Let f be a function that has a continuous derivative over the interval $[a,b]$, and let $f(a) = f(b) = 0$. Prove that
$$\max\limits_{a\leq x\leq b} \left|f'(x)\right| \geq \frac{4}{(b-a)^2}\int_a^b\left|f(x)\right|dx$$
I know that I should use the Mean Value Theorem to prove this, but I have no idea where to start with…
Could anyone explain how to prove this inequality?
Thank you!
Best Answer
I can prove the result with the restriction $f(x)\geq0$ on $[0,1]$.
WLOG we may take $a=0$, $b=1$ since applying the result to $g(x)=f((b-a)x+a)$ proves the result in the general case.
The idea is that the condition $\max\limits_{0\leq x\leq 1} \left|f'(x)\right| <K$ and $f(0)=f(1)=0$ implies that $f$ lies in the triangle $\{(0,0),(1/2,k/2),(1,0)\}$ which has area $K/4$ and hence $\int_{0}^{1}f(t)dt<K/4. $
Proof: $f'(x)\leq\left|f'(x)\right| <K$ implies by integrating both sides from $0$ to $x<1/2$ that $f(x)<Kx$ for $x\in[0,1/2]$. This is valid as $f$ has a continuous derivative. Integrating again gives $\int_0^xf(t)dt<Kx^2/2.$ and hence $\int_0^{1/2}f(t)dt<K/8.$ Repeating for $g(x)=f(1-x)$ gives $\int_{1/2}^{1}f(t)dt<K/8.$ Add to obtain $\int_{0}^{1}f(t)dt<K/4.\blacksquare$
Setting $K=\frac{1}{4}\int_0^1f(x)dx$ we have $\int_0^1f(x)dx<K/4=\frac{1}{4}\max\limits_{0\leq x\leq 1} \left|f'(x)\right|$ and hence $$\max\limits_{0\leq x\leq 1} \left|f'(x)\right|>\frac{1}{4}\int_0^1f(x)dx.$$ This is a contradiction. Therefore $\max\limits_{0\leq x\leq 1} \left|f'(x)\right| \geq K=\frac{1}{4}\int_0^1|f(x)|dx.$ This proves the result on the assumption that $f(x)\geq0$ on $[0,1]$.