Find maximum $\int^{1}_{0}\{f(x)\}^3dx$ I would appreciate if somebody could help me with the following problem:
Question: Find maximum $\int^{1}_{0}\{f(x)\}^3dx$ when 
(1). $f(x) : \text{conti-and}  \int^{1}_{0}f(x)dx=0$
(2). $-1\leq f(x)\leq 1 (x \in [0,1] )$
I tried  but couldn’t get it that way.
 A: The maximum doesn't exist, the supremum does, it is $\frac14$.
Consider the polynomial
$$P(t) = (1-t)\left(t+\frac12\right)^2 = \frac14 + \frac34 t - t^3$$
It is clear $P(t) \ge 0$ for all $t \le 1$.
Let $g : [0,1] \to (-\infty,1]$ be any piecewise continuous function which satisfies 
$\int_0^1 g(x) dx = 0$, we have
$$\int_0^1 P(g(x)) dx \ge 0\quad\implies\quad\frac14 \ge \int_0^1 g(x)^3 dx$$
The limit $\frac14$ is achievable. One example is the function
$$g_{eg}(x) = \begin{cases}1,&x \in [0,\frac13]\\-\frac12,& x \in (\frac13,1]\end{cases}$$
Let us call a function $f$ admissible if it satisfies the requirement in question. i.e
$f : [0,1] \to [-1,1]$ is continuous and $\int_0^1 f(x) dx = 0$. It is clear any admissible
$f$ satisfies the requirement of $g$ above. This means
$$\frac14 \ge \int_0^1 f(x)^3 dx$$
Furthermore, it is trivial to construct a sequence of admissible $f$ which
converges pointwise to the piecewise function $g_{eg}$ above. This implies
$$\frac14 = \sup \left\{\; \int_0^1 f(x)^3 dx \;:\; f \text{ admissible} \;\right\}$$
Let $f$ be any admissible function with $\int_0^1 f(x)^3 > 0$, it is clear


*

*There is a $c \in [0,1]$ such that $f(c) > 0$. 

*Since $\int_0^1 f(x) dx = 0$, there is also a $d \in [0,1]$ such that $f(d) < 0$. 

*By IVT, there must be an $e$ between $c$ and $d$ such that $f(e) = 0$.

*Since $P(f(e)) = P(0) > 0$, this forces  $\int_0^1 P(f(x)) dx > 0$ and hence $\frac14 > \int_0^1 f(x)^3 dx$.


Conclusion: there is no admissible $f$ which achieves the supremum $\frac14$.
