How to prove the convexity of complicated derivatives?

Question

How to prove the following function is convex ?

$F(x) = \frac{a}{x \ln \left(1+ \frac{b}{x}\right)}$ such that $a,b > 0$

I found that taking the second derivative is too complicated. Then, I tried to use the definition and the approximation of $\ln(1+b/x) \approx b/x$ however I found that the function is just a constant $a/b$ at the end.

Any suggestion? Thanks for your help in advance.

Answer 1

Convexity of $F(x)$ further requires $x > 0$, which I will assume.

Convexity is easily shown by following the rules of Disciplined Convex Programming (DCP) to constructively show convexity.

In particular, using the CVX functions rel_entr, which is convex, thereby making -rel_entr concave, and inv_pos, which is a convex function of its nonnegative concave argument. we have

a/(x*log(1+b/x)) = a*inv_pos(-rel_entr(x,x+b)) which proves convexity.

Indeed, this would be jointly convex in $x, b$ if $b$ were also a nonnegative variable.

As a bonus, the function has been reformulated in a way allowing entry into CVX, thereby facilitating numerical optimization.

The CVX Users' Guide may be consulted fro definitions of these functions.