Persistence of $C^1$ Inertial Manifolds Under Small Random Perturbations

Abstract

In this paper, we study a class of semilinear parabolic equation and its perturbed system driven by a random force. Such driving noise is assumed to be a regular approximation to the white noise and satisfy certain properties. We show that the $C^1$ inertial manifold structure is persisted under such perturbation in the sense that inertial manifolds of the perturbed system are converging to those of the original system as the perturbation tends to zero.

Appendix: Some Proofs

We give two proofs related to the hypotheses on the noise given in the Introduction. First we show that (1.8)

$$\begin{aligned} \lim _{t \rightarrow -\infty } \frac{1}{t} \Big [\int _0^t \zeta _\delta (\theta _s \omega ) d s-\omega (t) \Big ]=0~\text{ uniformly } \text{ for }~\delta \in (0,1] \end{aligned}$$

holds for $\zeta _\delta (\theta _t\omega )={\mathcal {G}}_{\delta }(\theta _t \omega )=\frac{1}{\delta }(\omega (t+\delta )-\omega (t))$.

Proof

We observe that

$$\begin{aligned} \left| \int _{0}^{t} {\mathcal {G}}_{\delta }(\theta _s \omega ) d s -\omega (t)\right| \le \left| \int _{0}^{\delta } \frac{\omega (s)}{\delta } d s \right| +\left| \int _{t}^{t+\delta } \frac{\omega (s)-\omega (t)}{\delta } d s \right| . \end{aligned}$$

It then follows that for all $\delta \in (0,1]$,

$$\begin{aligned} \begin{aligned} \left| \frac{1}{t}\Big [\int _{0}^{t} {\mathcal {G}}_{\delta }(\theta _s \omega ) d s -\omega (t) \Big ]\right| \le&\,\, \frac{1}{|t|}\left| \int _{0}^{\delta } \frac{\omega (s)}{\delta } d s \right| +\frac{|\omega (t^*)|}{|t|} \\ \le&\,\, \frac{1}{|t|}\sup _{s \in [0,1]}{|\omega (s)|} +\frac{|\omega (t^*)|}{|t|}, \end{aligned} \end{aligned}$$

where $t^*=t^*(\omega ) \in (t, t+\delta )$. Along with (1.4), for any $\zeta >0$, there exists a $T_1=T_1(\omega )>0$ such that for every $t<-T_1$, we have that

$$\begin{aligned} \Big |\frac{\omega (t)}{t} \Big | < \frac{\zeta }{4}, \end{aligned}$$

which yields that for all $t \le -T_1-2$ and $\delta \in (0,1]$, we have that

$$\begin{aligned} \Big |\frac{\omega (t^*)}{t^*} \Big | < \frac{\zeta }{4}. \end{aligned}$$

On the other hand, there exists a $T_2=T_2(\omega )>1$ such that for all $t \le -T_2$ and $\delta \in (0,1]$,

$$\begin{aligned} \frac{1}{|t|}\sup _{s \in [0,1]}{|\omega (s)|}<\frac{\zeta }{2},\,\,\, \frac{|t^*|}{|t|} \le \frac{|t|+1}{|t|}<2. \end{aligned}$$

Taking $T_3=\max \{T_1+2, T_2\}$, for $t \le -T_3$ and $\delta \in (0,1]$ we get that

$$\begin{aligned} \begin{aligned} \left| \frac{1}{t}\Big [\int _{0}^{t} {\mathcal {G}}_{\delta }(\theta _s \omega ) d s -\omega (t) \Big ]\right| <\zeta , \end{aligned} \end{aligned}$$

which implies the desired result. $\square$

Secondly, based on the fact that there exists a $\theta _t$-invariant subset of full measure $\varOmega _1$ such that

$$\begin{aligned} \lim _{t \rightarrow \pm \infty } \frac{\omega (t)}{t}=0,~\forall \omega \in \varOmega _1, \end{aligned}$$

(A.1)

we justify the properties of the stationary solution to the random differential equation

$$\begin{aligned} {\dot{x}} =-x + \varPhi _\delta (\theta _t \omega ). \end{aligned}$$

