# 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

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

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 $$a\ge 0$$ 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}

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 $$L^1(0,1)$$.

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

• I do not know how to use those equations (My math knowledge is slightly more than that of a high schooler). Please elaborate further Apr 29 '20 at 7:14
• After further reading on this, you cannot use a lagrangian because it is not given that $f(x)$ is a continuous and differentiable function for $x \in [0,1]$ May 23 '20 at 13:55
• Also, how does E-L even work when $f'$ is not present? Oct 10 at 12:10

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

• For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Jul 12 '20 at 10:43
• ok sir @JoséCarlosSantos I will make sure to use mathjax from the next time! Jul 12 '20 at 10:44
• @VenkatAmith $f^3(x)-1 = (f(x)-1)(f^2(x) + f(x) + 1)$. Why did you factor it as $(f(x)-1)(af^2(x) + bf(x) + c)$? Jul 12 '20 at 16:29
• @AniruddhaDeb, Firstly I knew $f(x)$ only 1 as its root so to find the remaining second-degree equation, I took the general equation$at^2+bt+c$,i.e $t=f(x)$ Jul 12 '20 at 16:50