# Subtracting a constant from log-concave function preserves log-concavity, if the difference is positive

I am trying to work out a question from 'Convex Optimization - Boyd' . Specifically, exercise 3.48:

Show that if $f : \mathbb R^n \to \mathbb R$ is log-concave and $a > 0$, then the function $g = f - a$ is log-concave, where $\operatorname{dom} g= \{x \in \operatorname{dom} f | f(x) > a\}?$

This is my attempt so far:

By assumption, we have $$\frac{d^2\log f}{dx^2}=\frac{f''f-(f')^2}{f^2}\leq0.$$ Considering $g=f-a$, we have $$\frac{d^2\log g}{dx^2} = \frac{f''(f-a)-(f')^2}{(f-a)^2}=\frac{f''f-(f')^2}{(f-a)^2}-\frac{f''a}{(f-a)^2}.$$ If $f''>0$, then we are complete. I'm not sure how to handle the alternative situation though, can anyone help?

• Is $f$ assumed differentiable? Aug 22 '14 at 21:40
• It is not, as far as the exercise statement is concerned. Jan 29 '16 at 8:59

You correctly compute $$\frac{d^2\log g}{dx^2} = \frac{f''(f-a)-(f')^2}{(f-a)^2}.$$ When $a=0$, we know this quantity is non-positive, so the numerator is non-positive. Focus just on the numerator. What happens to the numerator when $a$ increases from $0$? Can you say whether this expression remains non-positive? Note that by assumption, $f-a >0$. It may be useful to consider the cases when $f''>0$ and $f''\le 0$ separately.
And by the way, a complete proof would take account of the fact that $f$ is a function of values in $\mathbb{R}^n$, not just $\mathbb{R}$. That's really not a big issue here, though, since $f$ is log-concave if for all $x_0,v\in\mathbb{R}^n$, the function $\tilde{f}:\mathbb{R}\to\mathbb{R}$ is log-concave, where $$\tilde{f}(t) \equiv f(x_0 + v t).$$ This is just a simple trick for converting questions of concavity in $n$ dimensions into equivalent questions in one dimension by considering how $f$ changes along arbitrary directions $v$.
• ... Actually, it's blindingly obvious. Either $f''>0$, in which case (as I stated) our result is satisfied, or $f''<0$, in which case $f''(f-a)<0\implies f''(f-a) - (f')^2 < 0$, which gives us our result. I guess it was my separating the final equatio that made it slightly confusing. Aug 22 '14 at 21:33