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Let $f$ be a function that has a continuous derivative over the interval $[a,b]$ and let $f(a)=f(b)=0$. Prove that $$\max\:\left|f'\left(x\right)\right|\ge \frac{4}{\left(b-a\right)^2}\int _a^b\:\left|f\left(x\right)\right|dx$$

I proceeded by splitting up the integral into 2 parts i.e from $a$ to $(a+b)/2$ and from $(a+b)/2$ to $b$, but got stuck, please help?

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    Interesting problem. It is true when the graph of $f$ is a triangle with basis $[a,b]$, so I suppose it must be true in general. – 2017-02-22
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    To prove it in general, can we take any random function and satisfy this equation, since i have tried numerous ways to prove the above mentioned problem – 2017-02-22
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    Let $\max|f'(x)|=c$. See if you can show that $-c(x-a)\leq f(x)\leq c(x-a)$ and $-c(b-x)\leq f(x)\leq c(b-x)$. Try using Fundamental Theorem of Calculus on $[a,x]$ and $[x,b]$. – 2017-02-22
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    i have tried doing it, but didn't get the desired result – 2017-02-22

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Note that

$$\frac{4}{(b-a)^2}\int_a^b |f(x)| \, dx = \frac{4}{(b-a)^2}\int_a^{(a+b)/2} |f(x)| \, dx + \frac{4}{(b-a)^2}\int_{(a+b)/2}^b |f(x)| \, dx $$

By the MVT, for $x \in [a, (a+b)/2]$

$$|f(x) - f(a)| = |f(x)| = |f'(\xi_1)||x - a| \leqslant \max|f'|(x - a)$$

For $x \in [(a+b)/2, b]$

$$|f(x) - f(b)| = |f(x)| = |f'(\xi_2)||x - b| \leqslant \max|f'|(b - x)$$

Now substitute and integrate.

$$\frac{4}{(b-a)^2}\int_a^b |f(x)| \, dx \leqslant \frac{4 \max |f'|}{(b-a)^2}\left( \int_a^{(a+b)/2} (x-a) \, dx + \frac{4}{(b-a)^2}\int_{(a+b)/2}^b (b-x) \, dx\right)$$

You should be able to finish by showing the sum of the two integrals equals $(b-a)^2/4.$

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    thanks a lot, i should have thought about applying MVT – 2017-02-22
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    @LalitPathak: You're welcome. You had the right idea. – 2017-02-22