I came along the statement that for $x \geq z$, if $U(x)$ is a renewal function, there exists a constant $K$ such that \begin{align} U(x) - U(x-z) \leq U(z) \leq K (z+1). \end{align} This is not clear to me. For instance, Blackwell's renewal theorem gives $U(x) - U(x-z) = O(1)$, but not the above... It seems to be a rather basic statement. References are also welcome.
1 Answer
Renewal functions are subadditive. For a reference, this article (in Example 1) points to [Dal, Section 4], [Fell, Ch XI]:
[Dal] D. J. Daley, Upper bounds for the renewal function via Fourier methods, Annals of Probability 6 (1987), 876-884. MR0494547
[Fel] W. Feller, An introduction to probability theory and its applications, Volume II, 2nd ed., Wiley, 1971. MR0270403
This gives the first inequality by the very definition, and the other one by a simple estimate $$U(z) = U(1 + 1 + \ldots + 1 + (z - \lfloor z\rfloor)) \leqslant \lfloor z\rfloor U(1) + \sup\limits_{[0,1]} U.$$