2
$\begingroup$

Please help me with the following problem. Let $f$ be a differentiable (therefore continuous) on $[0,2]$ so that the three followings are satisfied:

$$\forall x \in [ 0,2 ],\quad \left \lvert f'(x) \right \rvert \leq 1,$$

$$ \left\vert\int_{0}^{2} f(x)dx\right\vert \leq 1.$$

$$ f(0)=f(2)=1$$

Does such function exist?

I think not but I haven't proved it yet.

I have tried Rolle and Langrange theorem, also Taylor formula, they might be useful but I am not there yet. Please help me or suggest some potential idea.

Thank you.

  • 2
    Take $f=0$, and it satisfies your conditions. – 2017-01-08
  • 0
    My mistake, there is one more demanding, sorry – 2017-01-08

2 Answers 2

3

Note that the mean value theorem shows that $f(x) \ge |1-x|$, and since $\int_0^2 f(x) dx \le 1$ and $\int_0^2 |1-x| dx = 1$, we must have $f(x) = |1-x|$, which is impossible since $f$ is differentiable.

  • 0
    @MartinR: Thanks for catching that, cut and paste accident. – 2017-01-08
3

From $|f'(x)| \le 1$, we have $f(x) \ge f(0) - \int_0^x dx = 1-x$. And similarly, $f(x)\ge f(2) - \int_2^x dx = x-1$. Define $g(x)=\max(1-x,x-1)=|x-1|$. We have $f\ge g$, so $\int_0^2 f \ge \int_0^2 g = 1$. But since the integral is bounded by $1$, we get $\int_0^2 f = 1$ and $f=g$. But this is impossible because $g$ is not differentiable.