Let
\(\mathbb{V}\) be the set of
\({\mathbb{F}}^{W^{0}}\)-adapted, square-integrable semimartingales on
\([0, T]\). Thus for any
\((V_{t})_{0 \leq t \leq T} \in \mathbb{V}\), there exist two
\({\mathbb{F}}^{W^{0}}\)-adapted and square-integrable processes
\((V_{1,t})_{0 \leq t \leq T}\) and
\((V_{2,t})_{0 \leq t \leq T}\) such that
$$\begin{aligned} V_{t} = V_{0} + \int _{0}^{t}v_{1,s}\,ds + \int _{0}^{t}v_{2,s}\,dW_{s}^{0} \end{aligned}$$
(4.7)
for
\(0 \leq t \leq T\). The processes of the above form for which
\((v_{1,t})_{0 \leq t \leq T}\) and
\((v_{2,t})_{0 \leq t \leq T}\) are simple processes, that is,
$$\begin{aligned} v_{i,t} = F_{i}\mathbb{I}_{ [t_{1}, t_{2} ]}(t) \end{aligned}$$
(4.8)
for all
\(0 \leq t \leq T\) and
\(i \in \{1, 2\}\), with each
\(F_{i}\) being
\(\mathcal{F}_{t_{1}}^{W^{0}}\)-measurable, span a linear subspace
\(\tilde{\mathbb{V}}\) which is dense in
\(\mathbb{V}\) for the
\(L^{2}\)-norm. By using the boundedness of
\(h\) and then the estimate (
2.3), for any
\(p > 0\) and any
\(T > 0\), we obtain
$$ \int _{0}^{T}\big\Vert h^{p} (\sigma _{\frac{t}{\epsilon }}^{1,1} )u^{ \epsilon }(t, \cdot )\big\Vert _{L_{\sigma , \mathcal{C}}^{2}( \mathbb{R}_{+}\times \Omega )}^{2}\,dt \leq TC^{2p}\left \Vert u_{0} \right \Vert _{L^{2}(\mathbb{R}_{+})}^{2}. $$
(4.9)
It follows that any sequence
\(\epsilon _{n} \rightarrow 0{+}\) always has a subsequence
\((\epsilon _{k_{n}})_{n \in \mathbb{N}}\) such that
\(h^{p}(\sigma _{\frac{\cdot }{\epsilon }}^{1,1})u^{\epsilon _{k_{n}}}( \cdot , \cdot )\) converges weakly to some
\(u_{p}(\cdot , \cdot )\) in
\(L_{\sigma , \mathcal{C}}^{2}([0, T] \times \mathbb{R}_{+} \times \Omega )\) for
\(p \in \{1, 2 \}\). Testing (
2.2) against an arbitrary smooth and compactly supported function
\(f : \mathbb{R}_{+} \to \mathbb{R}\), using Itô’s formula for the product of
\(\int _{\mathbb{R}_{+}}u^{\epsilon }(\cdot , x)f(x)\,dx\) with a process
\(V \in \tilde{\mathbb{V}}\) having the form (
4.7), (
4.8) and finally taking expectations, we find
$$\begin{aligned} & \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{t}\int _{\mathbb{R}_{+}}u^{ \epsilon }(t, x)f(x)\,dx\right ] \\ & = \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{0}\int _{\mathbb{R}_{+}}u_{0}(x)f(x) \,dx\right ] + r\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s} \int _{\mathbb{R}_{+}}u^{\epsilon }(s, x)f'(x)\,dx\right ]\,ds \\ & \phantom{=:} - \int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}} \bigg[V_{s}\int _{ \mathbb{R}_{+}}\frac{h^{2} (\sigma _{\frac{s}{\epsilon }}^{1,1} )}{2}u^{ \epsilon }(s, x)f'(x)\,dx\bigg]\,ds \\ & \phantom{=:} +\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\bigg[V_{s}\int _{ \mathbb{R}_{+}}\frac{h^{2} (\sigma _{\frac{s}{\epsilon }}^{1,1} )}{2} u^{ \epsilon }(s, x)f''(x)\,dx\bigg]\,ds \\ &\phantom{=:}+\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\bigg[v_{1,s}\int _{ \mathbb{R}_{+}} u^{\epsilon }(s, x)f(x)\,dx\bigg]\,ds \\ & \phantom{=:} + \rho _{1,1}\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}} \bigg[v_{2,s} \int _{\mathbb{R}_{+}}h (\sigma _{\frac{s}{\epsilon }}^{1,1} )u^{ \epsilon }(s, x)f'(x)\,dx \bigg]\,ds \end{aligned}$$
(4.10)
for all
\(t \leq T\). Upon setting
\(\epsilon = \epsilon _{k_{n}}\) and taking
\(n \rightarrow \infty \), the weak convergence results above yield
