f is a distribution, does the right derivative, written here $f'$, always exists on the domain?

Question

I am reading the paper from Tehranchi: https://arxiv.org/abs/1701.03897

Page 20, he mentions that for a density $f$, the right derivative always exists on its domain (the definition of $f$ is given page 18 for anyone who would like to check). Put it briefly, We only assume that $f$ is a continuous density.

I am baffled by the fact that the right derivative would exist. I know that densities are cadlag, but I didn't know that the right derivative always exists.

Can anyone confirm me this, and prove it ? If I am missing some hypothesis (which I highly doubt but maybe?), can you please mention it, that would be great.

Answer 1

I think the paper is assuming that $f$ is log-concave throughout this section. (See on page 19 where Proposition 3.2.3 is invoked.) So, by definition $\log\circ f$ is concave and so right differentiable which implies $f$ is also right differentiable (by the chain rule).

Answer 2

Consider a continuous function $f_n$ on $[\frac 1{n+1} ,\frac 1 n]$ such that $f_n$ vanishes at the end points, is linear in $[\frac 1{n+1} ,c_n]$ as well as $[c_n ,\frac 1 n]$ where $c_n$ is the mid-point of $[\frac 1{n+1} ,\frac 1 n]$ and $f(c_n)=1$. Putting these together we get a continuous function with $\int_0^{1}f(x)dx <\infty$. Let $f$ be a suitable constant multiple of this functions so that $\int_0^{1} f(x)dx=1$. Then $f$ is a continuous density function and the right-hand derivative at $0$ does not exist.