# Calculus - Integral - existence problem

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 $$x \in [a,b]$$, $$f'(x). Use that to find boundary for $$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. 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

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

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.