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We are given a function $f$, $f$ is integrable (in the riemann sense) in $[a,b]$ and also $f'$ is a continuous, and $f(a)=f(b)=0$. Prove that there exists a point $c$ such that $|f'(c)| \geq \frac{4}{(b-a)^2}\cdot\int_{a}^{b}f(x)dx$.
[Hint: Let $M = \frac{4}{(b-a)^2}\cdot\int_{a}^{b}f(x)dx$ and assume for contradiction that for all $x \in [a,b]$, $f'(x)<M$. Use that to find boundary for $f$ in each of the intervals $[a,(a+b)/2]$ and $[(a+b)/2,b]$, using the mean value theorem, and use that to find a boundary for $\int_{a}^{b}f(x)dx$. Use that to get a contradiction.
Here is what I did. Suppose the opposite. Take $x \in [a,(a+b)/2]$. Same thing is true for $[(a+b)/2,b]$.
$f'$ is continuous so define $u=maxf'=M-\delta<M$. We get the following: $$|f(x)|=|f(x)-f(a)|=|\frac{f(x)-f(a)}{x-a}\cdot(x-a)|=|f'(c)||x-a|\leq (M-\delta)\cdot(b-a)/2$$
Now, there exists $x$ such that: $|\int_{a}^{b}f|=|f(x)|\cdot(b-a)\leq(M-\delta)\cdot(b-a)^2/2$
Plugging $\int_{a}^{b}f=M(b-a)^2/4$ we get $M\geq 2\delta$, and I'm not sure how this is a contradiction. How can I complete the proof? Thanks

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The case $f=0$ is trivial. Otherwise, without loss of generality, assume $a=0.$ If the claim is false, and if $x\in [0,b],$ then

$\tag1 -Mx<\int^x_0f'(t)dt<Mx\quad \text{and}\quad -M(b-x)<\int^{b}_xf'(t)dt<M(b-x).$

The first integral above is $f(x)$ and the second is $-f(x).$ Therefore, we have

$\tag 2-2M\cdot \frac{b^2}{8}<\int^{b/2}_0f(x)dx+\int^{b}_{b/2}f(x)dx<2M\cdot \frac{b^2}{8}$ which is

$\tag3 -2M\cdot \frac{b^2}{8}<\int^{b}_0f(x)dx<2M\cdot \frac{b^2}{8}$

Plugging in for $M$ we get

$\tag4 -\int^{b}_0f(x)dx<\int^{b}_0f(x)dx<\int^{b}_0f(x)dx,$

which is impossible.

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