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‎Let ‎$‎f‎$ ‎be a‎ ‎differentiable ‎real ‎function ‎on ‎‎$‎[1, +\infty)‎$ such that ‎‎‎$‎f^\prime‎‎$ ‎is ‎convex.‎

Now,‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎ ‎what ‎can ‎be ‎said ‎about ‎‎$‎f‎$ ‎?‎ ‎

I know that, if ‎$‎f‎$ ‎is ‎differentiable ‎on ‎‎$‎(a, b)‎$‎, then ‎$‎f‎$ ‎is ‎convex ‎if ‎and ‎only ‎if ‎‎$‎f^\prime‎‎$ ‎is ‎increasing.

Also, ‎if ‎$‎f^{‎\prime‎‎\prime‎}‎$ ‎exists ‎on ‎‎$‎(a, b)‎$‎, then ‎$‎f‎$ ‎is ‎convex ‎if ‎and ‎only ‎if ‎‎$‎f^{‎\prime‎‎\prime‎}‎\geq ‎0‎$‎.‎ ‎‎‎‎‎‎‎‎‎‎‎

‎

koohyar eslami
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  • In what sense do you want something said about $f$? if $f'$ is convex, $f$ is probably going to look reverse-sigmoid, or similar to $f''$ Think about whether $f$ will be increasing or decreasing - why do you think so? – axolotl Oct 07 '18 at 05:53
  • I said in general, but can i associate with concave or increasing(decreasing)? – koohyar eslami Oct 07 '18 at 06:08
  • If $f'$ is convex, then $f''$ is increasing and $f'''$ is positive. We do not have a standard name for what this means about $f$ itself, because this property has not (so far) been useful for anything. – GEdgar Oct 07 '18 at 12:23
  • can it be conditions which can be talked about $f$? – koohyar eslami Oct 07 '18 at 13:21
  • Or, in addition to the assume the question, is there any other condition which can be talked about $f$? – koohyar eslami Oct 07 '18 at 13:29
  • I think I already answered you: $f$ is probably going to look reverse-sigmoid, or similar to $f′′$ Think about whether $f$ will be increasing or decreasing - why do you think so? – axolotl Oct 07 '18 at 18:55
  • how can I prove it? – koohyar eslami Oct 08 '18 at 05:25

1 Answers1

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If $f'$ is convex on $\left]1,+\infty\right[$, then the running average $$g\colon \left]1,+\infty \right[ \to \mathbb{R} : t \mapsto \frac{1}{t-1}\int_{1}^{t}f'(x)dx$$ is convex. (Hint: you may use the technique in Running average of a convex function is convex to prove this). Since, for every $t >1$, we have $g(t) = (f(t)-f(1))/(t-1)$, one may conclude that $$\frac{f(t)-f(1)}{t-1} \text{ is convex on } \left]1,+\infty\right[.$$ Moreover, if $f(1)>0$, then $f(1)/(t-1)$ is convex (second-order test), and thus, you get $f(t)/(t-1)$ is convex.

weirdo
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