Maximum value of $\int_0 ^1 f(x)(x^2) dx - \int_0^1 x(f(x)^2) dx $

Question

Let $f: [0,1] \to \mathbb R$ be a continuous function. What is the maximum value of the following functional?

$$\displaystyle \int_0 ^1 f(x)(x^2) \,\mathrm d x - \int_0^1 x(f(x)^2) \,\mathrm d x $$

Unable to understand how to solve this problem. I have thought about it analytically and think that the following points are important:

• $x^2 \le x$ for $x\in [0,1]$
• $(f(x) )^2 x ^2 \le x(f(x))^2$

How do I proceed from here?

Answer 1

We have $$ \begin{aligned} \int_0^1 f(x) \cdot x^2 \,dx - \int_0^1 f^2(x) x \,dx &\leq \sqrt{\left(\int_0^1 f^2(x) x \,dx \right)\left(\int_0^1 x^3 \,dx\right)} - \int_0^1 f^2(x) x \,dx \\ &= \frac{1}{2}\sqrt{\int_0^1 f^2(x) x \,dx } - \int_0^1 f^2(x) x \,dx\\ &\leq \frac{1}{16} + \int_0^1 f^2(x) x \,dx - \int_0^1 f^2(x) x \,dx\\ &= \frac{1}{16}, \end{aligned} $$ where the first inequality follows from the Cauchy-Schwarz inequality, and the second inequality follows from the AM-GM inequality. It can be verified that all inequalities become equalities when $f(x) = x/2$.

Answer 2

This sort of hand wavy, but I think it works.

Let $g(s,t) = st(s-t)$ and find its max and min on $[0,1]\times [0,1].$ The max turns out to be at $(1,1/2),$ which is on the boundary of the domain. When we need this result, we can scale it to any square $[0,a]\times [0,a].$

Now take your expression, and push the integrals together to get

$$\int_0^1 x^2f(x) - x(f(x)^2) \; dx = \int_0^1 xf(x)(x-f(x)) \; dx.$$

Rescale $f(x)$ so that its range is $[0,1]$. (I suppose we just divide $f(x)$ by its maximum value on $[0,1].$)

The integrand is of the form $st(s-t)$ so to make it as large as possible we want $t=s/2$ by from the result above. So I think that shows that $f(x) = x/2$ and the integrand is $x(x/2)(x/2)$ making the max $1/16$.

Answer 3

$$I=\int_{0}^{1}x\left(xf(x)-(f(x))^{2}\right)\,dx$$

$$I=\int_{0}^{1}x\left(-\left((f(x))^{2}-xf(x)+\frac{x^{2}}4\right)+\frac{x^{2}}4\right)\,dx$$

$$I=\int_{0}^{1}x\left(\frac{x^{2}}4-\left(f(x)-\frac{x}{2}\right)^{2}\right)\,dx$$

$$\le\int_{0}^{1}\frac{x^{3}}4\,dx=\frac1{16}$$

Following along in the proof, we clearly have equality in this integral if we let $\displaystyle f(x)=\frac{x}2.$

Answer 4

The Lagrangian is

$$\mathcal L (x,y) := x^2 y - x y^2$$

Since there is no $y'$ in $\mathcal L$, the Euler-Lagrange equation is simply

$$\partial_y \mathcal L (x,y) = x^2 - 2 x y = x ( x - 2 y) = 0$$

and, thus, $y = \frac{x}{2}$ is critical. Let $y = \frac{x}{2} + \varepsilon (x)$. Hence,

$$\int_0^1 \mathcal L \left( x, \frac{x}{2} + \varepsilon (x) \right) \, \mathrm d x = \cdots = \underbrace{\frac 14 \int_0^1 x^3 \,\mathrm d x}_{=\frac{1}{16}} - \underbrace{\int_0^1 x \left( \varepsilon (x) \right)^2 \, \mathrm d x}_{\geq 0} \leq \frac{1}{16}$$

Thus, $y = \frac{x}{2}$ corresponds to a maximum.