(A.2)

As we stated earlier, here we consider the probability space $(\varOmega _1, {\mathcal {F}}_1, {\mathbb {P}})$, where ${\mathcal {F}}_1$ is the trace algebra of $\varOmega _1$. For simplicity, this space is still denoted by $(\varOmega , {\mathcal {F}}, {\mathbb {P}}).$

Proposition A.1

For any $T_1, T_2 \in {\mathbb {R}}$ with $T_1<T_2$ and $\delta >0$, the following statements hold:

• (1) The random variable

  $$\begin{aligned} x_{\delta }^*(\omega )=\int _{-\infty }^{0} e^r \varPhi _\delta (\theta _r \omega )d r, \quad \omega \in \varOmega , \end{aligned}$$

  exists and generates a stationary solution of (A.2) given by

  $$\begin{aligned} \varOmega \times {\mathbb {R}}\ni (\omega ,t)\rightarrow x_{\delta }^*(\theta _t\omega ) =\int _{-\infty }^{0} e^r \varPhi _\delta (\theta _{r+t} \omega )d r. \end{aligned}$$

  The mapping $t\rightarrow x_\delta ^*(\theta _t\omega )$ is continuous.

• (2) On $\varOmega$ we have

  $$\begin{aligned} \lim _{t\rightarrow \pm \infty }\frac{|x_{\delta }^*(\theta _t\omega )|}{|t|}= & {} 0, \quad \lim _{t\rightarrow \pm \infty }\frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )\;d r=0, \end{aligned}$$

  uniformly with respect to $\delta \in (0, 1]$.

• (3) For every $\omega \in \varOmega$,

  $$\begin{aligned} \lim _{\delta \rightarrow 0^+} \sup _{t\in [T_1, T_2]} |x_\delta ^*(\theta _{t} \omega )-x(\theta _{t} \omega )|=0, \end{aligned}$$

  where $x(\theta _t \omega )=-\int _{-\infty }^0 e^r \theta _t \omega (r) dr$ is a stationary solution of stochastic differential equation

  $$\begin{aligned} dx=-xdt+dW. \end{aligned}$$

  (A.3)

Proof

Let $\omega \in \varOmega$. We first show (1) holds. Recalling that for each $\delta >0$,

$$\begin{aligned} \varPhi _\delta (\theta _t\omega )=-\frac{1}{\delta ^2}\int _0^\delta {\dot{\phi }}(\frac{s}{\delta })\theta _t\omega (s)ds. \end{aligned}$$

by (A.1) we have $x_{\delta }^*(\omega )=\alpha \int _{-\infty }^{0} e^r \varPhi _\delta (\theta _r \omega )d r$ is well-defined. Note that $x_\delta ^*(\theta _t\omega )$ can be written as

$$\begin{aligned} x_{\delta }^*(\theta _t\omega ) =\int _{-\infty }^{t} e^{r-t} \varPhi _\delta (\theta _{r} \omega )d r. \end{aligned}$$

Then, $x_\delta ^*(\theta _t\omega )$ satisfies equation (A.2) and is a stationary process.

(2) We observe that

$$\begin{aligned} \varPhi _\delta (\theta _{r} \omega )= & {} -\frac{1}{\delta ^2} \int _{0}^{\delta } {\dot{\phi }}(\frac{s}{\delta })\omega (r+s)ds. \end{aligned}$$

Along with Fubini’s theorem, we get that

$$\begin{aligned} \begin{aligned} x_\delta ^*(\theta _t \omega ) \,\, =&\,\, \int _{-\infty }^t e^{r-t} \Big (-\frac{1}{\delta ^2} \int _{0}^{\delta } {\dot{\phi }}(\frac{s}{\delta })\omega (r+s)ds \Big ) dr \\ \,\, =&\,\, \int _{0}^{\delta } -\frac{1}{\delta ^2} {\dot{\phi }}(\frac{s}{\delta }) \big [\int _{-\infty }^t e^{r-t}\omega (r+s) dr \big ]ds. \end{aligned} \end{aligned}$$

