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.



Renewal functions inequalities
Renewal functions are subadditive. For a reference, [this article](https://www.jstor.org/stable/20535671) (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 $$U(z) = U(1 + 1 + \ldots + 1 + (z - \lfloor z\rfloor)) \leqslant \lfloor z\rfloor U(1) + \sup\limits\_{[0,1]} U.$$


