A functional inequality involving maximum and minimum

Question

I'm trying to show this inequality $\int_{-1}^{1} |\frac{d}{dx}f(x)|dx\ge \max_{x\in[-1,1]} f(x)-\min_{x\in[-1,1]} f(x)$ for differentiable $f$.

I've tried $$ \int_{-1}^{1} |\frac{d}{dx}f(x)|dx\ge |\int_{-1}^{1} \frac{d}{dx}f(x)dx|=|f(1)-f(-1)|, $$ using fundamental theorem of Calculus, but it seems in vain. Any reference, suggestion, idea, or comment is welcome. Thank you!

Answer 1

Almost there.

In fact, note that $f$ is differentiable hence continuous, therefore there are points $a,b$ such that $f(a) = \sup_{[-1,1]} f(x)$ and $f(b) = \inf_{[-1,1]} f(x)$. WLOG let $a \leq b$. Then :

But : $$ \int_{-1}^1 |f_x|dx \geq \int_{a}^b |f_x|dx = \left|\int_{a}^b f_x dx\right| \geq |f(a) - f(b)| $$

The first inequality follows from the fact that we are integrating a non-negative function over a smaller interval, the second inequality from FTC.

Note : Even if $a,b$ did not exist, we could still have taken approximating sequences $f(a_n) \to \sup f, f( b_n) \to \sup f$ and worked by taking $a_n,b_n$ in the inequality.

Answer 2

Note that $\int_{-1}^{1} |f_x| dx\geq \int_a^{b} |f_x| dx\geq |\int_a^{b} f_x dx|=|f(b)-f(a)|$ for all $a, b \in [-1,1]$ with $ a<b$. The inequality also holds for $a \geq b$. Can you finish the proof ?

Answer 3

@LHC and @Kabo Murphy. This maybe is not the best way to prove this. Note that if one considers $f(x) = x^2$ for $x \in [-1, 1]$ then $|f(1) - f(-1)| = |1 - 1|= 0$ meanwhile $ \max_{f} - \min_f = 1 - 0= 1$.