A (classical) bound on the jumps of a cadlag function: $\left|\Delta f(t)\right|\leq 2\,\sup_{0\leq s\leq t}\left|f(s)\right|$.

Question

For a càdlàg function $f:[0,\infty)\rightarrow\mathbb{R}$ I need to prove the following intuitive bound

$$ \left|\Delta f(t)\right|=\left|f(t)-f(t-)\right|\leq 2\,\sup_{0\leq s\leq t}\left|f(s)\right| $$

The idea is that

$$ \left|\Delta f(t)\right|\leq \left|f(t)\right|+\left|f(t-)\right|\leq \sup_{0\leq s\leq t}\left|f(s)\right|+\sup_{0\leq s\leq t}\left|f(s-)\right| $$

But I miss to prove that $$\sup_{0\leq s\leq t}\left|f(s-)\right|=\sup_{0\leq s\leq t}\left|f(s)\right|$$

Any suggestion is welcome.

Answer 1

For $0 \le \tau < t$ is $$ \begin{align} |f(t-)| &\le |f(t-) - f(\tau)| + |f(\tau)| \\ &\le \underbrace{|f(t-) - f(\tau)|}_{\to 0 \text{ for } \tau \nearrow t } + \sup_{0\leq s\leq t}|f(s)| \end{align} $$ so that $$ |f(t-)| \le \sup_{0\leq s\leq t}|f(s)| \, . $$ Also, obviously, $$ |f(t)| \le \sup_{0\leq s\leq t}|f(s)| \, , $$ and therefore $$ |\Delta f(t)| \le |f(t)| + |f(t-)| \le 2 \sup_{0\leq s\leq t}|f(s)| \, . $$

Remark: We used only the existence of the left limit, not the right continuity.

Answer 2

For cadlag (right-continuous with left limits) $f$, you can show:

$$|Δf(t)| = |f(t) - f(t-)| ≤ 2\sup_{ 0≤s≤t} |f(s)|$$

Proof:

Since $f$ is cadlag, $f(t-) = f(t)$ for all $t$. So $$ \begin{split}\sup_{0≤s≤t} |f(s-)| = \sup_{0≤s≤t} |f(s)|&\qquad \text{(swap suprema)}\\ |Δf(t)| ≤ |f(t)| + |f(t-)| ≤ 2 \sup_{0≤s≤t} |f(s)|&\qquad\text{(use step 2)} \end{split}$$

The key point is that for cadlag functions, the left and right limits agree at each point, so you can swap the two suprema. The end result is that the difference between left/right limits is bounded by twice the maximum value on the given subinterval.