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Hey guys I've been stuck on this problem for a while and I've still got no headway into how I'm supposed to do this. Any tips or guiding points would be helpful.

Let $f$ be a differentiable function such that $f′$ is continuous on $[0,1]$, and $M$ the maximum value of $|f′(x)|$ on $[0,1]$. Prove that if $f(0) = f(1) = 0$, then

$$\int_0^1 |f(x)| dx\leq\frac{M}{4}$$

I've come to a conclusion via Rolle's theorem that there exists a c between 0 and 1 where f'(c) = 0 and that would be a local extrema. But nothing past that other than the fact that the max value of M is 1.

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  • $\begingroup$ Try to figure out which function will give an equality (Actually, there’s no &C^1$ function s.t. equality holds.) $\endgroup$ – Seewoo Lee Apr 26 '18 at 10:36
  • $\begingroup$ hmm is there a way to illustrate what you're saying? because I don't really understand $\endgroup$ – Jonathan Low Apr 26 '18 at 10:47
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By the mean value theorem, for all $x\in (0,1)$, it is $\displaystyle{\frac{f(x)}{x}=\frac{f(x)-f(0)}{x-0}=f'(c_x)}$ for some $c_x\in(0,x)$ and similarly $\displaystyle{\frac{f(x)}{1-x}}=\frac{f(x)-f(1)}{1-x}=f'(d_x)$ for some $d_x\in(x,1)$. Therefore, since $f(x)=xf'(c_x)$ and $f(x)=(1-x)f'(d_x)$, by taking absolute values we get that for all $x\in(0,1)$, it is $|f(x)|\leq Mx$ and $|f(x)|\leq M(1-x)$. Hence $|f(x)|\leq M\min\{x,1-x\}:=Mg(x)$. $g(x)$ is a piecewise function, equal to $x$ on $[0,\frac{1}{2}]$ and equal to $1-x$ on $[\frac{1}{2},1]$.

By integrating the preceding inequality, one gets $$\int_0^1|f(x)|dx\leq M \bigg{(}\int_0^\frac{1}{2}xdx+\int_\frac{1}{2}^1(1-x)dx\bigg{)}=\frac{M}{4}$$

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    $\begingroup$ Thanks so much! Took awhile to understand the last part on g(x) being a piecewise function but it's a clear now. Thanks! $\endgroup$ – Jonathan Low Apr 26 '18 at 14:11

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