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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.

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up vote 6 down vote accepted

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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+1, this is right on. –  Eric Naslund Jun 11 '11 at 0:45
    
Missed the trick in the first line. Nice answer. Thank you. :) –  Qiang Li Jun 11 '11 at 5:01
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