How do I find the maximum value of $\int_0^1 (f(x))^3 dx$ given that
- $\int_0^1 f(x) dx = 0$
- $ -1 \leq f(x) \leq 1$
- Domain of $f = \mathbb R$
- $f$ is a real-valued function
For some values of $ x \space\epsilon \space (0,1)$, f(x) has to alternate between being positive and negative in a way that results in $(1)$.
But given that $|f(x)| \leq 1$, for positive $f(x)$, $(f(x))^3$ has to be lesser than $f(x)$ and vice-versa. So even though positive values of $f(x)$ are diminished when cubed, the negative values, on the other hand, increase in value. How do I use all of these ideas to move towards a solution?
The answers given in Find maximum value of $\int_{0}^{1}\left(f(x)\right)^3dx$ seem out of reach to me. I am a stranger to Lebesgue integrals considering my calculus syllabus is restricted to high school calculus. I found this question in a practice question bank for a college entrance exam.
This is more of an elementary/high school level approach to the problem. Consider the expression $$\int_0^1 \left(f(x)-1\right){\left(f(x)+\frac{1}{2}\right)}^2 \; \mathrm{d}x$$Since $f(x) \leq 1$, the expression above is less than or equal to $0$. Expanding the integrand yields $$\int_0^1 f(x)^3-\frac{3}{4}f(x)-\frac{1}{4} \; \mathrm{d}x=\int_0^1 f(x)^3 \; \mathrm{d}x-\frac{3}{4}\int_0^1 f(x) \; \mathrm{d}x -\frac{1}{4} \int_0^1 \; \mathrm{d}x $$ $$= \int_0^1 f(x)^3 \; \mathrm{d}x -\frac{1}{4}$$ And because the expression equals zero when $f(x)=1$ or $f(x)=-\frac{1}{2}$, $\frac{1}{4}$ is the maximum value.