Examples of Increasing, concave $f(x)$, strictly positive with $\vert f''(x)\vert$ increasing

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?

Thanks

• 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. Commented Feb 7, 2020 at 0:16

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 $$f(x)=3x-x^3$$.
• I think your answer also generalizes to show that it cannot happen on a half line like $(0,\infty)$ or $(-\infty, 0)$ either. Commented Feb 7, 2020 at 0:06
• @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 $(0,\infty)$. Commented Feb 7, 2020 at 0:58
• 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. Commented Feb 7, 2020 at 4:31