Is the logarithm of increasing (not necessarily strictly) function is concave?

Question

I need some help. We know log(x) is a concave function. Now, if f(x) is an increasing function, can we say that log(f(x)) is concave?

Any help regarding this problem is really appreciated. Thanks.

Answer 1

No.

Take any non-concave increasing function $g$ and take the function $f(x)=e^{g(x)}$

Then:

• $f(x)$ is an increasing function, because $x\mapsto e^x$ is increasing, and $g$ is increasing.
• $x\mapsto \log(f(x))$ is actually the function $g$, so it is not concave.

Answer 2

Assume as a contradiction that your claim were true, and specialize it to the case that $g:(0,\infty)\to(0,\infty)$ is $C^2$. Since $$(f\circ g)''=g''\cdot f'\circ g+(g')^2\cdot f''\circ g$$

This would mean that (when $f=\ln$ and $g$ is increasing on $(0,\infty)$, positive and $C^2$), $$\frac{g''(x)}{g(x)}-\frac{(g'(x))^2}{(g(x))^2}\le 0\qquad\forall x>0\\ g''(x)g(x)\le (g'(x))^2\qquad\forall x>0$$

However, this fails for several functions. One is, for instance, $g(x)=x^2+x+1$. In fact the condition $$2x^2+2x+2\le 4x^2+4x+1$$ is not satisfied for small positive values of $x$.

And, in fact, $\ln(x^2+x+1)$ is not concave on $(0,\infty)$: its positive inflection point is at $x=\frac{\sqrt3-1}2$.

Notice that, qualitatively, the sign of $g'$ is of no consequence when you consider the sign of $(f\circ g)'$. Therefore, one would expect the condition of being increasing alone to be equally uninformative when it comes to determining if $f\circ g$ is concave or convex.