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Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a càdlàg function (i.e. everywhere right-continuous and has left limits everywhere). Then is it true it can be decomposed as $$f(t)=f^c(t)+\sum_{0<s\leq t}\Delta f(s)$$ where $f^c(t)$ is continuous with $f^c(0)=f(0)$ and the jumps $\Delta f(s)=f(s)-f(s-)$.

Amira
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    Is it true if it has bounded variation on each bounded interval. Otherwise, the sum of the jumps does not make sense. – Christophe Leuridan Feb 22 '23 at 13:55
  • @ChristopheLeuridan It’s certainly not obvious to me how you can get an example where the jumps fail to be absolutely summable! Failure to have bounded variation can easily come from just the continuous part (e.g. typical sample path of Brownian motion). – Julian Newman Jul 31 '23 at 13:45
  • @ChristopheLeuridan Okay, I see that an example where the jumps fail to be absolutely summable is when the graph of $f(t)$ on $[0,1)$ consists of an infinite sequence of flat steps whose heights are the partial sums of $\sum_n \frac{(-1)^n}{n}$. Still, I wonder whether for every cadlag function, the jumps can still be considered summable under some appropriate weaker notion of summability than absolute summability; see my question https://mathoverflow.net/questions/451903/ – Julian Newman Aug 01 '23 at 21:45

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A decomposition of the kind that you're asking for will not be possible for general càdlàg functions, as I will show further below.

Positive answer for the bounded-variation case:

As in Christophe Leuridan's comment, the setting in which you naturally get a decomposability result of the type that you're after is càdlàg functions of locally bounded variation. In this case, a sketch-proof of the decomposition is as follows:

By combining the answers to Bounded variation, difference of two increasing functions and Total variation of a cadlag function, you can write $f = F_1-F_2$ where $F_1$ and $F_2$ are nondecreasing càdlàg functions. For $i=1,2$, we have that $$ \sum_{0<s\leq t} \Delta F_i(s) \leq F_i(t)-F_i(0) < \infty, $$ and letting $$ F_i^c(t) = F_i(t) - \sum_{0<s\leq t} \Delta F_i(s), $$ we have that for any $[a,b] \subset [0,\infty)$, \begin{align*} |F_i^c(b) - F_i^c(a)| &= \left|F_i(b) - F_i(a) - \sum_{a < t \leq b} \Delta F_i(s) \right| \\ &= \left|F_i(b-) - F_i(a) - \sum_{a < t < b} \Delta F_i(s) \right| \\ &\leq F_i(b-)-F_i(a), \end{align*} implying that $F_i^c$ is continuous.

Counter-example for the general case:

For every integer $n \geq 1$, let $H_n$ denote the $n$-th harmonic number $H_n := \sum_{k=1}^n \frac{1}{k}$. Given any decreasing homeomorphism $h \colon [0,1) \to (-\infty,0]$, we can define a càdlàg function $f \colon [0,\infty) \to [0,1]$ by $$ f(x) = \begin{cases} h(x) + \min\!\big( \{H_n\}_{n \geq 1} \cap (-h(x),\infty) \big) & x \in [0,1) \\ 0 & x \in [1,\infty). \end{cases} $$ The set of discontinuity points of this function can be enumerated as an increasing sequence $0<s_2<s_3<s_4<\ldots$ tending to $1$; and if we also define $s_1:=0$, then we have the following:

  • for all $n \geq 1$, $f$ is continuous and strictly decreasing on $[s_n,s_{n+1})$ and maps $[s_n,s_{n+1})$ onto $(0,\frac{1}{n}]$.

Hence, on $[0,1)$, the decomposition you ask for does exist; the continuous part $f^c$ is given by $f^c(x)=h(x)+1$. However, $f^c(x) \to -\infty$ as $x \nearrow 1$, so the decomposition cannot be extended beyond $[0,1)$.

Remark:

Càdlàg functions can, in fact, even get far crazier than the above. For example, as in the answer to https://mathoverflow.net/questions/451903/, it's possible to have a a càdlàg function $f \colon [0,\infty) \to \mathbb{R}$ such that in every interval of strictly positive width, the sum of the positive jumps is infinite and the sum of the negative jumps is infinite. And, slightly modifying that example, it's possible to have a càdlàg function $f \colon [0,\infty) \to \mathbb{R}$ such that all the jumps are positive and in every interval of strictly positive width, the sum of the jumps is infinite.

Julian Newman
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