Lecture 3. Riemann-Stieltjes integrals

Let $y:[0,T] \to \mathbb{R}^{e \times d}$ be a piecewise continuous path and $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ . It is well-known that we can integrate $y$ against $x$ by using the Riemann–Stieltjes integral which is a natural extension of the Riemann integral. The idea is to use the Riemann sums
$\sum_{k=0}^{n-1} y(t_k) (x(t_{k+1})-x(t_k)),$
where $\Pi=\{ 0 =t_0 < t_1 < \cdots < t_n =T \}$ . It is easy to prove that, when the mesh of the subdivision $\Pi$ goes to 0, the Riemann sums converge to a limit which is independent from the sequence of subdivisions that was chosen. The limit is then denoted $\int_0^T y(t) dx(t)$ and called the Riemann-Stieltjes integral of $y$ against $x$ . Since $x$ has a bounded variation, it is easy to see that, more generally,
$\sum_{k=0}^{n-1} y(\xi_k) (x(t_{k+1})-x(t_k)),$
with $t_k \le \xi_k \le t_{k+1}$ would also converge to $\int_0^T y(t) dx(t)$ . If
$x(t)=x(0)+\int_0^t g(s) ds$
is an absolutely continuous path, then it is not difficult to prove that we have
$\int_0^T y(t) dx(t) =\int_0^T y(t) g(t) dt,$
where the integral on the right hand side is understood in Riemann’s sense.

We have
$\left\| \sum_{k=0}^{n-1} y(t_k) (x(t_{k+1})-x(t_k))\right\|$
$\le \sum_{k=0}^{n-1} \| y(t_k)\| \| (x(t_{k+1})-x(t_k))\|$
$\le \sum_{k=0}^{n-1} \| y(t_k)\| \| (x(t_{k+1})-x(t_k))\|$
$\le \sum_{k=0}^{n-1} \| y(t_k)\| \| x \|_{1-var,[t_k,t_{k+1}]}.$
Thus, by taking the limit when the mesh of the subdivision goes to 0, we obtain the estimate
$\left\| \int_0^T y(t) dx(t) \right\| \le \int_0^T \| y(t) \| \| dx(t) \| \le \| y \|_{\infty, [0,T]} \| x \|_{1-var,[0,T]},$
where $\int_0^T \| y(t) \| \| dx(t) \|$ is the notation for the Riemann-Stieltjes integral of $\| y \|$ against the bounded variation path $l(t)= \| x \|_{1-var,[0,t]}$ . We can also estimate the Riemann-Stieltjes integral in the 1-variation distance. We collect the following estimate for later use:

Proposition: Let $y,y':[0,T] \to \mathbb{R}^{e \times d}$ be a piecewise continuous path and $x,x' \in C^{1-var} ([0,T], \mathbb{R}^d)$ . We have
$\left\| \int_0^{\cdot} y'(t) dx'(t)-\int_0^{\cdot} y(t) dx(t) \right\|_{1-var,[0,T]} \le \| x \|_{1-var,[0,T]} \| y-y' \|_{\infty, [0,T]} + \| y' \|_{\infty, [0,T]} \| x -x'\|_{1-var,[0,T]}.$

The Riemann-Stieltjes satisfies the usual rules of calculus, for instance the integration by parts formula takes the following form
Proposition: Let $y \in C^{1-var} ([0,T], \mathbb{R}^{e \times d} )$ and $x\in C^{1-var} ([0,T], \mathbb{R}^d)$ .
$\int_0^T y(t) dx(t)+\int_0^T dy(t) x(t)=y(T)x(T) -y(0)x(0).$

We also have the following change of variable formula:

Proposition: Let $x\in C^{1-var} ([0,T], \mathbb{R}^d)$ and let $\Phi: \mathbb{R}^d \to \mathbb{R}^e$ be a $C^1$ map. We have
$\Phi (x(T)) =\Phi (x(0)) + \int_0^T \Phi'(x(t)) dx(t).$

Proof: From the mean value theorem
$\Phi (x(T)) -\Phi (x(0))=\sum_{k=0}^{n-1} (\Phi (x(t_{k+1})) -\Phi (x(t_k)))=\sum_{k=0}^{n-1}\Phi'(x_{\xi_k}) (x(t_{k+1}) -x(t_k)),$
with $t_k \le \xi_k \le t_{k+1}$ . The result is then obtained by taking the limit when the mesh of the subdivision goes to 0 $\square$

We finally state a classical analysis lemma, Gronwall’s lemma, which provides a wonderful tool to estimate solutions of differential equations.

Proposition: Let $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ and let $\Phi: [0,T] \to [0,\infty)$ be a bounded measurable function. If,
$\Phi(t) \le A+B\int_0^t \Phi(s) \| d x(s)\|, \quad 0 \le t \le T,$
for some $A,B \ge 0$ , then
$\Phi(t) \le A \exp (B \| x \|_{1-var,[0,t]} )\quad 0 \le t \le T.$

Proof: Iterating the inequality
$\Phi(t) \le A+B\int_0^t \Phi(s) \| d x(s)\|$
$N$ times, we get
$\Phi(t) \le A+\sum_{k=1} ^n AB^{k} \int_0^ t \int_0^{t_1} \cdots \int_0^{t_{k-1}} \| d x(t_k)\| \cdots \| dx(t_1) \| +R_n(t),$
where $R_n(t)$ is a remainder term that goes to 0 when $n \to \infty$ . Observing that
$\int_0^ t \int_0^{t_1} \cdots \int_0^{t_{k-1}} \| d x(t_k)\| \cdots \| dx(t_1) \|=\frac{ \| x \|^k_{1-var,[0,t]} }{k!}$
and sending $n$ to $\infty$ finishes the proof $\square$