Taking the integral transformation $u=r+s$ and $v=\frac{s}{\delta }$ successively, we have that

$$\begin{aligned} x_\delta ^*(\theta _t \omega )= & {} \frac{-1}{\delta } \int _0^1 {\dot{\phi }}(v) \big [\int _{-\infty }^{t+\delta v} e^{u-\delta v-t}\omega (u) du \big ]dv. \end{aligned}$$

By integration by parts, we further get that

$$\begin{aligned} \begin{aligned} x_\delta ^*(\theta _t \omega ) =&\, \int _0^1 \phi (v) \big [-\int _{-\infty }^{t+\delta v} e^{u-\delta v-t}\omega (u) du+\omega (t+\delta v) \big ]dv \\ =&\, \int _0^1 \phi (v) \big [-\int _{-\infty }^{0} e^{u}\omega (u+\delta v+t) du+\omega (t+\delta v) \big ]dv \\ =&\, \int _0^1 \phi (v) \big [\int _{-\infty }^{0} e^{u} \Big (\omega (t+\delta v)-\omega (u+\delta v+t) \Big ) du\big ]dv. \end{aligned} \end{aligned}$$

(A.4)

We first show that $\frac{|x_{\delta }^*(\theta _t\omega )|}{|t|}\rightarrow 0$ as $t \rightarrow -\infty$, uniformly for $\delta \in (0, 1]$. (A.1) implies that for any $\varepsilon >0$, there exists a positive constant ${\hat{T}}_0={\hat{T}}(\omega , \varepsilon )$ such that for all $|s| \ge {\hat{T}}_0$,

$$\begin{aligned} |\omega (s)| \le \varepsilon |s|. \end{aligned}$$

(A.5)

Then we have for $t\le -{\hat{T}}_0-1$ and $\delta \in (0, 1]$,

$$\begin{aligned} \frac{|x_\delta ^*(\theta _t \omega )|}{|t|}\le & {} \frac{1}{|t|} \int _0^1 \phi (v)\Big [\int _{-\infty }^0 e^{u} (|\omega (t+\delta v)|+|\omega (u+\delta v+t)|)du \Big ] dv \\\le & {} \frac{\varepsilon }{|t|}\int _{-\infty }^0 e^{u}(2|t|+2-u)du \\= & {} 2 \varepsilon +\frac{3\varepsilon }{|t|}, \end{aligned}$$

where (A.5) is used in the second inequality. Clearly, there exists a positive constant ${\hat{T}}_1$ such that as $|t| \ge {\hat{T}}_1$, $\frac{1}{|t|} < 1.$ Choosing ${\hat{T}}_2=\max \{{\hat{T}}_0, {\hat{T}}_1\}$, as $t \le -{\hat{T}}_2$, for all $\delta \in (0,1]$, we have that

$$\begin{aligned} \frac{|x_\delta ^*(\theta _t \omega )|}{|t|}<5\varepsilon . \end{aligned}$$

Hence, $\frac{|x_{\delta }^*(\theta _t\omega )|}{|t|}\rightarrow 0$ as $t \rightarrow -\infty$, uniformly for $\delta \in (0, 1]$.

Next we show that $\frac{|x_{\delta }^*(\theta _t\omega )|}{|t|}\rightarrow 0$ as $t \rightarrow +\infty$, uniformly for $\delta \in (0, 1]$. Using (A.1) again, we find that there exists ${\hat{T}}_3 = {\hat{T}}_3(\omega )>0$ such that for $|t| \ge {\hat{T}}_3$,

$$\begin{aligned} |\omega (t)| \le |t|. \end{aligned}$$

Meanwhile, by the continuity of $\omega \in \varOmega$, there exists $M(\omega )>0$ such that for $|t| \le {\hat{T}}_3$

$$\begin{aligned} |\omega (t)| \le M(\omega ). \end{aligned}$$

Then we get for all $t \in {\mathbb {R}}$,

$$\begin{aligned} |\omega (t)| \le M(\omega )+|t|. \end{aligned}$$

(A.6)

