Is anyone able to provide a(n) example(s) of strictly positive, increasing, concave function where the second derivative becomes larger (in absolute value) as $x$ increases?

Or, if this is not possible, can someone provide some intuition or a proof/link to explain?


  • $\begingroup$ I don't think you can have strictly positive and $|f''|$ increasing since that would mean that $f''$ is decreasing ($f$ is concave so $f'' \le 0$) and therefore that $f'$ becomes negative at some point on any interval containing a positive ray. $\endgroup$ Feb 7, 2020 at 0:16

1 Answer 1


Not on all of $(-\infty,\infty)$, no. In fact, if $f$ is increasing and concave on $(-\infty,\infty)$ then it must satisfy $\lim_{x\to-\infty}f(x)=-\infty$.

To see this, note that since $f$ is increasing, its derivative exists at some $x_0\in(-\infty,\infty)$. Since $f'(x_0)>0$, the tangent line $T(x)$ to $f$ at $x_0$ has positive slope, so that $T(x)\to-\infty$ as $x\to-\infty$. However due to concavity of $f$ we have $f(x)\leq T(x)$ for all $x\in(-\infty,\infty)$, and hence $f(x)\to -\infty$ as $x\to-\infty$ as claimed.

It's also impossible to find an example on a positive ray, although for different reasons. Without loss of generality we may suppose $f:(0,\infty)\to(-\infty,\infty)$ is concave and that $f''$ exists with $|f''(x)|$ increasing. Since $f''(x)<0$, that means $f''$ is decreasing. In particular, there are $a,\delta>0$ such that $f'(x)<-\delta$ for all $x\geq a$. Note that there is $C\in(-\infty,\infty)$ such that $f'(x)=C+\int_a^xf''(t)\;dt\leq C+\delta a-\delta x\to -\infty$ as $x\to\infty$. Hence $f$ is eventually negative.

If you want $f$ defined on an interval, that means the interval must be bounded. This is quite easy to do. For instance let $f:(0,1)\to(0,\infty)$ be defined by $f(x)=3x-x^3$.

  • $\begingroup$ I think your answer also generalizes to show that it cannot happen on a half line like $(0,\infty)$ or $(-\infty, 0)$ either. $\endgroup$
    – angryavian
    Feb 7, 2020 at 0:06
  • $\begingroup$ @angryavian You need a different argument for that (see my edits). But if all you want is $f$ increasing, concave, and positive, then you can take $f(x)=\log(x+1)$ restricted to $(0,\infty)$. $\endgroup$
    – Ben W
    Feb 7, 2020 at 0:58
  • $\begingroup$ Thanks. This is a nice answer. Two quick follow up questions if you have a moment: 1) For the paragraph on the positive ray, is the WLOG referring to the domain of $f$ being $(0,\infty)$ (as opposed to say $(5,\infty)$? Because I don't think we need WLOG for the rest of that sentence (concave, increasing absolute value and second derivative exists), as those are our assumptions. 2). $f'(x) = C +\int f''(t)dt)$ -- is this trivial? i.e., just define C to be the difference between the integral and the first derivative. $\endgroup$
    – user106860
    Feb 7, 2020 at 4:31

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