The Picard Map on Adapted Continuous Processes

From an SDE to an Operator

An SDE can be read as an equation for an entire process, rather than as a separate equation at each time. This viewpoint turns the equation into an operator: supply a candidate process to the drift and diffusion coefficients, integrate those coefficients, and obtain a new process. The operator’s natural setting records two requirements at once: the candidate must depend only on information available by the current time, and its paths must be continuous.

Fix a finite horizon \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying an \(m\)-dimensional Brownian motion \(\mathbf{B}_t\) relative to the filtration. Consider the \(d\)-dimensional SDE with initial value \(\xi\):

Here \(\mu:[0,T]\times\mathbb{R}^d\to\mathbb{R}^d\) is the drift coefficient, and \(\sigma:[0,T]\times\mathbb{R}^d\to\mathbb{R}^{d\times m}\) is the diffusion coefficient. The initial value \(\xi\) is \(\mathcal{F}_0\)-measurable. A solution on this stochastic basis is a continuous adapted process satisfying the corresponding integral equation, with the equality holding almost surely for all times (equivalently, indistinguishably when both sides have continuous paths).

The supremum is taken along each path before expectation. Thus this norm controls the whole trajectory in mean square; it is stronger than requiring only \(\mathbb{E}|X_t|^2<\infty\) separately at each fixed time. Adaptedness means \(X_t\) is \(\mathcal{F}_t\)-measurable for each \(t\). It encodes nonanticipation: the input at time \(t\) cannot use future Brownian information.

Definition and Domain of the Picard Map

For a candidate \(X\), substitute \(X_s\) into both coefficients and integrate from \(0\) to \(t\). The result is the Picard map, denoted by \(\Phi\). The drift integral is an ordinary time integral. The stochastic integral is the Ito integral, not the Stratonovich integral.

Progressive measurability is the condition that makes an integrand eligible for the Ito integral; it requires that its restriction to \([0,t]\times\Omega\) be measurable with respect to the product of the Borel sets on \([0,t]\) and \(\mathcal{F}_t\), for every \(t\). For Borel coefficient functions \(\mu,\sigma\) and continuous adapted \(X\), the coefficient processes are progressively measurable. To obtain a map into \(\mathcal{S}^2_T\), rather than merely a continuous adapted output, we also need a finite second moment of the output’s path supremum. The following result gives a convenient sufficient condition.

Proof. Borel measurability of the coefficients and progressive measurability of \(X\) make \(\mu(s,X_s)\) and \(\sigma(s,X_s)\) progressively measurable. Boundedness of \(\sigma\) gives \(\mathbb{E}\int_0^T\|\sigma(s,X_s)\|_F^2\,ds<\infty\), so the Ito integral exists for an integrand in the class \(\mathcal{L}^2\) of progressively measurable processes satisfying that condition. Boundedness of \(\mu\) makes its time integral finite. Write \(M_t=\int_0^t\sigma(s,X_s)\,d\mathbf{B}_s\). The Ito integral has a continuous adapted version. The standard Doob \(L^2\) maximal inequality, \(\mathbb{E}\sup_{t\le T}|M_t|^2\le4\mathbb{E}|M_T|^2\), and the Ito isometry imply \(\mathbb{E}\sup_{t\le T}|M_t|^2\le4\mathbb{E}\int_0^T\|\sigma(s,X_s)\|_F^2\,ds<\infty\). The maximal inequality is a standard result; its proof is not needed for the definition of the map. If \(|\mu|\le C_\mu\), then \(\sup_{t\le T}|\int_0^t\mu(s,X_s)\,ds|^2\le T^2C_\mu^2\). Finally, \(|a+b+c|^2\le3(|a|^2+|b|^2+|c|^2)\) bounds the squared supremum of \(\Phi(X)\) by three times the sum of the corresponding squared suprema for \(\xi\), the drift integral, and \(M\). Each has finite expectation, so \(\Phi(X)\in\mathcal{S}^2_T(\mathbb{R}^d)\). \(\square\)

Bounded coefficients are a sufficient condition for this proposition, not a requirement in the definition of the operator. More generally, one can define \(\Phi(X)\) whenever the coefficient processes have the required measurability and integrability and the resulting output belongs to the chosen process space. The domain can therefore be smaller than all of \(\mathcal{S}^2_T\). Conditions that guarantee the map is defined for every input are useful, but they do not by themselves imply that the map has a fixed point.

What Fixed Points Mean

The operator organizes the integral equation into the form \(\Phi(X)=X\). This gives an exact correspondence, provided the integrals exist and equalities of processes are understood up to indistinguishability.

Proof. If \(X\) is a fixed point, then, for every \(t\in[0,T]\), \(X_t=\Phi(X)_t=\xi+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,d\mathbf{B}_s\), with the equality holding indistinguishably. This is the integral form of the SDE. The process is adapted to the given filtration and uses the Brownian motion on the given basis, so it is a strong solution. Conversely, if \(X\) is a strong solution, its defining integral equation is precisely \(X_t=\Phi(X)_t\) for every \(t\), up to indistinguishability. Hence \(X\) is a fixed point. \(\square\)

This proposition does not assert that a fixed point exists or is unique. Those questions require additional arguments about the coefficients and the operator. In particular, being well defined on a process space is not the same as being a contraction in its norm.

Worked Examples

Common Mistake: Forgetting the Time Integrals

A common error is to replace the operator by the coefficients evaluated at the input, without integrating them. For the constant-coefficient example \(\xi=1\), \(\mu=2\), \(\sigma=3\), the incorrect expression \(1+2+3B_t\) treats the drift as if it accumulated only once, regardless of the time \(t\). At \(t=2\), it gives \(3+3B_2\), while the correct expression is \(1+\int_0^2 2\,ds+\int_0^2 3\,dB_s=5+3B_2\). The correction is to integrate the drift over elapsed time and the diffusion against Brownian motion.

Another distinction matters when analyzing this construction: the norm \(\|X\|_{\mathcal{S}^2_T}\) measures the input process, whereas the map \(\Phi\) is determined by the coefficients and integrals. Even if \(\Phi\) sends an admissible input to an admissible output, it need not satisfy a contraction estimate such as \(\|\Phi(X)-\Phi(Y)\|_{\mathcal{S}^2_T}\le c\|X-Y\|_{\mathcal{S}^2_T}\) with \(c<1\). Such an estimate is a separate property, not part of the definition of the Picard map.