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I would like to know for what kind of functions, the inequality $$\frac{f^\prime(x)}{f(x)}\leq \frac{1}{b-a}$$ hold for all $x\in[a,b]$. $f^\prime(x)$ is derivative w.r.t. $x$.

Is there some general rule? (E.g. $f(x)$ must be convex, etc.) Thanks.

EDIT: Maybe an "inverse" question. For which functions does it never hold?

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It depends heavily on $b-a$. If $b-a$ is very large, then you are demanding that $f(x)$ must be much larger than $f'(x)$, while if $b-a$ is small, there is hardly any demand, appart from the fact that, obviously, $f$ must have no roots on $[a,b]$.

Nothing in your question, however, demands that the functions need to be convex or concave. In fact, for every function $f$, the function $g(x) = f(x) + M$ will satisfy your inequality for some large enough value of $M$.

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  • $\begingroup$ But it only has to hold for $x\in [a,b]$ so if $b-a$ is small it has to hold for fewer values of $x$.... $\endgroup$ – JPi Jun 5 '14 at 12:33
  • $\begingroup$ @JPi Yes, exactly. I am interested only in functions with domain $[a,b]$. $\endgroup$ – pisoir Jun 5 '14 at 12:34
  • $\begingroup$ @Jpi sure, you are correct, but I don't see your point. It still holds that convex, concave and other functions can satisfy the demand written. $\endgroup$ – 5xum Jun 5 '14 at 12:35
  • $\begingroup$ I was thinking about using the Mean value theorem, but I can't come up with anything reasonable,i.e., $f^\prime(c)=\frac{f(b)-f(a)}{b-a}$. $\endgroup$ – pisoir Jun 5 '14 at 12:36
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    $\begingroup$ For $x\in[a,b]$, $\frac{f^\prime(x)}{f(x)}=\frac{f(b)-f(a)}{f(x)(b-a)}\leq\frac{1}{b-a}$ $\Rightarrow$ $\frac{f(b)-f(a)}{f(x)}\leq 1$ $\endgroup$ – pisoir Jun 5 '14 at 12:59

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