Immuno-epidemiology of a population structured by immune status: a mathematical study of waning immunity and immune system boosting

Abstract

When the body gets infected by a pathogen the immune system develops pathogen-specific immunity. Induced immunity decays in time and years after recovery the host might become susceptible again. Exposure to the pathogen in the environment boosts the immune system thus prolonging the time in which a recovered individual is immune. Such an interplay of within host processes and population dynamics poses significant challenges in rigorous mathematical modeling of immuno-epidemiology. We propose a framework to model SIRS dynamics, monitoring the immune status of individuals and including both waning immunity and immune system boosting. Our model is formulated as a system of two ordinary differential equations (ODEs) coupled with a PDE. After showing existence and uniqueness of a classical solution, we investigate the local and the global asymptotic stability of the unique disease-free stationary solution. Under particular assumptions on the general model, we can recover known examples such as large systems of ODEs for SIRWS dynamics, as well as SIRS with constant delay.

Proof of Theorem 1

In the following we show the continuous differentiability of the map $Q$, which is necessary to have existence and uniqueness of a classical solution of the abstract Cauchy problem (8). Continuous differentiability of $Q$ can be shown in two steps: (a) First we determine the existence of the operator $DQ(x;w)$, for all $x,\,w \in X$, defined by

$$\begin{aligned} DQ(x;w):=\lim \limits _{h \rightarrow 0}\frac{Q(x+hw)-Q(x)}{h}. \end{aligned}$$

(b) Second we show that the operator $DQ(x;\cdot )$ is continuous in $x$, that is

$$\begin{aligned} \lim \Vert DQ(x;\cdot ) -DQ(y;\cdot )\Vert _{OP}=0 \qquad \text{ for } \qquad \left\| {x-y}\right\| _X\rightarrow 0, \end{aligned}$$

where $\Vert \cdot \Vert _{OP}$ is the operator norm.

For simplicity of notation we write

$$\begin{aligned} DQ_1(x; w)&: = P_1(x;w)-P_2(x;w),\\ DQ_2(x; w)&: = P_2(x;w),\\ DQ_3(x; w)&: = -P_3(x;w)+P_4(x;w), \end{aligned}$$

where

$$\begin{aligned} P_1(x;w)&:=\,\lim \limits _{h \rightarrow 0}\frac{b(\hat{x}+h \hat{w})-b(\hat{x})}{h},\\ P_2(x;w)&:=\,\lim \limits _{h \rightarrow 0}\frac{1}{h} \beta \left( \frac{(x_1+hw_1)(x_2+hw_2)}{\hat{x}+h \hat{w}}-\frac{x_1x_2}{\hat{x}}\right) ,\\ P_3(x;w)&:=\,\lim \limits _{h \rightarrow 0}\frac{1}{h} \beta \left( \frac{(x_2+hw_2)(x_3(z)+hw_3(z))}{\hat{x}+h\hat{w}} - \frac{x_2x_3(z)}{\hat{x}}\right) ,\\ P_4(x;w)&:=\,\lim \limits _{h \rightarrow 0}\frac{1}{h} \beta \left( \frac{(x_2+hw_2)\int _{z_{min}}^{z} (x_3(u)+hw_3(u))\,p(z,u)\,du}{\hat{x}+h\hat{w}} \right. \\&\quad \left. -\frac{x_2\int _{z_{min}}^{z} x_3(u)p(z,u)\,du}{\hat{x}}\right) . \end{aligned}$$

Proof of (a)

We compute the limit for the first component of $Q_1$,

$$\begin{aligned} P_1(x;w)= & {} \lim \limits _{h \rightarrow 0}\frac{1}{h} \biggl ( b\left( \hat{x}+h \hat{w}\right) -b\left( \hat{x}+hw_1+hw_2\right) \biggr )\\&+ \lim \limits _{h \rightarrow 0}\frac{1}{h} \biggl ( b\left( \hat{x}+hw_1+hw_2\right) -b\left( \hat{x}+hw_1\right) \biggr )\\&+\lim \limits _{h \rightarrow 0}\frac{1}{h} \bigl ( b\left( \hat{x}+hw_1\right) -b\left( \hat{x}\right) \bigr )\\= & {} \lim \limits _{h \rightarrow 0}\frac{b\left( \hat{x}+h \hat{w}\right) -b\left( \hat{x}+hw_1+hw_2\right) }{h \int _{z_{min}}^{z_{max}}w_3(u)\,du} \int _{z_{min}}^{z_{max}}w_3(u)\,du\\&+\lim \limits _{h \rightarrow 0}\frac{b\left( \hat{x}+hw_1+hw_2\right) -b\left( \hat{x}+hw_1\right) }{h w_2}\,w_2\\&+ \lim \limits _{h \rightarrow 0}\frac{b\left( \hat{x}+hw_1\right) -b\left( \hat{x}\right) }{h\,w_1} \,w_1\\= & {} b'(\hat{x}) \hat{w}. \end{aligned}$$