$$\begin{aligned} & \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{t}\int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\right ] \\ & = \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{0}\int _{\mathbb{R}_{+}}u_{0}(x)f(x) \,dx\right ] + r\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s} \int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x)\,dx\right ]\,ds \\ & \phantom{=:} - \frac{1}{2}\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s} \int _{\mathbb{R}_{+}}u_{2}(s, x)f'(x)\,dx\right ]\,ds \\ & \phantom{=:} + \frac{1}{2}\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s} \int _{\mathbb{R}_{+}} u_{2}(s, x)f''(x)\,dx\right ]\,ds \\ &\phantom{=:} +\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [v_{1,s}\int _{ \mathbb{R}_{+}} u^{*}(s, x)f(x)\,dx\right ]\,ds \\ & \phantom{=:}+ \rho _{1,1}\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [v_{2,s} \int _{\mathbb{R}_{+}}u_{1}(s, x)f'(x)\,dx\right ]\,ds \end{aligned}$$
(4.11)
for all
\(0 \leq t \leq T\). The convergence of the terms on the right-hand side of (
4.10) holds pointwise in
\(t\), while the term on the left-hand side converges weakly. Since we can easily find uniform bounds for all terms in (
4.10) (by using (
4.9)), dominated convergence implies that all the weak limits coincide with the corresponding pointwise limits, which gives (
4.11) as a limit of (
4.10) both weakly and pointwise in
\(t\). It is clear then that
\(\mathbb{E}_{\sigma , \mathcal{C}}[V_{t}\int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx]\) is differentiable in
\(t\) (in a
\(W^{1,1}\)-sense). Next, we can check that the expectation
\(\mathbb{E}_{\sigma , \mathcal{C}}[v_{i,t}\int _{\mathbb{R}_{+}}u^{ \epsilon _{k_{n}}}(t, x)f(x)\,dx]\) converges to
\(\mathbb{E}_{\sigma , \mathcal{C}}[v_{i,t}\int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx]\) for both
\(i = 1\) and
\(i = 2\), both weakly and pointwise in
\(t \in [0, T]\), while the limits are also differentiable in
\(t\) everywhere except in the two jump points
\(t_{1}\) and
\(t_{2}\). This follows because everything is zero outside
\([t_{1}, t_{2}]\), while both
\(v_{1}\) and
\(v_{2}\) are constant in
\(t\) and thus of the form (
4.7), (
4.8) if we restrict to that interval. Subtracting from each term of (
4.10) the same term but with
\(u^{\epsilon }\) replaced by
\(u^{*}\) and then adding it back, we can rewrite this identity as
$$\begin{aligned} & \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{t}\int _{\mathbb{R}_{+}}u^{ \epsilon }(t, x)f(x)\,dx\right ] \\ & = \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{0}\int _{\mathbb{R}_{+}}u_{0}(x)f(x) \,dx\right ] + r\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s} \int _{\mathbb{R}_{+}}u^{\epsilon }(s, x)f'(x)\,dx\right ]\,ds \\ & \phantom{=:} - \int _{0}^{t}\frac{h^{2}(\sigma _{\frac{s}{\epsilon }}^{1,1})}{2} \bigg(\mathbb{E}_{\sigma , \mathcal{C}}\bigg[V_{s}\int _{\mathbb{R}_{+}}u^{ \epsilon }(s, x)f'(x)\,dx\bigg] \\ & \phantom{=:} \qquad \qquad \qquad \quad - \mathbb{E}_{\sigma , \mathcal{C}} \bigg[V_{s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x)\,dx\bigg]\bigg) \,ds \\ & \phantom{=:} - \int _{0}^{t}\frac{h^{2}(\sigma _{\frac{s}{\epsilon }}^{1,1})}{2} \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x)\,dx\right ]\,ds \\ &\phantom{=:} + \int _{0}^{t}\frac{h^{2} (\sigma _{\frac{s}{\epsilon }}^{1,1} )}{2} \bigg(\mathbb{E}_{\sigma , \mathcal{C}}\bigg[V_{s}\int _{\mathbb{R}_{+}}u^{ \epsilon }(s, x)f''(x)\,dx\bigg] \\ & \phantom{=:} \qquad \qquad \qquad \quad - \mathbb{E}_{\sigma , \mathcal{C}} \bigg[V_{s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f''(x)\,dx\bigg]\bigg) \,ds \\ & \phantom{=:} + \int _{0}^{t}\frac{h^{2} (\sigma _{\frac{s}{\epsilon }}^{1,1} )}{2} \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f''(x)\,dx\right ]\,ds \\ & \phantom{=:} +\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [v_{1,s}\int _{ \mathbb{R}_{+}} u^{\epsilon }(s, x)f(x)\,dx\right ]\,ds \\ & \phantom{=:} + \rho _{1,1}\int _{0}^{t}h (\sigma _{\frac{s}{\epsilon }}^{1,1} ) \bigg(\mathbb{E}_{\sigma , \mathcal{C}}\bigg[v_{2,s}\int _{ \mathbb{R}_{+}}u^{\epsilon }(s, x)f'(x)\,dx\bigg] \\ & \phantom{=:} \qquad \qquad \qquad \qquad \,\, - \mathbb{E}_{\sigma , \mathcal{C}}\bigg[v_{2,s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x)\,dx \bigg]\bigg)\,ds \\ &\phantom{=:} + \rho _{1,1}\int _{0}^{t}h (\sigma _{\frac{s}{\epsilon }}^{1,1} ) \mathbb{E}_{\sigma , \mathcal{C}}\left [v_{2,s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x)\,dx\right ]\,ds. \end{aligned}$$
(4.12)
Then we have
$$\begin{aligned} & \bigg|\int _{0}^{t}h (\sigma _{\frac{s}{\epsilon }}^{1,1} )\bigg(\mathbb{E}_{\sigma , \mathcal{C}}\left [v_{2,s}\int _{ \mathbb{R}_{+}}u^{\epsilon }(s, x)f'(x)\,dx\right ] \\ & \qquad \qquad \qquad - \mathbb{E}_{\sigma , \mathcal{C}}\bigg[v_{2,s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x)\,dx \bigg]\bigg)\,ds \bigg| \\ & \leq C\int _{0}^{t}\bigg|\mathbb{E}_{\sigma , \mathcal{C}}\left [v_{2,s} \int _{\mathbb{R}_{+}}u^{\epsilon }(s, x)f'(x)\,dx\right ] \\ & \qquad\quad \quad \! - \mathbb{E}_{\sigma , \mathcal{C}}\bigg[v_{2,s} \int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x)\,dx\bigg]\bigg|\,ds, \end{aligned}$$
which tends to zero (when
\(\epsilon = \epsilon _{k_{n}}\) and
\(n \rightarrow \infty \)) by dominated convergence, since the quantity inside the last integral converges to zero pointwise and can be dominated by using (
4.9). The same argument is used to show that the fourth and sixth terms in (
4.12) also tend to zero along the same subsequence. Finally, for any term of the form
$$\begin{aligned} \int _{0}^{t}h^{p} (\sigma _{\frac{s}{\epsilon }}^{1,1} )\mathbb{E}_{ \sigma , \mathcal{C}}\left [V_{s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f^{(m)}(x) \,dx\right ]\,ds \end{aligned}$$
for
\(p, m \in \{0, 1, 2\}\), we recall the differentiability of the second factor inside the
\(ds\)-integral (which was mentioned earlier) and then integrate by parts to write it as
$$\begin{aligned} & \int _{0}^{t}h^{p} (\sigma _{\frac{w}{\epsilon }}^{1,1} )\,dw \bigg(\mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s}\int _{\mathbb{R}_{+}}u^{*}(t, x)f^{(m)}(x)\,dx\right ]\bigg) \\ & -\int _{0}^{t}\int _{0}^{s}h^{p} (\sigma _{\frac{w}{\epsilon }}^{1,1} )\,dw\bigg(\mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s}\int _{ \mathbb{R}_{+}}u^{*}(s, x)f^{(m)}(x)\,dx\right ]\bigg)'ds \end{aligned}$$
which converges by the positive recurrence property to the quantity
$$\begin{aligned} & t\mathbb{E} [h^{p} (\sigma ^{1,1,1,*} ) \, | \, \mathcal{C} ] \bigg( \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s}\int _{\mathbb{R}_{+}}u^{*}(t, x)f^{(m)}(x)\,dx\right ]\bigg) \\ & -\int _{0}^{t}s\mathbb{E} [h^{p} (\sigma ^{1,1,1,*} ) \, | \, \mathcal{C} ]\bigg(\mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s}\int _{ \mathbb{R}_{+}}u^{*}(s, x)f^{(m)}(x)\,dx\right ]\bigg)'ds. \end{aligned}$$
Using integration by parts once more, this last expression is equal to
$$\begin{aligned} \mathbb{E} [h^{p} (\sigma ^{1,1,1,*} ) \, | \, \mathcal{C} ]\int _{0}^{t} \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f^{(m)}(x)\,dx\right ]\,ds. \end{aligned}$$
This last convergence result also holds if we replace
\(V\) by
\(v_{1}\) or
\(v_{2}\), as we can show by following exactly the same steps in the subinterval
\([t_{1}, t_{2}]\) (where
\(v_{i}\) is supported for
\(i \in \{1, 2\}\) and where we have differentiability that allows integration by parts).
