Given that $\int_0^1{f’(x)dx}=0$ and $\int_0^1{xf’(x)dx}=-1$ , find the maximum value of $\int_0^1{|f’(x)|dx}.$ (Also given- $f(0)=f(1)=0$, and $f$ is differentiable.)
A friend of mine actually came up with this question, and we don’t know if it’s possible to solve or not. We have tried using simple modulus inequalities for integrals, but this gets us nowhere.
Any help would be greatly appreciated.
Best Answer
The integral $\int_0^1|f'(x)|dx$ is unbounded in $X=\{f\in C^0[0,1]|f(0)=f(1)=0\}$. Here is the proof:
(1) Any function $f\in X$ satisfies the first condition: $\int_0^1f'(x)dx=f(1)-f(0)=0.$
(2) Integration by parts shows that $\int_0^1 xf'(x)dx=-1$ is equivalent to $\int_0^1 f(x)dx = 1$.
(3) Now, for $a<\frac{1}{2}$ let $f_a(x):[0,1]\to \Bbb{R}$ be defined as $f_a(x)=\frac{x}{a^2}$ for $x\in[0,a]$, $f_a(x)=\frac{2a-x}{a^2}$ for $x\in(a,2a]$, and $f_a(x)=0$ for $x\in(2a,1]$.
It's straightforward to verify that (a) $f_a\in X$, (b) $\int_0^1f_a(x)dx=1$, and (c) $\int_0^1|f'_a(x)|dx=\frac{2}{a}$. Therefore, by choosing $a$ sufficiently small we can make the latter integral arbitrarily large. $\square$