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let $f(x)$ have twice differentiable on $[a,b]$,and such $$f(x)\cdot f''(x)<0$$ show that $$\dfrac{1}{b-a}\int_{a}^{b}|f(x)|dx>\dfrac{1}{2}|f(a)+f(b)|$$

I only know and can prove follow this inequality $$\dfrac{1}{b-a}\int_{a}^{b}f(x)dx\ge\dfrac{1}{2}(f(a)+f(b))$$ where $f''(x)<0$ But My problem can't use this methods,Thank you

I try to $$\dfrac{1}{b-a}\int_{a}^{b}|f(x)|dx\ge\dfrac{1}{b-a}\left|\int_{a}^{b}f(x)dx\right|$$ $$\Longleftrightarrow \left|\int_{a}^{b}f(x)dx\right|>\dfrac{b-a}{2}|f(a)+f(b)|$$ and this can't usefull

Thank you very much!

math110
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  • $f$ cannot vanish and $|f|$ is strictly concave. – WimC Dec 27 '13 at 08:53
  • Strike. The following demonstrates that strict negativity is necessary: (Is this even true?) Consider $f(x)=\sin(x)$ on $[a,b]=[-\tfrac\pi3,\tfrac\pi3]$. The claim would be that $\frac{3}{4\pi}>\frac{\sqrt3}2$. – Lutz Lehmann Dec 27 '13 at 08:57
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    your example does not satisfy the conditions, because $f(x).f''(x)<0$ but in your example $f(0)f''(0)=0$. – Farshad Nahangi Dec 27 '13 at 09:03

1 Answers1

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By the conditions over $f$, $f(x)\neq 0$ for all $x\in [a,b]$. Therefore, we should have that $f(x)>0$ or $f(x)<0$ for all $x\in [a,b]$. Assume, without loss of generality, that $f(x)>0$. Then, we have that $f''(x)<0$ and then, $f$ is concave on $[a,b]$. This implies that for every $t\in [0,1]$, we have $$ f(ta+(1-t)b)\geq tf(a)+ (1-t)f(b) $$ then, by the monotony of the integral, $$ \int_0^1f(ta+(1-t)b) dt \geq \int_0^1 (tf(a)+ (1-t)f(b))dt $$ The right hand side of the inequality is $(f(a)+f(b))/2$. If we made the change of variable $x=ta+(1-t)b$, then $dt= (1/(a-b))dx$ and it follows that $$ \int_0^1f(ta+(1-t)b) dt= \frac{1}{a-b}\int_b^a f(x)dx= \frac{1}{b-a}\int_a^b f(x)dx $$ That is what we wanted because in this case, $f=|f|$ If $f(x)<0$, then $-f$ satisfies the same conditions and $-f(x)>0$. Then we have that $$ \frac{1}{b-a}\int_a^b -f(x)dx\geq \frac {-f(a)-f(b)}{2} $$ and again we obtain the result because in this case, $-f=|f|$.

Brandon
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