# Maximizing symmetric functions on the unit cube

Let $f:[0,1]^n \to \mathbb{R}$. We say that $f$ is symmetric, if for every permutation $\sigma \in S_n$ and every $(x_1,..,x_n) \in [0,1]^n$ we have that $$f(x_1,..,x_n) = f(x_{\sigma(1)},...,x_{\sigma(n)})$$

I am interested in the following question. Denote by $x_* = \text{argmax}_{x\in [0,1]^n}(f(x))$. Must it be the case that $x_*$ is a diagonal element? that it, there exists $0 \leq c \leq 1$ such that $x_* = (c,...,c)$?

If so, is there a short and intuitive explanation?

If not, any easy counterexample?

Thanks.

No, define $f(x) = \begin{cases} 1,& x \in \{e_k\} \\ 0,& \text{otherwise} \end{cases}$.
Here is a continuous one: $f(x) = \|x- {1 \over \sqrt{n}} \langle e, x\rangle e \|_2$, where $e=(1,...1)^T$. Then $f$ is zero on the diagonal, but $f(x)>0$ off the diagonal.
To see that that latter is permutation invariant, let $P$ be a permutation matrix, then since $Pe=P^T e = e$, and $\|Py\|_2 = \|y\|_2$, we have \begin{eqnarray} f(Px) &=& \|Px- {1 \over \sqrt{n}} \langle e, Px\rangle e \|_2 \\ &=& \|Px- {1 \over \sqrt{n}} \langle P^T e, x\rangle P e \|_2 \\ &=& \|Px- {1 \over \sqrt{n}} \langle P^T e, x\rangle P e \|_2 \\ &=& \|Px- {1 \over \sqrt{n}} \langle e, x\rangle P e \|_2 \\ &=& \|x- {1 \over \sqrt{n}} \langle e, x\rangle e \|_2 \\ &=& f(x) \end{eqnarray}