For the second term in $Q_1(x)$ we have

$$\begin{aligned} P_2(x;w)= & {} \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{(x_1+hw_1)(x_2+hw_2)}{\hat{x}+h \hat{w}}- \frac{(x_1+hw_1)(x_2+hw_2)}{\hat{x} + h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\right) \\&+ \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{(x_1+hw_1)(x_2+hw_2)}{\hat{x} + h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du} - \frac{x_1(x_2+hw_2)}{\hat{x} + h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\right) \\&+ \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{x_1(x_2+hw_2)}{\hat{x} + h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du} - \frac{x_1(x_2+hw_2)}{\hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\right) \\&+ \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{x_1(x_2+hw_2)}{\hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du} -\frac{x_1x_2}{\hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\right) \\&+ \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{x_1x_2}{\hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du} -\frac{x_1x_2}{\hat{x}}\right) \\= & {} -\beta \lim \limits _{h \rightarrow 0} \frac{(x_1+hw_1)(x_2+hw_2)}{\left( \hat{x}+h \hat{w}\right) \left( \hat{x} + h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du\right) }\,w_1\\&+ \beta \lim \limits _{h \rightarrow 0} \frac{x_2+hw_2}{\hat{x} + h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\,w_1\\&- \beta \lim \limits _{h \rightarrow 0} \frac{x_1(x_2+hw_2)}{\left( \hat{x} + h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du\right) \left( \hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du\right) }\,w_2\\&+ \beta \lim \limits _{h \rightarrow 0} \frac{x_1}{\hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\,w_2\\&- \beta \lim \limits _{h \rightarrow 0} \frac{x_1x_2}{\left( \hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du\right) \hat{x}}\,\int _{z_{min}}^{z_{max}}w_3(u)\,du\\= & {} \beta \left[ \frac{x_2(\hat{x}-x_1)}{{\hat{x}}^2}\,w_1 + \frac{x_1(\hat{x} -x_2)}{{\hat{x}}^2} \,w_2 -\frac{x_1x_2}{{\hat{x}}^2}\,\int _{z_{min}}^{z_{max}}w_3(u)\,du\right] . \end{aligned}$$

The first term in $Q_3(x)$:

$$\begin{aligned} P_3(x;w)= & {} \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{(x_2+hw_2)(x_3(z)+hw_3(z))}{\hat{x}+h\hat{w}}- \frac{(x_2+hw_2)(x_3(z)+hw_3(z))}{\hat{x}+h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\right) \\&+ \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{(x_2+hw_2)(x_3(z)+hw_3(z))}{\hat{x}+h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du} - \frac{x_2(x_3(z)+hw_3(z))}{\hat{x}+h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\right) \\&+ \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{x_2(x_3(z)+hw_3(z))}{\hat{x}+ w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du} -\frac{x_2(x_3(z)+hw_3(z))}{\hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\right) \\&+ \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{x_2(x_3(z)+hw_3(z))}{\hat{x}+h\int _{z_{min}}^{z_{max}}w_3(u)\,du} - \frac{x_2x_3(z)}{\hat{x}+h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\right) \\&+ \beta \lim \limits _{h \rightarrow 0}\frac{1}{h} \left( \frac{x_2x_3(z)}{\hat{x}+h\int _{z_{min}}^{z_{max}}w_3(u)\,du} - \frac{x_2x_3(z)}{\hat{x}}\right) . \end{aligned}$$

