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 RiemannStieltjes 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

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