By (A.4), we have

$$\begin{aligned} \frac{x_\delta ^*(\theta _t \omega )}{t}= & {} \frac{1}{t}\int _0^1 \phi (v) \big [\int _{-\infty }^{-T_*} e^{u} \Big (\omega (t+\delta v)-\omega (u+\delta v+t) \Big ) du\big ]dv \\&+\frac{1}{t}\int _0^1 \phi (v) \big [\int _{-T_*}^{0} e^{u} \Big (\omega (t+\delta v)-\omega (u+\delta v+t) \Big ) du\big ]dv, \end{aligned}$$

where $T_*$ is a positive number to be specified later. For the first integral on the right hand side, we observe that for all $\delta \in (0,1]$,

$$\begin{aligned}&\Big |\frac{1}{t}\int _0^1 \phi (v) \big [\int _{-\infty }^{-T_*} e^{u} \Big (\omega (t+\delta v)-\omega (u+\delta v+t) \Big ) du\big ]dv \Big | \\&\quad \le \frac{1}{|t|}\int _{-\infty }^{-T_*} e^{u} (2M(\omega )+2|t|+2-u)du \\&\quad \le \frac{1}{|t|}\int _{-\infty }^{0} e^{u}(2M(\omega )+2-u) du+2\int _{-\infty }^{-T_*} e^{u} du =\frac{2 M(\omega )+3}{|t|}+2 e^{-T_*}. \end{aligned}$$

We choose a sufficiently large $T_*>0$ such that $2 e^{-T_*}<\varepsilon$ and note that there exists a ${\hat{T}}_4={\hat{T}}_4(\omega , \varepsilon )>0$ such that as $|t|>{\hat{T}}_4$, $\frac{2 M(\omega )+3}{|t|}<\varepsilon$. Thus as $t>{\hat{T}}_4$, for all $\delta \in (0,1]$ we have that

$$\begin{aligned} \Big |\frac{1}{t}\int _0^1 \phi (v) \big [\int _{-\infty }^{-T_*} e^{u} \Big (\omega (t+\delta v)-\omega (u+\delta v+t) \Big ) du\big ]dv \Big |<2 \varepsilon . \end{aligned}$$

(A.7)

On the other hand, we note that for $\delta \in (0, 1]$, $|t+\delta v|>|t|-|\delta v|>|t|-1$ and $|u+\delta v+t|>|t|-|u+\delta v|>|t|-(T_*+1)$ on $u \in [-T_*, 0]$. It then follows that for $t>T_*+1+{\hat{T}}_0$ and $\delta \in (0, 1]$,

$$\begin{aligned} \begin{aligned} \Big |\frac{1}{t}\int _0^1 \phi (v) \big [\int _{-T_*}^{0} e^{u} \Big (\omega (t+\delta v)-\omega (u+\delta v+t) \Big ) du\big ]dv \Big | \le&\,\, \frac{\varepsilon }{|t|}\int _{-T_*}^0 e^{u} (2|t|+2-u)dr \\ \le&\,\, \frac{\varepsilon }{|t|}\int _{-\infty }^0 e^{u} (2|t|+2-u)dr \\ \le&\,\, 2\varepsilon + \frac{3 \varepsilon }{|t|}. \end{aligned} \end{aligned}$$

(A.8)

Combing (A.7) with (A.8), as $t>\max \{{\hat{T}}_4,T_*+1+{\hat{T}}_0, {\hat{T}}_1 \}$, we find that for all $\delta \in (0, 1]$,

$$\begin{aligned} \frac{|x_\delta ^*(\theta _t \omega )|}{|t|}<7 \varepsilon . \end{aligned}$$

This implies that $\frac{|x_{\delta }^*(\theta _t\omega )|}{|t|}\rightarrow 0$ as $t \rightarrow +\infty$, uniformly for $\delta \in (0, 1]$.