Hence we have

$$\begin{aligned} P_3(x;w)= & {} -\beta \lim \limits _{h \rightarrow 0} \frac{(x_2+hw_2)(x_3(z)+hw_3(z))}{\left( \hat{x} +h\hat{w}\right) \left( \hat{x}+h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du\right) }\,w_1\\&+ \beta \lim \limits _{h \rightarrow 0}\frac{x_3(z)+hw_3(z)}{\hat{x}+h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\,w_2\\&-\beta \lim \limits _{h \rightarrow 0} \frac{x_2(x_3(z)+hw_3(z))}{\left( \hat{x}+h w_2+h\int _{z_{min}}^{z_{max}}w_3(u)\,du\right) \left( \hat{x}+h\int _{z_{min}}^{z_{max}}w_3(u)\,du\right) }\,w_2\\&+ \beta \lim \limits _{h \rightarrow 0}\frac{x_2}{\hat{x}+h\int _{z_{min}}^{z_{max}}w_3(u)\,du}\,w_3(z)\\&- \beta \lim \limits _{h \rightarrow 0}\frac{x_2x_3(z)}{\hat{x}\left( \hat{x} +h\int _{z_{min}}^{z_{max}}w_3(u)\,du\right) }\,\int _{z_{min}}^{z_{max}}w_3(u)\,du\\= & {} \beta \biggl [-\frac{x_2x_3(z)}{{\hat{x}}^2}\,w_1+\frac{x_3(z)(\hat{x} -x_2)}{{\hat{x}}^2}\,w_2 + \frac{x_2}{\hat{x}}\,w_3(z)\\&- \frac{x_2x_3(z)}{{\hat{x}}^2}\,\int _{z_{min}}^{z_{max}}w_3(u)\,du\biggr ]. \end{aligned}$$

Analogously, compute the last term in $Q_3(x)$,

$$\begin{aligned} P_4(x;w)= & {} \beta \lim \limits _{h \rightarrow 0}\frac{\int _{z_{min}}^{z} (x_3(u)+hw_3(u))\,p(z,u)\,du}{\hat{x}+h\hat{w}} \,w_2\\&- \beta \lim \limits _{h \rightarrow 0} \frac{x_2\int _{z_{min}}^{z} (x_3(u)+hw_3(u))\,p(z,u)\,du}{\left( \hat{x} +h\hat{w}\right) \left( \hat{x}+h w_2+ h \int _{z_{min}}^{z_{max}}w_3(u)\,du\right) }\,w_1\\&+ \beta \lim \limits _{h \rightarrow 0} \frac{x_2}{\hat{x}+h w_2+ h \int _{z_{min}}^{z_{max}}w_3(u)\,du}\,\int _{z_{min}}^{z} w_3(u)\,p(z,u)\,du\\&-\beta \lim \limits _{h \rightarrow 0}\frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{\left( \hat{x}+h w_2+ h \int _{z_{min}}^{z_{max}}w_3(u)\,du\right) \left( \hat{x}+ h \int _{z_{min}}^{z_{max}}w_3(u)\,du\right) }\,w_2\\&- \beta \lim \limits _{h \rightarrow 0} \frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{\left( \hat{x}+ h \int _{z_{min}}^{z_{max}}w_3(u)\,du\right) \hat{x}}\,\int _{z_{min}}^{z_{max}}w_3(u)\,du \\= & {} \beta \left[ - \frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}^2}\,w_1 + \frac{(\hat{x} -x_2)\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}^2}\,w_2\right. \\&\quad \left. + \frac{x_2\left( \hat{x} -\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du\right) }{{\hat{x}}^2}\int _{z_{min}}^{z_{max}}w_3(u)\,du \right] . \end{aligned}$$

To prove that the operator $DQ(x;\cdot )$ is continuous in $x$ we consider the norm $\Vert DQ(x;\cdot ) -DQ(y;\cdot ) \Vert _{OP}$, that is

$$\begin{aligned}&\underset{ \left\| {w}\right\| _X\le 1 }{\sup } \left\| {DQ(x;w) -DQ(y;w)}\right\| _X\\&\quad = \underset{ \left\| {w}\right\| _X\le 1 }{\sup } \biggl \{|P_1(x;w)\!-\!P_1(y;w)-P_2(x;w)\!+\!P_2(y;w)| \!+\! |P_2(x;w)-P_2(y;w)|\\&\quad \quad + \int _{z_{min}}^{z_{max}} |(P_3(y;w)-P_3(x;w)+P_4(x;w)-P_4(y;w))(z)|\,dz\biggr \}, \end{aligned}$$

and show that

$$\begin{aligned} \Vert DQ(x;\cdot ) -DQ(y;\cdot ) \Vert _{OP} \rightarrow 0, \qquad \text{ for } \quad \left\| {x- y}\right\| _X\rightarrow 0. \end{aligned}$$

We estimate the operator norm as follows

$$\begin{aligned} \underset{ \left\| {w}\right\| _X\le 1 }{\sup } \left\| {DQ(x;w) -DQ(y;w)}\right\| _X\le \underset{ \left\| {w}\right\| _X\le 1 }{\sup } \sum \limits _{j=1}^{4} T_j(x,y;w), \end{aligned}$$

with

$$\begin{aligned} T_1(x,y;w)&= |P_1(x;w)-P_1(y;w)|,\\ T_2(x,y;w)&= 2|P_2(x;w)-P_2(y;w)|,\\ T_3(x,y;w)&= \int _{z_{min}}^{z_{max}} |(P_3(x;w)-P_3(y;w))(z)|\,dz,\\ T_4(x,y;w)&=\int _{z_{min}}^{z_{max}}|(P_4(x;w)-P_4(y;w))(z)|\,dz. \end{aligned}$$

