Maximize $L^p$ norm over sphere For $p \in \mathbb{R}$, consider the function
$$F_p(\lambda_1, \dots, \lambda_n) =  \lambda_1^p + \dots + \lambda_n^p.$$
My goal is to maximize this function under the constraints that
$$ \lambda_1^2 + \dots + \lambda_n^2 = 1, ~~~\text{and}~~~ \lambda_1 + \dots + \lambda_n = 0.$$
I am mainly interested in the case $p \in \mathbb{N}$, $p \geq 3$. In that case $F_p$ is a polynomial.
Remarks


*

*If $p=3$, one can show that the maximum is achieved at the vector $(a, -b, \dots, -b)$, where $a, b>0$ are fixed by the constraints. A working hypothesis is that this is true for any $p$. I do not know how to treat $p = 4, 5, \dots$ though.

*The reason is that for $p=3$, one can use the method of Lagrange multipliers to get that each $\lambda_j$ satisfies the same quadratic equation, hence the maximizer consists of vectors $(\lambda_1, \dots, \lambda_n)$, where $\lambda_j = a$ for $k$ indices $j$ and $\lambda_j = b$ for $n-k$ indices $j$. The optimal $k$ can then be easily determined. For general $p \in \mathbb{N}$, one gets that the $\lambda_j$ satisfy a polynomial equation of degree $p-1$, hence take one of $p-1$ values. Already for $p=4$, this makes matters quite hard.

*Of course, maximizing a polynomial over a sphere is very hard in general, but this is a very basic, very explicit polynomial, so one could hope to have an explicit exact solution.

*The problem is equivalent to maximizing $$\tilde{F}_p(A) = \mathrm{tr}(A^p)$$ over the space of symmetric trace-free matrices of norm one, hence to maximize an $O(n)$-invariant polynomial over an irreducible $O(n)$-representation.


Related questions


*

*If $p$ is odd, minimizing and maximizing is the same thing, but if $p$ is even, one could also ask to minimize the value of $F_p$.

*What if one drops the second constraint? [\edit: This is uninteresting: The maximum is one, taken at a unit vector.]

 A: Proof for the case $p$ odd:
The Lagrange-multiplier equation yields
$$
p\lambda_i^{p-1}+a\lambda_i+b=0
$$
for some $a,b \in \mathbb{R}$. If $p>1$ is odd the LHS is a strictly convex function in $\lambda_i$ and can be zero at at most two different real values.
Assume now we have $n_1, n_2 \in \mathbb{N}$ times $\lambda_1, \lambda_2$ then we want to maximize
$$
\sum_{i=1}^2 n_i\lambda^p_i
$$
with the constraints
$$
\sum_{i=1}^2 n_i\lambda^2_i=1, \sum_{i=1}^2 n_i\lambda_i=0,
\sum_{i=1}^2 n_i=n
$$
or equivalently that
$$
n_1=\frac{n\lambda_ 2}{ \lambda_2-\lambda_1},\\ n_2=\frac{n\lambda_ 1}{ \lambda_1-\lambda_2},\\ n\lambda_1 \lambda_2=-1.
$$
WLOG $\lambda_1>\lambda_2$. Since $\lambda_1 \lambda_2<0$ we have $\lambda_1>0>\lambda_2$. Because $n_1 \geq1$ we have $(n-1)\lambda_2+\lambda_1\leq 0$
and hence
$$
\lambda_1^2\leq -(n-1)\lambda_1\lambda_2
=\frac{n-1}{n}.$$
Hence
$$
\lambda_1\leq (n-1)^{1/2}n^{-1/2}, \lambda_2=-1/(n\lambda_1) \leq -(n-1)^{-1/2}n^{-1/2}.
$$
We then have
$$
\sum_{i=1}^2 n_i\lambda^p_i= \frac{\lambda_1^{p-1}-\lambda_2^{p-1}}{\lambda_1-\lambda_2}\leq \frac{(n-1)^{(p-1)/2}n^{-(p-1)/2} -(n-1)^{-(p-1)/2}n^{-(p-1)/2} }{  (n-1)^{-1/2}n^{1/2}}=((n-1)^{p/2}-(n-1)^{-(p-2)/2})n^{-p/2}
$$
Equality holds as conjectured by the OP.
Proof for $p$ even:
If $p$ is even the LHS of the Lagrange multiplier equation is strictly convex on the positive halfline and strictly concave on the negative half line. On both half lines we can have at most two solutions. Furthermore we have in total an odd number of solutions. Hence for $p$ even we will have at most three different valus of $\lambda$. Now if we do something similar as in the case $p$ odd we end up with
$$
n_2=\frac{1-\lambda_1\lambda_3 n}{(\lambda_2-\lambda_1)(\lambda_2-\lambda_3)}, ...
$$
where  WLOG $\lambda_1>\lambda_2>\lambda_3$
. Since $n_2>0$ we have $\lambda_1\lambda_3>0$ Hence all three $\lambda 's$ have the same sign and thus they can not all satisfy the Lagrange multiplier equation.
A: Too long for a comment. No simple answer is expected in view of the results of math experiment done with Mathematica:
Maximize[{x1^4 + x2^4 + x3^4 + x4^4, x1 + x2 + x3 + x4 == 0 && x1^2 + x2^2 + x3^2 + x4^2 == 1}, 
{x1, x2,    x3, x4}] // ToRadicals


$$\left\{\frac{7}{12},\left\{\text{x1}\to \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{3}},\text{x2}\to -\frac{\sqrt{3}}{2},\text{x3}\to \frac{1}{2 \sqrt{3}},\text{x4}\to \frac{1}{2 \sqrt{3}}\right\}\right\} $$

Maximize[{x1^2 + x2^2 + x3^2 + x4^2 x5^2 + x6^2, x1 + x2 + x3 + x4 + x5 + x6 == 0 && 
   x1^2 + x2^2 + x3^2 + x4^2 x5^2 + x6^2 == 1}, {x1, x2, x3, x4, x5,x6}]


$$\left\{1,\left\{\text{x1}\to \frac{\sqrt{10165}+5}{1024}-\frac{5}{512},\text{x2}\to -\frac{507}{512},\text{x3}\to 0,\text{x4}\to 0,\text{x5}\to 1,\text{x6}\to \frac{-\sqrt{10165}-5}{1024}\right\}\right\}$$

Maximize[{x1^6 + x2^6 + x3^6, x1 + x2 + x3 == 0 && x1^2 + x2^2 + x3^2 == 1}, {x1, x2, x3}]


$ \left\{\frac{11}{36},\left\{\text{x1}\to \sqrt{\frac{2}{3}}-\frac{1}{\sqrt{6}},\text{x2}\to -\sqrt{\frac{2}{3}},\text{x3}\to \frac{1}{\sqrt{6}}\right\}\right\}$

