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

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

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

Any hints/solutions are appreciated.

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 $$age 0$$ by $$int_{0}^{1}f(x)=0.$$

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

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 $$f^+(x)^3le f^+(x)$$
hence we have $$int_{A^+}f^+(x)^3le int_{A^+}f^+(x) = a.$$
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 $$int_{0}^{1}f^3(x)=int_{A^+}f^+(x)^3-int_{A^-}f^-(x)^3le a-frac{a^3}{m(A^-)^2}le a-frac{a^3}{(1-m(A^+))^2}le a-frac{a^3}{(1-a)^2}$$
Since $$2a=int_{A^-}f^-+int_{A^+}f^+le 1$$, so $$ale 1/2.$$ So by a simple computation $$a-frac{a^3}{(1-a)^2}le frac{1}{4}quad ain[0,1/2].$$ 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)=1,0le xle frac{1}{3},f(x)=-frac{1}{2},frac{1}{3}

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:
$$sum_{i=1}^nalpha_i=0, quad -1leqalpha_ileq 1, 1leq ileq n.$$
Similarly, the objective function becomes
$$F(alpha) = frac{1}{n}sum_{i=1}^nalpha_i^3$$
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 $$L^1(0,1)$$.

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