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

Question

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.

Answer 1

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