# Maximizing an integral

A function $$f:\mathbb{R}\rightarrow [-1,1]$$, has its integral vanish over the interval $$x\in[1,3]$$.

Find the maximum value of the integral $$\displaystyle\int_{1}^{3}\dfrac{f(x)}{x}\mathrm{d}x$$

I tried to throw the Euler-Lagrange equation at this functional but got some non-sensical condition that said $$x=\mathrm{constant}$$.

My teacher gave a non constructive proof that f is actually a non-continuous function, and proved the upper bound.

My question is:

1. What does this condition mean? What should you do in the case the EL equation yields no function?
2. Is there any motivation for considering a discontinuous function?
• Using Euler Lagrange imposes some differentiability conditions on $f$. Without having checked I would speculate that the solution is a 'bang-bang' solution, so I would guess $+1$ on $[1,2]$ and $-1$ on $[2,3]$. – copper.hat Feb 10 at 5:43
• You're right. Can you explain why? And what you mean by 'bang-bang'? – Archimedesprinciple Feb 10 at 5:46
• It means that the control ($f$) is at a limit most of the time. The constraints here are $\int f = 0$, $-1 \le f(x) \le 1$. – copper.hat Feb 10 at 5:53

There are three constraints on $$f$$, $$-1 \le f(x) \le 1$$ and $$\int f = 0$$. This combination of constraints makes this slightly different than a classical optimal control setting.

Intuitively, since $$x \mapsto {1 \over x}$$ is decreasing, we expect $$f$$ to have value $$+1$$ and then transition to $$-1$$. Since we have the zero average constraint the transition would occur at $$t=2$$.

Let $$f^*$$ be $$1$$ on $$[1,2]$$ and $$-1$$ on $$(2,3]$$ and let $$f$$ satisfy the constraints. We have $$\int_1^2 f+ \int_2^3 f = 0$$ and $$\int_1^2 f^*+ \int_2^3 f^* = 0$$, so $$\int_1^2 (f^*-f) = -\int_2^3 (f^*-f)$$, or equivalently $$\int_1^2 (1-f) = \int_2^3 (1+f)$$.

Since $${1 \over x} \ge {1 \over 2} \ge {1 \over y}$$ for $$x \in [1,2], y \in [2,3]$$, we have $$\int_1^2 {(1-f(x)) \over x}dx \ge \int_1^2 {(1-f(x)) \over 2}dx = \int_2^3 {(1+f(x)) \over 2}dx \ge \int_2^3 {(1+f(x)) \over x}dx$$ from which we get $$\int_1^3 {f^*(x) \over x} dx \ge \int_1^3 {f(x) \over x} dx$$.

Alternative:

Here is a slightly different approach that might offer some more intuition (or just moves the complication somewhere else).

Let $$C= \{f:[1,3]\to [-1,1] | f \text{ is measurable}, \int f = 0 \}$$, and consider $$C \subset L^2[1,3]$$. It is straightforward to show that $$C$$ is convex & compact (in $$L^2$$) and the function $$c(f)= \int_1^3 {f(x) \over x}dx$$ is continuous and linear. Hence there is a minimiser $$f^*$$ because of compactness.

Finally, we can show that if $$f \in C$$, then if the measure of $$\{ x \in [1,2] |f(x) < 1 \}$$ is non zero, there is some $$f' \in C$$ such that $$c(f') > c(f)$$. Hence the function $$1_{[1,2]}-1_{(2,3]}$$ (or any ae. equivalent function) is a minimiser.

Another way of doing this would be to note that since $$c$$ is linear, we can take a minimiser $$f^*$$ to be an extreme point of $$C$$ and we can show that the extreme points of $$C$$ are the measurable zero average functions that satisfy $$f(x) \in \{-1,+1\}$$ for ae. $$x$$. In particular, you can see a simple form of a 'bang bang' control, and the problem reduces to determining the 'on set' (where $$f(x) = 1$$).