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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.]
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  • $\begingroup$ Concerning your second related question (droping the constraint $\lambda_1+\dots+\lambda_n = 0$): Isn't the $p$-norm on the $2$-unit sphere for $p\ge 2$ maximized by the unit vectors $e_k$ since we have $\|x\|_p \le \|x\|_2$ for each $x$ with equality for $x = e_k$? $\endgroup$ – Jochen Glueck Mar 9 at 9:35
  • $\begingroup$ Yup, you are right, I did not think this through. Added in the post above. $\endgroup$ – Matthias Ludewig Mar 11 at 0:42
  • $\begingroup$ I don't know if this helps, or indeed was motivating your conjecture, but vectors of the form in your first bullet point are precisely those obtained by taking $e_j - Pe_j$ where $P$ is the orthogonal projection onto the hyperplane defined by your second constraint, and then renormalizing to get a unit vector in $\ell_2^n$ $\endgroup$ – Yemon Choi Mar 13 at 19:34
  • $\begingroup$ The polynomial equation is $p\lamda^{p-1}+a\lamda+c=0$.The left hand side is either strictly convex or strictly concave hence you have at most two different real $\lamdas$. Then there are only finitely many cases left to analyze. $\endgroup$ – user35593 Mar 14 at 21:06
  • $\begingroup$ For $p$ and $n$ even we have $\sum_i \lambda_i^p \geq n^{1-p/2} (\sum_i \lambda_i^2)^{p/2}\geq n^{1-p/2}$ with equality if $\lambda_i=(-1)^i/\sqrt{n}$. $\endgroup$ – user100927 Mar 19 at 7:24
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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 $\lambda_1>\lambda_2$ 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.

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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\}$

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  • $\begingroup$ Only the first one is interesting: It seems to show that my guess that for all $p$, the minimizer is of the form $(a, -b, \dots, -b)$ is false for $p=4$. The second one is $p=2$, where the function $F_2$ is constant, and the minimizer from the output is a weird one that doesn't even satisfy the constraints. The third example looks at the case $p=3$, where I already know the answer. Maybe you could edit your answer? $\endgroup$ – Matthias Ludewig Mar 10 at 12:46
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    $\begingroup$ Also, did you notice that in the third example, you have the typo of x3^4 instead of x3^3? $\endgroup$ – Matthias Ludewig Mar 13 at 6:26
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    $\begingroup$ @MatthiasLudewig The first one actually supports your guess, because $\displaystyle\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{3}} = \frac{1}{2\sqrt{3}}$. I would love to say I spotted this but actually I only noticed after squaring the LHS by hand and getting $\frac{1}{12}$ $\endgroup$ – Yemon Choi Mar 13 at 19:31
  • $\begingroup$ @Matthias Ludewig: Thank you, fixed. $\endgroup$ – user64494 Mar 19 at 8:10

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