Find maximum value of $\int_{0}^{1}\left(f(x)\right)^3dx$ Find maximum possible value of $\int_{0}^{1}\left(f(x)\right)^3dx$ given that $-1\leq f(x)\leq 1$ and that $\int_{0}^{1}f(x)dx=0$.
My attempt:
I tried to guess such functions which could satisfy all the above conditions but could not arrive at any conclusion.
 A: Let $f\colon [0,1]\to [-1,1]$ be a measurable function
satisfying $\int_0^1 f = 0$, let
$$
E := \{x\in [0,1]:\ f(x) < 0\},
$$
and denote by $\sigma\in [0,1]$ the Lebesgue measure of $E$.
It is not restrictive to assume $\sigma > 0$, since if $\sigma = 0$ we have $f=0$ a.e., which is not a maximizer. 
Moreover, let
$$
f^+ := (|f|+f)/2,
\qquad
f^- := (|f| - f)/2,
$$
so that $f^+, f^-\geq 0$, $f=f^+ - f^-$, $|f| = f^++f^-$. Moreover the conditions
$\int_0^1 (f^+-f^-) = 0$, $0\leq \int_0^1(f^++f^-) \leq 2$ give
$$
A := \int_0^1 f^+ = \int_0^1 f^- \in [0, 1/2],
\qquad
A = \int_0^1 f^+ \leq 1-\sigma.
$$
Since $\int_0^1 f = 0$ and $|f|\leq 1$, we have that
$$
\begin{split}
\int_0^1 f^3 & = \int_0^1 (f^3+f) = \int_0^1 f(1+f^2)
= \int_0^1 (f^+ - f^-)(1+f^2)
\\ & = \int_0^1 f^+(1+f^2) - \int_0^1 f^-(1+f^2)
\leq 2 \int_0^1 f^+ -\int_0^1 f^- - \int_0^1 (f^-)^3.
\end{split}
$$
Since $0 = \int_0^1 f = \int_0^1 f^+ - \int_0^1 f^-$, and, by Jensen's inequality,
$$
\int_0^1 (f^-)^3 = \int_E (f^-)^3 \geq \frac{1}{\sigma^2} \left(\int_E f^-\right)^3 = \frac{1}{\sigma^2}\left(\int_0^1 f^+\right)^3,
$$ 
we get
$$
\int_0^1 f^3 \leq A - \frac{A^3}{\sigma^2} =: g(\sigma, A).
$$
We are led to maximize the function $g$ in the region 
$$
B := \{(\sigma, A):\ \sigma \in (0, 1],\ 0\leq A\leq 1-\sigma\}.
$$
The function $g$ is continuous in $B$ and $g \to -\infty$ as $\sigma\to 0^+$,
hence $g$ admits a maximizer in $B$.
Since $\frac{\partial g}{\partial \sigma} = 2A^3 / \sigma^3 > 0$, the maximizer must lie on the boundary of $B$ (excluding the segment with $\sigma = 0$).
A quick check shows that the maximizer is obtained for on the side $A = 1-\sigma$ for $\sigma = 2/3$, hence
$$
\max_B g = \frac{1}{4}
$$
and, finally,
$$
\int_0^1 f^3 \leq \frac{1}{4}.
$$
On the other hand, every function $f$ with $f=-1/2$ on a set $E\subset[0,1]$ of measure $2/3$ and $f=1$ on $[0,1]\setminus E$ gives $\int_0^1 f^3 = 1/4$, and so is a maximizer for the original problem.
A: Consider:
$$f(x)=\begin{cases} 1, a\le x\le 1 \\ -c, 0\le x<a \end{cases}$$
The constraint:
$$1-a=-ac \Rightarrow c=\frac{a-1}{a}.$$
The objective function:
$$S(a,c)=1-a+ac^3 \to max$$
Solution:
$$S(a)=1-a+a\left(\frac{a-1}{a}\right)^3=-2+\frac{3}{a}-\frac{1}{a^2}.$$
$$S'=-\frac{3}{a^2}+\frac{2}{a^3}=0 \Rightarrow a=\frac{2}{3}.$$
$$S''(2/3)<0.$$
Hence:
$$S(2/3)=\frac{1}{4} (max).$$
A: A somewhat more conceptual approach is the following.  
The set of points 
$S=\{(\int_0^1f(x)\,dx,\int_0^1(f(x))^3\,dx): |f(x)|\le 1 \text{ on } [0,1]\}\subset\mathbb R^2$, is exactly the set of points representable as $(E[X],E[X^3])$ for a random variable $X$ for which $P(|X|\le1)=1$.  This, in turn, is the convex hull of the  set $C=\{(x,x^3):|x|\le 1\}$, and the desired answer is the $y$-coordinate of the intersection of the $y$ axis with the upper envelope of $S$.  
A sketch should make it obvious that the upper envelope of $S$ is the union of a piece of $C$ stretching from $(-1,-1)$ to $p$, and of the line segment connecting $p$ to $(1,1)$, where $p$ is the unique point of tangency to $C$ of a line passing through $(1,1)$.  Simple calculus verifies that $p=(-1/2,-1/8)$.  The line segment from $p$ to $(1,1)$ intersects the $y$ axis at $(0,1/4)$, from which the answer $1/4$ is read off.