We now show that $\frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )d r\rightarrow 0$ as $t \rightarrow -\infty$, uniformly for $\delta \in (0, 1]$. By using (A.4) and Fubini’s theorem, we have that

$$\begin{aligned} \begin{aligned}&\; \frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )d r \\&\quad = \; \frac{1}{t} \int _0^t \Big [\int _0^1 \phi (v) \Big (-\int _{-\infty }^{r+\delta v} e^{u-\delta v-r}\omega (u) du+\omega (r+\delta v) \Big )dv\Big ]dr \\&\quad = \; \frac{1}{t} \int _0^1 \phi (v) \Big \{-\int _t^0 \Big [-\int _{-\infty }^{r+\delta v} e^{u-\delta v-r}\omega (u) du +\omega (r+\delta v) \Big ] dr\Big \} dv. \end{aligned} \end{aligned}$$

(A.9)

Here we first restrict $t<-1$. By applying Fubini’s theorem again, we find that

$$\begin{aligned}&\int _t^0 \Big [\int _{-\infty }^{r+\delta v} e^{u-\delta v-r}\omega (u) du \Big ]dr \\&\quad = \int _{-\infty }^{t+\delta v} du \Big [\int _t^0 e^{u-\delta v-r}\omega (u) dr\Big ] + \int _{t+\delta v}^{\delta v} du \Big [\int _{u-\delta v}^0 e^{u-\delta v-r}\omega (u) dr\Big ] \\&\quad = \int _{-\infty }^{t+\delta v} \omega (u)\Big [e^{u-\delta v-t}-e^{u-\delta v}\Big ]du + \int _{t+\delta v}^{\delta v} \omega (u) \Big [1-e^{u-\delta v}\Big ] du \\&\quad = \int _{-\infty }^{t} \omega (u+\delta v)\Big [e^{u-t}-e^{u}\Big ]du + \int _{t}^{0} \omega (u+\delta v) \Big [1-e^{u}\Big ] du. \end{aligned}$$

Then, as $t<-1$ we have that

$$\begin{aligned}&\frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )d r \\&\quad = \frac{1}{t} \int _0^1 \phi (v) \Big \{\int _{-\infty }^t \omega (u+\delta v) e^{u-t} du- \int _{-\infty }^0 \omega (u+\delta v)e^u du \Big \} dv \\&\quad = \frac{1}{t} \int _0^1 \phi (v) \Big \{\int _{-\infty }^0 e^{u} (\omega (u+t+\delta v)-\omega (u+\delta v)) du\Big \} dv. \end{aligned}$$

By using (A.5) and (A.6), for $t\le -{\hat{T}}_0-1$ and $\delta \in (0, 1]$ we have that

$$\begin{aligned} \frac{|\int _0^t x_{\delta }^*(\theta _r \omega )d r|}{|t|}\le & {} \frac{1}{|t|} \int _0^1 \phi (v)\Big [\int _{-\infty }^0 e^{u} (|\omega (u+\delta v+t)|+|\omega (u+\delta v)|)du \Big ] dv \\\le & {} \frac{\varepsilon }{|t|}\int _{-\infty }^0 e^{u}(|t|+1-u)du+\frac{1}{|t|}\int _{-\infty }^0 e^{u}(M(\omega )-u+1) du \\= & {} \varepsilon +\frac{2\varepsilon }{|t|}+\frac{M(\omega )+2}{|t|}. \end{aligned}$$

Clearly, as $t<\max \{-{\hat{T}}_1, -{\hat{T}}_4, -{\hat{T}}_0-1\}$, for all $\delta \in (0, 1]$, we find that $|\frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )d r|<4\varepsilon$ and hence $\frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )\;d r\rightarrow 0$ as $t \rightarrow -\infty$, uniformly for $\delta \in (0, 1]$.

