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

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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 $f(x)$:$$f(x)=1,0le xle frac{1}{3},f(x)=-frac{1}{2},frac{1}{3}<xle 1. $$

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 $mathscr{F}(f)=int_{-1}^1 F(x,f(x),f'(x))dx$.

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