When does the sum of Lipschitz constants over intervals of length 1 converge?

Question

Given a function $f$ that is Lipschitz-continuous on every unit interval $[i,i+1]$, $i\in\mathbb{Z}$ with some Lipschitz constant $K_i$, what kind of conditions must be met for $$ \sum_{i\in\mathbb{Z}}K_i $$ to converge? In particular I'm thinking of $f$ as a probability density.
This sum seems to converge for most any conventional distribution I can think of: Normal, Exponential, Chi-squared, etc. ... Any density that I can come up with as a counterexample has to pull quite a stunt by having infinitely many, smaller and smaller spikes - rather pathological.
Is there some natural assumption one could state to rule these kinds of behaviors out?

Answer 1

A sufficient condition is that $\int_\mathbb{R} |f'|$ and $\int_\mathbb{R} |f''|$ both converge. Indeed, the Lipschitz constant of a differentiable function is the supremum of its derivative in absolute value. Note that $$ |f'(a)-f'(b)|\le \int_{i}^{ i+1} |f''|\quad \forall a, b \in [i, i+1] $$ and that there exists $x\in [i, i+1]$ such that $f'(x) = \int_{i}^{i+1} f'(t)\,dt$. Combining these, we get $$ \sup_{[i, i+1]} |f'| \le \left|\int_{i}^{i+1} f'(t)\,dt \right| + \int_{i}^{ i+1} |f''(t)|\,dt \le \int_{i}^{ i+1}(|f'(t)|+ |f''(t)|)\,dt $$ as claimed.

A non-artificial density function satisfies the above, because $f$ tends to $0$ and both $f'$ and $f''$ have constant sign at infinity (they don't oscillate like a trigonometric function). These properties imply the convergence of both $\int f'$ and $\int f''$.