In order to show $\frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )d r\rightarrow 0$ as $t \rightarrow +\infty$, uniformly for $\delta \in (0, 1]$, we rewrite (A.9) as

$$\begin{aligned} \frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )\;d r= & {} \frac{1}{t} \int _0^1 \phi (v) \Big \{\int _0^t \Big [-\int _{-\infty }^{r+\delta v} e^{u-\delta v-r}\omega (u) du +\omega (r+\delta v) \Big ] dr\Big \} dv. \end{aligned}$$

By applying Fubini’s theorem, for $t>0$ we investigate that

$$\begin{aligned}&\int _0^t \Big [\int _{-\infty }^{r+\delta v} e^{u-\delta v-r}\omega (u) du \Big ]dr \\&\quad = \int _{-\infty }^{\delta v} du \Big [\int _0^t e^{u-\delta v-r}\omega (u) dr\Big ] + \int _{\delta v}^{t+\delta v} du \Big [\int _{u-\delta v}^t e^{u-\delta v-r}\omega (u) dr\Big ] \\&\quad = \int _{-\infty }^{\delta v} \omega (u)\Big [e^{u-\delta v}-e^{u-\delta v-t}\Big ]du + \int _{\delta v}^{t+\delta v} \omega (u) \Big [1-e^{u-\delta v-t}\Big ] du \\&\quad = \int _{-\infty }^{0} \omega (u+\delta v)\Big [e^{u}-e^{u-t}\Big ]du + \int _{0}^{t} \omega (u+\delta v) \Big [1-e^{u-t}\Big ] du. \end{aligned}$$

Then, for $t>0$ we have that

$$\begin{aligned} \frac{1}{t}\int _0^t z_{\delta }^*(\theta _r \omega )\;d r= & {} \frac{1}{t} \int _0^1 \phi (v) \Big \{-\int _{-\infty }^0 e^u \omega (u+\delta v) du \Big \} dv \\&+\frac{1}{t} \int _0^1 \phi (v) \Big \{\int _{-\infty }^t e^{u-t} \omega (u+\delta v) du \Big \} dv \\= & {} \frac{1}{t} \int _0^1 \phi (v) \Big \{\int _{-\infty }^0 e^{u}[\omega (u+t+\delta v)-\omega (u+\delta v)] du \Big \} dv. \end{aligned}$$

Recall that we have chosen a $T_{*}>0$ such that $2 e^{-T_*}<\varepsilon$. Applying (A.6), as $t>{\hat{T}}_4$, for all $\delta \in (0, 1]$ we have that

$$\begin{aligned} \begin{aligned}&\; \Big |\frac{1}{t} \int _0^1 \phi (v) \Big \{\int _{-\infty }^{-T_*} e^{u}[\omega (u+t+\delta v)-\omega (u+\delta v)] du \Big \} dv \Big | \\&\quad \le \; \frac{1}{|t|}\int _{-\infty }^{-T_{*}} e^{u}(2M(\omega )-2u+|t|+2)du \\&\quad \le \; \frac{1}{|t|}\int _{-\infty }^{0} e^{u}(2M(\omega )-2u+2)d\tau +e^{-T_*}<\frac{2M(\omega )+4}{|t|}+\varepsilon <2 \varepsilon . \end{aligned} \end{aligned}$$

(A.10)

Moreover, since $|u+t+\delta v|>|t|-|u+\delta v|>|t|-(T_*+1)$ for all $u \in [-T_*, 0]$ and $\delta \in (0, 1]$, by (A.4) we have that for all $t>\max \{{\hat{T}}_0+T_*+1, {\hat{T}}_1, {\hat{T}}_4\}$ and $\delta \in (0, 1]$,

$$\begin{aligned} \begin{aligned}&\; \Big |\frac{1}{t} \int _0^1 \phi (v) \Big \{\int _{-T_*}^0 e^{u}[\omega (u+t+\delta v)-\omega (u+\delta v)] du \Big \} dv \Big | \\&\quad \le \; \frac{\varepsilon }{|t|} \int _{-T_*}^0 e^u (-u+|t|+1)du+ \frac{1}{|t|}\int _{-T_*}^0 e^u (M(\omega )-u+1) du \\&\quad \le \; \frac{\varepsilon }{|t|} \int _{-\infty }^0 e^u (-u+|t|+1)du+ \frac{1}{|t|}\int _{-\infty }^0 e^u (M(\omega )-u+1) du \\&\quad \le \; \frac{2\varepsilon }{|t|}+\varepsilon +\frac{M(\omega )+2}{|t|}<4\varepsilon . \end{aligned} \end{aligned}$$

