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Suppose $f\in C^2[a,b]$, $f(a)=f(b)=0$. Then for any $x\in [a, b]$: $\frac{f(x)}{(x-a)(b-x)}\le \frac{1}{b-a} \int_a^b{|f^{\prime\prime}(t)|dt}.$

Any help is appreciated.

1 Answers 1

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Fix $x\in ]a,b[$ once and for all. Notice that (b-a)\frac{f(x)}{(x-a)(b-x)}=\frac{f(x)-f(a)}{x-a}-\frac{f(b)-f(x)}{b-x}=f'(c)-f'(d) where I have used a corollary of Rolle's theorem to find $c\in ]a,x[$ and $d\in ]x,b[$ with f'(c)=\frac{f(x)-f(a)}{x-a} and f'(d)=\frac{f(b)-f(x)}{b-x}, and of course the fact that $f(a)=f(b)=0$. Finally, f'(c)-f'(d)=\int_c^d f'' and you get the inequality you want as a consequence. You can extend the inequality to $x=a$ or $b$ if you define the left hand side of your equation at these points as limits: the resulting function will be continuous and well defined, and the inequality remains true by continuity.

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    Missed the trick in the first line. Nice answer. Thank you. :)2011-06-11