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Suppose that $f(x)$ has twice order continuous derivative over $[a,b]$, $\forall x \in [a,b]:f''(x)<0$ and $f(a)=f(b)=0.$ Prove $$\displaystyle \int_a^b \left|\frac{f''(x)}{f(x)}\right|{\rm d}x>\frac{4}{b-a}.$$

Proof

Since $f''(x)<0$, then $f(x)$ is concave over $[a,b]$. Hence, $\forall x \in (a,b):f(x)>0$. Moreover, since $f(x)$ is continuous over $[a,b]$, it can reach its maximum value at some point $x=c \in (a,b)$,namely $\max\limits_{x \in [a,b]}f(x)=f(c)>0$ and $f'(c)=0$. Therefore $$\int_a^b \left|\frac{f''(x)}{f(x)}\right|{\rm d}x=-\int_a^b \frac{f''(x)}{f(x)}{\rm d}x> -\int_a^b \frac{f''(x)}{f(c)}{\rm d}x=\frac{f'(a)-f'(b)}{f(c)}.\tag{0}$$ By Taylor's formula, we obtain $$f(c)=f'(a)(c-a)+\frac{f''(\xi_1)}{2}(c-a)^2<f'(a)(c-a)\tag{1}$$ $$f(c)=f'(b)(c-b)+\frac{f''(\xi_2)}{2}(c-b)^2<f'(b)(c-b)\tag{2}$$ From $(1)$, we obtain $$\frac{f'(a)}{f(c)}>\frac{1}{c-a}.\tag{3}$$ From $(2)$, we obtain $$-\frac{f'(b)}{f(c)}>\frac{1}{b-c}.\tag{4}$$ By $(3)+(4)$ $$\frac{f'(a)-f'(b)}{f(c)}>\frac{1}{c-a}+\frac{1}{b-c}.\tag{5}$$ By HM-AM inequality $$\frac{2}{\dfrac{1}{c-a}+\dfrac{1}{b-c}}\leq \frac{(c-a)+(b-c)}{2}=\frac{b-a}{2}.\tag{6}$$ which implies $$\frac{1}{c-a}+\frac{1}{b-c}\geq \frac{4}{b-a}.\tag{7}$$ Combining $(0)$ and $(7)$, it follows that $$\int_a^b \left|\frac{f''(x)}{f(x)}\right|{\rm d}x>\frac{4}{b-a},$$ which is what we want to prove.

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    $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 8 at 14:07
  • $\begingroup$ @zhw. it is $f^{(2)}(x)$! $\endgroup$ – mengdie1982 May 8 at 14:58
  • $\begingroup$ Try again: Is $f''<0$ on all of $[a,b]?$ $\endgroup$ – zhw. May 8 at 15:03
  • $\begingroup$ @zhw u r right! $\endgroup$ – mengdie1982 May 8 at 15:09
  • $\begingroup$ There is a typo in (2): $(c-a)^2$ should be $(c-b)^2$. Apart from that your proof looks good to me. $\endgroup$ – Martin R May 8 at 15:15
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The integral equals $\infty$ in all cases.

Proof: From concavity we see $f'(a)>0.$ Since concave functions stay below their tangent lines, we have $f(x) \le f'(a)(x-a)$ for $x\in [a,b].$ We are also given that $|f''| \ge m$ on $[a,b]$ for some $m>0.$ Thus

$$\int_a^b \left |\frac{f''(x)}{f(x)}\right |\, dx \ge \int_a^b \frac{m}{f'(a)(x-a)}\, dx=\infty.$$

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  • $\begingroup$ I just wonder if the condition should actually be $f''(x) < 0$ on the open interval $(a, b)$ for the question to make sense. In that case $\sin(x)$ on $[0, \pi]$ would be an example where the integral is finite. $\endgroup$ – Martin R May 8 at 16:22
  • $\begingroup$ You're probably right @MartinR I did check with the OP on this however. $\endgroup$ – zhw. May 8 at 16:31

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