(A.11)

Combing (A.10) with (A.11), for all $t \ge \max \{{\hat{T}}_0+T_*+1, {\hat{T}}_1, {\hat{T}}_4 \}$ and $\delta \in (0, 1]$, we have

$$\begin{aligned} \Big |\frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )d r \Big |<6\varepsilon . \end{aligned}$$

It follows that $\frac{1}{t}\int _0^t x_{\delta }^*(\theta _r \omega )d r\rightarrow 0$ as $t \rightarrow +\infty$, uniformly for $\delta \in (0, 1]$.

(3) By (A.4) we investigate that

$$\begin{aligned} x_\delta ^*(\omega )= & {} \int _0^1 \phi (v) \big [\int _{-\infty }^{0} e^{u} \big (\omega (\delta v)-\omega (u+\delta v) \big ) du\big ]dv. \end{aligned}$$

Along with $x(\omega )=-\int _{-\infty }^0e^r\omega (r)d r$, we have

$$\begin{aligned} x_\delta ^*(\omega )-x(\omega )=\int _0^1 \phi (v) \big [\int _{-\infty }^{0} e^{u} \big (\omega (\delta v)-\omega (u+\delta v)+\omega (u) \big ) du\big ]dv. \end{aligned}$$

Note that $|e^{u} \big (\omega (\delta v)-\omega (u+\delta v)+\omega (u) \big ) | \le e^{u} (M(\omega )+2+2|u|)$ for $\delta \in (0, 1]$ and $v \in [0,1]$, and $\int _{-\infty }^0 e^{u} (M(\omega )+2+2|u|) du<+\infty$. Then, by using the Dominated Convergence Theorem, we get that

$$\begin{aligned} \lim _{\delta \rightarrow 0^+} x_\delta ^*(\omega )=x(\omega ). \end{aligned}$$

(A.12)

Recall that $x_\delta ^*(\theta _t \omega )$ and $x(\theta _t \omega )$ are solutions of equations (A.2) and (A.3), respectively. Then we can rewrite them as

$$\begin{aligned} x_\delta ^*(\theta _t \omega )= & {} x_\delta ^*(\omega ) - \int _{0}^{t} x_\delta ^*(\theta _s \omega ) d s+ \int _{0}^{t} \varPhi _\delta (\theta _r \omega ) d r, \\ x(\theta _t \omega )= & {} x(\omega ) - \int _{0}^{t} x(\theta _s \omega ) d s+ \omega (t), \end{aligned}$$

which together with Gronwall’s inequality implies that

$$\begin{aligned} |x_\delta ^*(\theta _t \omega )-x(\theta _t \omega )|\le & {} \left[ |x_\delta ^*(\omega )-x(\omega )|+\sup _{t \in [0, T]} \Big |\int _{0}^{t} \varPhi _\delta (\theta _r \omega ) d r-\omega (t) \Big | \right] e^{T} \end{aligned}$$

for all $t \in [0, T]$ and

$$\begin{aligned} |x_\delta ^*(\theta _t \omega )-x(\theta _t \omega )|\le & {} \left[ |x_\delta ^*(\omega )-x(\omega )|+\sup _{t \in [-T, 0]} \Big |\int _{0}^{t} \varPhi _\delta (\theta _r \omega ) d r-\omega (t) \Big | \right] e^{T} \end{aligned}$$

for all $t \in [-T, 0]$, where T is a positive constant such that $[T_1, T_2] \subset [-T, T]$. It then follows from (A.12) and (1.7) that

$$\begin{aligned} \lim _{\delta \rightarrow 0^+} \sup _{t \in [T_1, T_2]}\Vert x_\delta ^*(\theta _t \omega )-x(\theta _t\omega )\Vert =0. \end{aligned}$$

This completes the proof of this proposition. $\square$