**3. Solutions of the Problem (1), (2) and (4)**

Define the mapping *L*(*x*, *y*)=(*L*1*x*, *L*2*y*) where *L*1*x*, *L*2*y* are given by the following stochastic integral equations

$$L\_1 \mathbf{x}(t) \quad = \quad \mathbf{x}\_0 - \int\_0^\tau h\_1(\mathbf{s}, \mathbf{x}(\mathbf{s})) d\mathcal{W}(\mathbf{s}) + \int\_0^t f\_1(\mathbf{s}, \mathbf{y}(\boldsymbol{\phi}\_1(\mathbf{s}))) d\mathbf{s},\tag{17}$$

$$L\_2\mathfrak{z}(t) \quad = \quad \mathfrak{y} \mathbf{0} - \int\_0^\eta h\mathbf{\hat{z}}(\mathbf{s}, \mathbf{y}(\mathbf{s}))d\mathbf{s} + \int\_0^t f\mathbf{\hat{z}}(\mathbf{s}, \mathbf{x}(\boldsymbol{\phi}\mathbf{z}(\mathbf{s})))d\mathcal{W}(\mathbf{s}).\tag{18}$$

**Lemma 4.** *L* : *Q* → *Q*.

**Proof.** Let *x*, *y* ∈ *Q*, , then we obtain

$$\begin{split} \|\|L\_1\mathbf{x}(t)\|\|\_2 &\leq \quad \|\|\mathbf{x}\_0\|\|\_2 + \|\int\_0^\tau h\_1(s, \mathbf{x}(s)) dW(s)\|\_2 + \|\int\_0^t f\_1(s, y(\phi\_1(s))) ds\|\_2 \\ &\leq \quad \|\|\mathbf{x}\_0\|\|\_2 + \sqrt{\int\_0^\tau \|h\_1(s, \mathbf{x}(s))\|\_2^2 ds} + \int\_0^t \|f\_1(s, y(\phi\_1(s)))\|\_2 ds \\ &\leq \quad \|\|\mathbf{x}\_0\|\|\_2 + \sqrt{\int\_0^\tau (|k\_1(s)| + c\_1 \|\|\mathbf{x}(s)\|\_2)^2} \int\_0^t (|m\_1(s)| + b\_1 \|\|y(s)\|\_2) ds \\ &\leq \quad \|\|\mathbf{x}\_0\|\|\_2 + K\sqrt{T} + MT + c\sqrt{T} \|\|\mathbf{x}\|\|\_\mathcal{C} + bT \|\|y\|\_\mathcal{C}) \\ &\leq \quad \|\|\mathbf{x}\_0\|\|\_2 + \|\|y\_0\|\|\_2 + (K + M)T + 2rT(b + c) \end{split}$$

and

$$\begin{split} \|Ly(t)\|\_{2} &\leq \quad \|y\_{0}\|\_{2} + \|\int\_{0}^{\eta} h\_{2}(s,y(s))ds\|\_{2} \|\int\_{0}^{t} f z(s,x(\phi\_{2}(s)))dW(s)\|\_{2} \\ &\leq \quad \|y\_{0}\|\_{2} = \int\_{0}^{\eta} \|h\_{2}(s,y(s))\|\_{2}ds + \sqrt{\int\_{0}^{t} \|f\_{2}(s,x(\phi\_{2}(s)))\|\_{2}^{2}ds} \\ &\leq \quad \|y\_{0}\|\_{2} + \int\_{0}^{\eta} (|h\_{2}(s)| + c\_{2} \|y(s)\|\_{2})ds + \sqrt{\int\_{0}^{t} (|m\_{2}(t)| + b\_{2} \|x\|\_{2})^{2}ds} \\ &\leq \quad \|y\_{0}\|\_{2} + KT + M\sqrt{T} + cT\|y\|\_{\mathbb{C}} + b\sqrt{T}\|x\|\_{\mathbb{C}} \\ &\leq \quad \|y\_{0}\|\_{2} + (K + M)T + T(b + c) \|y\|\_{\mathbb{C}} + T(b + c)\|x\|\_{\mathbb{C}} \\ &\leq \quad \|x\_{0}\|\_{2} + \|y\_{0}\|\_{2} + (K + M)T + 2T(b + c). \end{split}$$

