Grönwall inequality for semimartingales

Question

I'm trying to prove the following version of the Grönwall inequality: Suppose that

$0 \leq A_t \leq \alpha + \int_0^t A_{s-}dC_s$

for a non-decreasing cadlag process $C$. Show that

$A_t \leq \alpha e^{C_t}$.

Proof attempt:

Inspired by other proofs of the Grönwall inequality, I'm trying to find an upper bound of $\int A_{s-}dC_s$ which does not depend on $A$, by utilizing the change of variables formula and integration by parts in a clever way. For instance, $e^{-C_t}$ satisfies by the change of variables formula (aka Ito formula)

$ e^{-C_t} = e^{-C_0} - \int_{0^+}^t e^{-C_{s-}}dC_s +\sum_{0 < s \leq t} \Big( \Delta e^{-C_s} + e^{-C_{s-}}\Delta C_s \Big).$

This gives, using integration by parts,

$e^{-C_t}\int_{0+}^t A_{s-}dC_s = \int_{0^+}^t e^{-C_{s-}}A_{s-}dC_s + \int_{0+}^t \int_{0+}^{s-} A_{u-}dC_u d(e^{C_s}) + [e^{-C},\int_{0+}^{\cdot} A_{s-}dC_s]_t.$

$= \int_{0^+}^t e^{C_{s-}}(A_{s-} - \int_{0+}^{s-} A_{u-}dC_u)dC_s + \sum_{0<s\leq t} \int_{0+}^{s-}A_{u-}dC_u \Big( \Delta e^{-C_s} + e^{-C_{s-}}\Delta C_s \Big) + [e^{-C},\int_{0+}^{\cdot} A_{s-}dC_s]_t.$

The first term on the rhs is bounded above by $\alpha\int_{0+}^t e^{-C_{s-}}dC_s$ by assumption, and therefore does not depend on $A$. However, I'm struggling to deal with the two other terms.. Any help or hints are greatly appreciated.

Edit: We should have $[e^{-C},\int_{0+}^{\cdot} A_{s-}dC_s] \leq 0$, since this is the quadratic covariation between a non-decreasing process and a non-increasing process, which should be non-increasing.

For reference, this is exercise 14, chapter V, p. 358 in Protter - Stochastic Integration and Differential Equations.

Answer 1

Yes, this is called "Stochastic Gronwall-type inequality" (see Lemma 2.1 "A new proof for comparison theorems for stochastic differential inequalities with respect to semimartingales", whose proof we alter.)

Let $N_t; U_t; B_t$ be one-dimensional processes such that $N$ is a semimartingale and, $B$ is a nondecreasing càdlàg process and for $t\geq s\geq 0$

$$U_{t}\leq N_{t}+\int_{s}^{t}U_{r}dB_{r}.$$

Then

$$U_{t}\leq(N_{s}e^{-B_{s}}+\int_{s}^{t}e^{-B_{r}}dN_{r}) e^{B_{t}}.$$

Proof

By Itô we have

$$N_t e^{-B_{t}}=N_{s}e^{-B_{s}}+\int_{s}^{t}e^{-B_{r}}dN_{r}-\int_{s}^{t}e^{-B_{r}}N_{r}dB_{r}$$

and

$$e^{-B_{t}}\int_{s}^{t}U_{r}dB_{r}=\int_{s}^{t}e^{-B_{r}}U_{r}dB_{r}-\int_{s}^{t}e^{-B_{r}}\left(\int_{s}^{r} U_{w}dB_{w}\right)dB_{r},$$

where the cross-variation term is zero because generally, the cross variation of a continuous semimartingale X and an adapted nonincreasing process Y is zero $[X,Y]=0$ (eg. see Quadratic Covariation of an Increasing Process with another Process is 0 where they do it for increasing but the proof is the same since we again get a telescoping sum)

Therefore,

$$U_t e^{-B_{t}}\leq N_{t}e^{-B_{t}}+e^{-B_{t}}\int_{s}^{t}U_{r}dB_{r}$$

$$=N_{s}e^{-B_{s}}+\int_{s}^{t}e^{-B_{r}}dN_{r}+\int_{s}^{t}e^{-B_{r}}\left( U_{r}-N_{r}-\int_{s}^{r} U_{w}dB_{w}\right)dB_{r}$$

the last term is nonpositive

$$\leq N_{s}e^{-B_{s}}+\int_{s}^{t}e^{-B_{r}}dN_{r}.$$

Therefore,

$$U_{t}\leq e^{B_t}\left(N_{s}e^{-B_{s}}+\int_{s}^{t}e^{-B_{r}}dN_{r} \right).$$

If $N_{t}\equiv \alpha$ and $s=0$, we get

$$U_{t}\leq \alpha e^{B_t}.$$