If we now set
\(\epsilon = \epsilon _{k_{n}}\) in (
4.12), take
\(n \rightarrow \infty \) and substitute all the above convergence results, we obtain
$$\begin{aligned} & \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{t}\int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\right ] \\ & = \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{0}\int _{\mathbb{R}_{+}}u_{0}(x)f(x) \,dx\right ] \\ & \phantom{=:}+ \bigg(r-\frac{\sigma _{2,1}^{2}}{2}\bigg)\int _{0}^{t}\mathbb{E}_{ \sigma , \mathcal{C}}\left [V_{s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x) \,dx\right ]\,ds \\ & \phantom{=:} + \frac{\sigma _{2,1}^{2}}{2}\int _{0}^{t} \mathbb{E}_{\sigma , \mathcal{C}}\left [V_{s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f''(x)\,dx \right ]\,ds \\ & \phantom{=:} +\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [v_{1,s}\int _{ \mathbb{R}_{+}} u^{*}(s, x)f(x)\,dx\right ]\,ds \\ &\phantom{=:} +\rho _{1,1}\sigma _{1,1}\int _{0}^{t}\mathbb{E}_{\sigma , \mathcal{C}}\left [v_{2,s}\int _{\mathbb{R}_{+}}u^{*}(s, x)f'(x)\,dx \right ]\,ds. \end{aligned}$$
(4.13)
Since
\(\tilde{\mathbb{V}}\) is dense in
\(\mathbb{V}\), for a fixed
\(t \leq T\), we can have (
4.13) for any square-integrable martingale
\((V_{s})_{0 \leq s \leq t}\), for which we have
\(v_{1,s} = 0\) for all
\(0 \leq s \leq t\). Next, we denote by
\(R_{u}(t, x)\) the right-hand side of (
2.4). Using then Itô’s formula for the product of
\(\int _{\mathbb{R}_{+}}R_{u}(s, x)f(x)\,dx\) with
\(V_{s}\) at
\(s = t\), subtracting
\(V_{t}\int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\) from both sides, taking expectations and finally substituting from (
4.13), we find that
$$\begin{aligned} \mathbb{E}_{\sigma , \mathcal{C}}\bigg[V_{t}\left (\int _{\mathbb{R}_{+}}R_{u}(t, x)f(x)\,dx - \int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\right )\bigg] = 0 \end{aligned}$$
for our fixed
\(t \leq T\). Using the martingale representation theorem,
\(V_{s}\) can be taken equal to
\(\mathbb{E}_{\sigma , \mathcal{C}} [\mathbb{I}_{\mathcal{E}_{s}} \, | \, \sigma (W_{s'}^{0}, s' \leq s ) ]\) for all
\(s \leq t\), where we define
$$\begin{aligned} \mathcal{E}_{t} = \left \{ \omega \in \Omega : \int _{\mathbb{R}_{+}}R_{u}(t, x)f(x)\,dx > \int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\right \} , \end{aligned}$$
and this implies
\(V_{t} = \mathbb{I}_{\mathcal{E}_{t}}\), allowing us to write
$$\begin{aligned} \mathbb{E}_{\sigma , \mathcal{C}}\left [\mathbb{I}_{\mathcal{E}_{t}} \left (\int _{\mathbb{R}_{+}}R_{u}(t, x)f(x)\,dx - \int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\right )\right ] = 0 \end{aligned}$$
for any
\(0 \leq t \leq T\). If we integrate the above over
\(t \in [0, T]\), we obtain that
$$\begin{aligned} \int _{0}^{T}\mathbb{E}_{\sigma , \mathcal{C}}\bigg[\mathbb{I}_{ \mathcal{E}_{t}}\left (\int _{\mathbb{R}_{+}}R_{u}(t, x)f(x)\,dx - \int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\right )\bigg]\,dt = 0, \end{aligned}$$
where the quantity inside the expectation is always nonnegative and becomes zero only when
\(\mathbb{I}_{\mathcal{E}_{t}} = 0\). This implies
\(\int _{\mathbb{R}_{+}}R_{u}(t, x)f(x)\,dx \leq \int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\) almost everywhere, and working in the same way with the indicator of the complement
\(\mathbb{I}_{\mathcal{E}_{t}^{c}}\), we can deduce the opposite inequality as well. Thus we must have
\(\int _{\mathbb{R}_{+}}R_{u}(t, x)f(x)\,dx = \int _{\mathbb{R}_{+}}u^{*}(t, x)f(x)\,dx\) almost everywhere, and since the function
\(f\) is an arbitrary smooth function with compact support, we can deduce that
\(R_{u}\) coincides with
\(u^{*}\) almost everywhere, which gives (
2.4).