Let f be a differentiable function on (0,1) inclusive, such that f(0)=0 and f(1)=1. If the derivative f' of f is also continuous on (0,1) inclusive, prove that

$$\int_0^1 |f'(x)-f(x)|dx \ge 1/e $$

My attempt: let $ h(x) = e^{-x} f(x)$

Then the integral becomes

$$\int_0^1 |e^x h'(x)|dx $$

which upon integration by parts reduces to

$$\left|1-\int_0^1 e^x h(x) dx\right| $$

How should I go about proving the inequality. Also, how do I remove the modulus sign if I am not certain of the domain for which the function is negative. Thanks.

  • $\begingroup$ Note that $1/e=h(1)-h(0)=\int_{0}^{1}h'(x)\,dx\leq \int_{0}^{1}|h'(x)|\,dx$ and you are done. $\endgroup$ – Paramanand Singh Jun 25 '17 at 8:18

If $f(x) = x$, then $f'(x) = 1$, so $f$ and $f'$ satisfy the hypotheses. But $\int_0^1 |f'(x)-f(x)|dx = \int_0^1 1-x dx = \frac{1}{2}$ which is not less than or equal to $\frac{1}{e}$, so the statement is false.

EDIT: Using your notation, $\int_0^1 |e^xh'(x)|dx \ge \int_0^1 |h'(x)|dx \ge \int_0^1 h'(x)dx = h(1)-h(0) = \frac{1}{e}$.

  • $\begingroup$ Gee, sorry, I made a mistake. It should be greater than or equal to. I will edit it now. $\endgroup$ – Anonymous Jun 25 '17 at 7:46
  • $\begingroup$ While I was typing my comment you gave an answer. +1 $\endgroup$ – Paramanand Singh Jun 25 '17 at 8:19
  • $\begingroup$ Got it. Thanks a lot. $\endgroup$ – Anonymous Jun 25 '17 at 8:30

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