Then we show the convergence to zero of the above sum. It is obvious that

$$\begin{aligned} \underset{ \left\| {w}\right\| _X\le 1 }{\sup } T_1(x,y;w) \; \rightarrow 0, \qquad \text{ for } \left\| {x- y}\right\| _X\rightarrow 0, \end{aligned}$$

as $b$ is continuously differentiable (see Assumption 1),

$$\begin{aligned} |P_1(x;w)-P_1(y;w)|&= |b'(\hat{x}) \hat{w} - b'(\hat{y}) \hat{w}|\;\le |b'(\hat{x}) - b'(\hat{y})| \left\| {w}\right\| _X. \end{aligned}$$

The term in $T_2(x,y;w)$ can be estimated as follows:

$$\begin{aligned} |P_2(x;w)-P_2(y;w)|\le & {} \underbrace{\beta \left| \left( \frac{x_2(\hat{x}-x_1)}{{\hat{x}}^2} -\frac{y_2(\hat{y}-y_1)}{{\hat{y}}^2}\right) \right| \,|w_1|}_{=:L_1(x,y;w)}\\&+\underbrace{\beta \left| \left( \frac{x_1(\hat{x} -x_2)}{{\hat{x}}^2}-\frac{y_1(\hat{y} -y_2)}{{\hat{y}}^2}\right) \right| \,|w_2|}_{=:L_2(x,y;w)}\\&+\underbrace{\beta \left| \left( \frac{x_1x_2}{{\hat{x}}^2}-\frac{y_1y_2}{{\hat{y}}^2}\right) \right| \,\int _{z_{min}}^{z_{max}}|w_3(u)|\,du}_{=:L_3(x,y;w)}. \end{aligned}$$

Since the addends of the last sum are all similar, we show convergence to zero for only one of them.

$$\begin{aligned}&\underset{ \left\| {w}\right\| _X\le 1 }{\sup }\;L_3(x,y;w)\\&\quad \le \beta \left| \frac{x_1x_2}{{\hat{x}}^2} -\frac{y_1y_2}{{\hat{y}}^2}\right| \\&\quad \le \beta \biggl ( \left| \frac{x_1x_2}{{\hat{x}}^2}- \frac{x_1x_2}{\hat{x}\hat{y}}\right| + \left| \frac{x_1x_2}{\hat{x} \hat{y}}-\frac{x_1y_2}{\hat{x} \hat{y}}\right| + \left| \frac{x_1y_2}{\hat{x} \hat{y}}-\frac{x_1y_2}{{\hat{y}}^2}\right| + \left| \frac{x_1y_2}{{\hat{y}}^2}-\frac{y_1y_2}{{\hat{y}}^2}\right| \biggr )\\&\quad \le \beta \biggl ( \left| \frac{x_1x_2(\hat{y}-\hat{x})}{{\hat{x}}^2\hat{y}}\right| + \left| \frac{x_1(x_2-y_2)}{\hat{x} \hat{y}}\right| + \left| \frac{x_1y_2(\hat{y}-\hat{x})}{\hat{x} {\hat{y}}^2}\right| + \left| \frac{(x_1-y_1)y_2}{{\hat{y}}^2}\right| \biggr )\\&\quad \le \frac{3\beta }{|\hat{y}|} \, \left\| {x-y}\right\| _X. \end{aligned}$$

It works analogously for the terms $L_1(x,y;w)$ and $L_2(x,y;w)$. Hence we have that

$$\begin{aligned} \underset{ \left\| {w}\right\| _X\le 1 }{\sup } T_2(x,y;w) \; \rightarrow 0, \qquad \text{ for } \left\| {x- y}\right\| _X\rightarrow 0. \end{aligned}$$

In a similar way one can estimate $T_3(x,y;w)$ and $T_4(x,y;w)$. We show the computation for $T_4$ which is the most challenging of the two, as it includes double integrals.