This implies that

$$\begin{aligned} \|L(\mathfrak{x},\mathfrak{y})\|\_{X} &= \| (L\_1 \mathfrak{x}, L\_2 \mathfrak{y}) \|\_{X} \\ &= \max\{ \|L\_1 \mathfrak{x}(t)\|\_{\mathbb{C}^\sigma} \|L\_2 \mathfrak{y}(t)\|\_{\mathbb{C}} \} \\ &\le \|\mathfrak{x}\_0\|\_{2} + \|\mathfrak{y}\_0\|\_{2} + (K+M)T + 2rT(b+c) = r \end{aligned}$$

where

$$r = \frac{||\varkappa\_0||\_2 + ||y\_0||\_2 + (K + M)T}{1 - T(b + c)}r$$

then the class {*L*(*x*, *y*)} is uniformly bounded and *L*(*x*, *y*) : *Q* → *Q*.

**Lemma 5.** *The class of function* {*L*(*x*, *y*)(*t*)} , *t* ∈ [0, *T*] *is equicontinuous.*

**Proof.** Let *x*, *y* ∈ *Q*, *t*1, *t*<sup>2</sup> ∈ [0, *T*] such that |*t*<sup>2</sup> − *t*1| < *δ*, then

$$\begin{aligned} \|\|L\_1 \mathbf{x}(t\_2) - L\_1 \mathbf{y}(t\_1)\|\|\_2 &= \|\int\_0^{t\_2} f\_1(s, \mathbf{y}(\boldsymbol{\phi}\_1(\mathbf{s}))) ds - \int\_0^{t\_1} f\_1(s, \mathbf{y}(\boldsymbol{\phi}\_1(\mathbf{s}))) ds\|\_2 \\ &\le \int\_{t\_1}^{t\_2} \|\|f\_1(s, \mathbf{y}(\boldsymbol{\phi}\_1(\mathbf{s})))\|\|\_2 ds \\ &\le \quad (M + b\|\|\boldsymbol{y}\|\|\_C)(t\_2 - t\_1) \end{aligned} \tag{19}$$

and

$$\begin{split} \|\|L\_2\mathbf{x}(t\_2) - L\_2\mathbf{x}(t\_1)\|\|\_2 &= \|\int\_0^{t\_2} f\_2(\mathbf{s}, \mathbf{x}(\boldsymbol{\varrho}\_2(\mathbf{s}))) d\mathcal{W}(\mathbf{s}) - \int\_0^{t\_1} f\_2(\mathbf{s}, \mathbf{x}(\boldsymbol{\varrho}\_2(\mathbf{s}))) d\mathcal{W}(\mathbf{s})\|\_2 \\ &\le \sqrt{\int\_{t\_1}^{t\_2} \|f\_2(\mathbf{s}, \mathbf{x}(\boldsymbol{\varrho}\_2(\mathbf{s})))\|\_2^2 ds} \\ &\le \quad (M + b\|\|\mathbf{x}\|\_{\mathcal{C}}) \sqrt{(t\_2 - t\_1)}. \end{split} \tag{20}$$

However,

$$\begin{aligned} L(\mathbf{x}(t\_2), \mathbf{y}(t\_2)) - L(\mathbf{x}(t\_1), \mathbf{y}(t\_1)) &= (L\_1 \mathbf{x}(t\_2), L\_2 \mathbf{y}(t\_2)) - (L\_1 \mathbf{x}(t\_1), L\_2 \mathbf{y}(t\_1)) \\ &= ((L\_1 \mathbf{x}(t\_2) - L\_1 \mathbf{x}(t\_1)), (L\_2 \mathbf{y}(t\_2) - L\_2 \mathbf{y}(t\_1))), \end{aligned}$$

then from (19) and (20), we deduce the equicontinuity of the class {*L*(*x*, *y*)(*t*)} on *Q*.
