Log-concave function changes when scalar is added

Question

A function $f$ is log-concave if $\log(f)$ is concave. Intuitively, one might guess that adding a scalar to a function would not affect properties like concavity, log-concavity, quasi-concavity etc., and I find it surprising this is not true.

For example, $f=\frac{1}{1+e^x}$ is known as a logistic function and is log concave, however $g=1+\frac{1}{1+e^x}$ is not. This "small" change can have important practical consequences - for example if we want to use optimization techniques to find a maximum.

Could someone kindly try to provide an intuitive explanation how adding a scalar changes such a fundamental property of the function?

Answer 1

Log-concavity is about multiplication. For example $f(x) = 1/(1+e^{-x})$ is log-concave, and $g(x) = 2/(1+e^{-x})$ is as well. Multiplying $f$ by a constant factor corresponds to adding a constant to $\log f$. But if you add a constant to $f$, that has a much bigger effect on $\log f$ in regions where $f$ is small than in regions where $f$ is large.

Answer 2

According to Boyd & Vandenberghe (2009, Convex Optimization, p. 105), a function $f(x)>0$ defined over $\mathbb{R}$ is log-concave iff $$f(x)f''(x)\leq (f'(x))^2$$ As shown here, while retaining log-concavity when subtracting the scalar, adding a scalar is more problematic. It requires that $f''(f+k)\leq(f')^2$ keeps true. While this might be true for low enough $k$, if you keep increasing $k$ you will get a point where the above inequality is reversed.