Maximize equation involving sum over subsets I am trying to find the function $f(x)$ the maximizes the following equation.
$$\sum_{S\subseteq \{1,\dots,n\}}(-1)^{n-|S|} \left(\sum_{i\in S}\int_{\frac{i-1}{n}}^{\frac{i}{n}}f(x)dx\right)^t$$
Where $t$ is a real constant greater than $1$, $0\leq f(x)\forall x\in\left[0,1\right]$ and $\int_0^1f(x)dx=1$.
From trying out different functions I've come to believe $f(x)=1$ is the answer, however I haven't been able to proof this result.
 A: This conjecture needs to at least be modified. While the conjecture holds true for $n=2$, the situation in $n=3$ is a little bit different. We will show this by brute force. Defining
$$(a,b,c):=\left(\int_{0}^{1/3}f(x)dx, \int_{1/3}^{2/3}f(x)dx, \int_{2/3}^{1}f(x)dx\right)$$
we note that under the constraints $a,b,c\geq 0$, $a+b+c=1$. Then the function can be written by subbing in $c=1-a-b$
$$L(a,b;t)=1+a^t+b^t+(1-a-b)^t-(1-a)^t-(1-b)^t-(a+b)^t$$
with the constraints $a,b\geq 0, a+b\leq 1$. We can now compute the eigenvalues of the Hessian at the point $(1/3,1/3)$. With some help from Mathematica we find that the eigenvalues are
$$(\lambda_1, \lambda_2)=\frac{3^{2-t}t(1-t)(2^t-4)}{4}(1,3)$$
Crucially, note that the eigenvalues always have the same sign. However the eigenvalues are positive (indicating a local maximum) only if $0<t<1, t>2$, and in fact we can see that for $t=1,2$ the function identically vanishes. When $1<t<2$ the maximum of the function above is located at the boundary and the maximum value attainable is $0$, when $(a,b)=(1,0)$.
We see that for $n=3$, $f(x)=1$ is not a maximizer for all $t$, and in fact it sometimes is a minimizer. Incidentally, this approach to the problem shows that the maximization procedure does not care about the details of the function $f$. There is no unique extremizers given that $f(x)$ can be zero in any arbitrary interval, in which case one can construct many families of examples of extremizers, by constructing piecewise non-zero functions. Even if $f$ is restricted to be non-zero, there are still many examples that satisfy the condition $a=b=c=1/3$. In this case however, the maximizer for $t\in(1,2)$ cannot be attained, but it still can be approximated with arbitrary precision.
The structure seems to persist to $n=4$ as well, where now the function vanishes at $t=1,2,3$ (this fact is intimately connected with the theory of symmetric polynomials, although I have yet to identify the link). This structure is very interesting, and if one can show that this functional reinterpreted as a maximization problem with constraints is concave/convex in those intervals of $t$ (an observation corroborated by plots for low $n$), one should be able to fully understand the maximizers. I will continue trying to do more work in this direction.
