Finding the maximum value of $\int_0^1 f^3(x)dx$

Question

Find the maximum value of $\int_0^1 f^3(x)dx$ given that $-1 \le f(x) \le 1$ and $\int_0^1 f(x)dx = 0$

I could not find a way to solve this problem. I tried to use the cauchy-schwarz inequality but could not proceed further

$$\int_0^1 f(x) \cdot f^2(x) dx \le \sqrt{\left(\int_0^1f^4(x)dx\right) \left( \int_0^1 f^2(x) dx\right)}$$

Any hints/solutions are appreciated.

Answer 1

Let $f^+(x)=\max\{0,f(x)\},f^-(x)=\max\{0,-f(x)\},$ and assume $$A^+=\{x|f^+(x)>0\},A^-=\{x|f^-(x)>0\},$$then $A^+\cap A^- = \emptyset$ $$\int_{A^+}f^+(x)=\int_{A^-}f^-(x)=a$$for some $a\ge 0$ by $\int_{0}^{1}f(x)=0.$

We have that $$a \le m(A^+),$$and,$$a\le m(A^-).$$

We now want to find the maximum of $$\int_{0}^{1}f^3(x)=\int_{A^+}f^+(x)^3-\int_{A^-}f^-(x)^3$$

So we just need to find the maximum of $\int_{A^+}f^+(x)^3$, and the minimum of $\int_{A^-}f^-(x)^3$.

For the first term, we have $f^+(x)\le 1$,So $$f^+(x)^3\le f^+(x)$$ hence we have $$\int_{A^+}f^+(x)^3\le \int_{A^+}f^+(x) = a.$$ and for the second term we have $$\frac{\int_{A^-}f^-(x)^3}{m(A^-)}\ge \left(\frac{\int_{A^-}f^-(x)}{m(A^-)}\right)^3=\left(\frac{a}{m(A^-)}\right)^3$$(You can prove it by Hölder's inequality)

So we have $$ \int_{0}^{1}f^3(x)=\int_{A^+}f^+(x)^3-\int_{A^-}f^-(x)^3\le a-\frac{a^3}{m(A^-)^2}\le a-\frac{a^3}{(1-m(A^+))^2}\le a-\frac{a^3}{(1-a)^2} $$ Since $2a=\int_{A^-}f^-+\int_{A^+}f^+\le 1$, so $a\le 1/2.$ So by a simple computation $$a-\frac{a^3}{(1-a)^2}\le \frac{1}{4}\quad a\in[0,1/2].$$ When $a=\frac{1}{3}$, it equals to $\frac{1}{4}.$

To show $\int_{0}^{1}f(x)^3$ can attain $\frac{1}{4},$ consider such $f(x)$:$$f(x)=1,0\le x\le \frac{1}{3},f(x)=-\frac{1}{2},\frac{1}{3}<x\le 1. $$

Answer 2

We solve the problem via approximation by simple functions on uniform partitions of $[0,1]$. Consider a partition of $[0,1]$ into $n$ parts with coefficients $\alpha_i$. Then the conditions on this simple function correspond to the conditions on the coefficients: $$ \sum_{i=1}^n\alpha_i=0, \quad -1\leq\alpha_i\leq 1, \ 1\leq i\leq n. $$ Similarly, the objective function becomes $$ F(\alpha) = \frac{1}{n}\sum_{i=1}^n\alpha_i^3 $$ Some numerical experiments inform the solution to this problem but I doubt it is that difficult to solve using symmetry and lagrange multipliers or something else. I believe that for a partition of $n$ intervals, the optimal coefficients are given by $\alpha_1=1$ and $\alpha_i = -1/(n-1)$, or any permutation of the indices. For $n>2$, the objective function then gives us $$ F(\alpha) = \frac{1}{n}\left(1-(n-1)\frac{1}{(n-1)^3}\right) = \frac{n-2}{(n-1)^2}, $$ which attains a maximum value of $1/4$ at $n=3$.

Simple functions on uniform partitions are dense in the space of simple functions, and simple functions are dense in $L^1(0,1)$. The functional is continuous on the considered domain, so a maximum on a dense subset corresponds to a maximum over the domain, which is a subset of $L^1(0,1)$.

Answer 3

Hint: You may use Eulero-Lagrange equations to the Lagrangian $F(x,z,p)=z^3$, considering the functional $\mathscr{F}(f)=\int_{-1}^1 F(x,f(x),f'(x))dx$.

Answer 4

I hope this is clear (i did it in a different approach)