij}$) and $x_i \in l^1$. We consider the following norm on $(l^1)^m$, $\| (x_1, \dots, x_m) \| = max_j \|x_j\|$

Clearly, considering $x_i = e_i ~ \forall i$ I get the inequality $\| A \| \ge \frac{1}{m!}$ but I haven't been able to get the other inequality so far.

How should I proceed?

• Just to be sure I understand: is $A:l^1\to \mathbb R$ (hence, the notation $A(x_1,\cdots,x_m)$ stands for $A((x_i\,:\,i\in\mathbb N))$ )? Or is $A:(l^1)^m\to \mathbb R$ ? – user228113 Dec 15 '15 at 13:51
• @G.Sassatelli The second one, $A: (l^1)^m \to \mathbb{R}$. – A. A. Dec 15 '15 at 13:52
• With what norm, then? There are several norms inducing the product topology, and $\|A\|$ depends on which one you pick. – user228113 Dec 15 '15 at 13:52
• @G.Sassatelli Sorry I forgot to specify it. I've just added to the post. – A. A. Dec 15 '15 at 14:39

Aside: It's not a linear operator (for $m > 1$), but a multilinear (and symmetric) operator.
By homogeneity, we can assume that $\lVert x_i\rVert \leqslant 1$ for all $i$. Then split the sum,
The base case $m = 1$ is immediately verified, and then by the induction hypothesis, the sums in parentheses are all bounded in absolute value by $1$, which yields the induction step. (There's a small explanation needed why the sums in the parentheses are instances of the $m-1$ case.)