For a function $f:[0,\infty)\rightarrow\overline{\mathbb{R}}$ that is a cadlag (right-continuous with left limits) define $$ T_n = \inf\{t\geq 0\mid \left|f(s)-f(0)\right|\geq n\}. $$ Suppose that $|\Delta f(t)|=|f(t)-f(t-)|\leq C$ for all $t$.
I need to prove that
$$ \sup_{0\leq s\leq T_n}|f(s)-f(0)|\leq n+C. $$
My attempt. Let $n\in\mathbb{N}$ be given. The idea is that, for all $t<T_n$ we have $|f(t)-f(0)|<n$. We cannot conclude that $\sup_{0\leq s\leq T_n}|f(s)-f(0)|\leq n$ because it could be that $|f(T_n)-f(0)|$ is strictly larger than $n$. Whence
\begin{eqnarray} \sup_{0\leq s\leq T_n}|f(s)-f(0)|&\leq& \max(n,|f(T_n)-f(0)|)\\ &\leq& n+|f(T_n)-f(0)|\\ &\leq& n+|f(T_n)-f(T_n-)|+|f(T_n-)-f(0)|\leq n+C+n=2\,n+C \end{eqnarray}
So my bound is not as tight as it should be. It is not super important, but since I have found in a book the bound $n+C$ I was wondering how to get it (or maybe my reasoning is somehow flawed).
Bibliographic note. The book I refer to is Jacod and Shiryaev's and the claimed inequality appears in the proof of Lemma I.4.24.
Let $0\leq s<T_n$; thus $|f(s)-f(0)| < n \leq n+C$ and the supremum inequality holds.
It remains to show that $|f(T_n)-f(0)| \leq n+C$. As the left limit jumps are bounded we know $$|f(T_n)-f(0)| \leq |f(T_n)-f(T_n-)|+|f(T_n-)-f(0)| \\\leq C+|f(T_n-)-f(0)|.$$ To complete this, we need that $|f(T_n-)-f(0)|\leq n$. This follows by taking the limit of $|f(s)-f(0)| < n$ as $s$ tends to $T_n$ from the left.