$$\begin{aligned}&T_4(x,y;w) \\&\quad \le \underbrace{{\beta \int _{z_{min}}^{z_{max}} \left| \frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}^2} -\frac{y_2\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{y}}^2}\right| \,|w_1|\,dz}}_{=:F_1(x,y;w)}\\&\qquad + \underbrace{\beta \int _{z_{min}}^{z_{max}} \left| \frac{(\hat{x} -x_2)\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}^2} -\frac{(\hat{y} -y_2)\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{y}}^2}\right| \,|w_2|\,dz}_{=:F_2(x,y;w)}\\&\qquad + \underbrace{\beta \int _{z_{min}}^{z_{max}} \left| \frac{x_2}{{\hat{x}}} -\frac{y_2}{{\hat{y}}} \right| \,\int _{z_{min}}^{z_{max}}|w_3(u)|\,du\,dz}_{=:F_3(x,y;w)}\\&\qquad + \underbrace{\beta \int _{z_{min}}^{z_{max}} \left| \frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}^2} -\frac{y_2\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{y}}^2} \right| \,\int _{z_{min}}^{z_{max}}|w_3(u)|\,du\,dz}_{=:F_4(x,y;w)}. \end{aligned}$$

Before proceeding to the next estimate, it is useful to observe that

$$\begin{aligned} \int _{z_{min}}^{z_{max}} \int _{z_{min}}^{z} x_3(v)p(z,v)\,dv\,dz&=\int _{z_{min}}^{z_{max}} \int _{v}^{z_{max}} x_3(v)p(z,v)\,dz\,dv\\&=\int _{z_{min}}^{z_{max}} x_3(v) \underbrace{\int _{v}^{z_{max}} p(z,v)\,dz}_{=1}\,dv\\&=\int _{z_{min}}^{z_{max}} x_3(v)\,dv. \end{aligned}$$

Let us now consider the last addend in $T_4(x,y;w)$. We have that

$$\begin{aligned}&\underset{ \left\| {w}\right\| _X\le 1 }{\sup }\; F_4(x,y;w)\\&\quad \le \beta \int _{z_{min}}^{z_{max}} \left| \frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}^2} -\frac{y_2\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{y}}^2}\right| \,dz\\&\quad \le \beta \int _{z_{min}}^{z_{max}} \left| \frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}^2} - \frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}\hat{y}}\right| \,dz\\&\qquad + \beta \int _{z_{min}}^{z_{max}} \left| \frac{x_2\int _{z_{min}}^{z} x_3(u)\,p(z,u)\,du}{{\hat{x}}\hat{y}} -\frac{x_2\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{x}}\hat{y}} \right| \,dz\\&\qquad + \beta \int _{z_{min}}^{z_{max}} \left| \frac{x_2\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{x}}\hat{y}} -\frac{x_2\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{y}}^2}\right| \,dz\\&\qquad + \beta \int _{z_{min}}^{z_{max}} \left| \frac{x_2\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{y}}^2} - \frac{y_2\int _{z_{min}}^{z} y_3(u)\,p(z,u)\,du}{{\hat{y}}^2}\right| \,dz. \end{aligned}$$

A similar computation as the one for $\underset{ \left\| {w}\right\| _X\le 1 }{\sup }\;L_3(x,y;w)$ yields

$$\begin{aligned} \underset{ \left\| {w}\right\| _X\le 1 }{\sup }\; F_4(x,y;w)\le & {} \beta \int _{z_{min}}^{z_{max}} \int _{z_{min}}^{z} |x_3(u)\,p(z,u)|\,du\,dz \, \left| \frac{x_2(\hat{y}-\hat{x})}{{\hat{x}}^2\hat{y}}\right| \\&+ \beta \int _{z_{min}}^{z_{max}} \int _{z_{min}}^{z} |x_3(u)\,p(z,u)|\,du\,dz\, \left| \frac{x_2(\hat{y}-\hat{x})}{\hat{x} {\hat{y}}^2}\right| \\&+ \beta \int _{z_{min}}^{z_{max}} \int _{z_{min}}^{z} |(x_3(u)-y_3(u))\,p(z,u)|\,du\;dz\, \left| \frac{x_2}{{\hat{x}}{\hat{y}}}\right| \\&+ \beta \int _{z_{min}}^{z_{max}} \int _{z_{min}}^{z} |y_3(u)\,p(z,u)|\,du\,dz\, \left| \frac{x_2-y_2}{{\hat{y}}^2}\right| \\\le & {} \frac{3\beta }{|\hat{y}|} \, \left\| {x-y}\right\| _X. \end{aligned}$$

Similar relations hold for the other terms $F_1(x,y;w)$, $F_2(x,y;w)$ and $F_3(x,y;w)$. It is then obvious that the norm $\Vert DQ(x;\cdot ) -DQ(y;\cdot ) \Vert _{OP}$ tends to zero, for $\left\| {x- y}\right\| _X$ going to zero, and the proof is complete. $\square$