>\alpha$, there exists a unique mild pathwise solution to Eq. (\[spde-uniq\]).

Note that the unique solution for given $u_0$ is independent of $\alpha.$ The proof of Theorem \[mixuniq\] will be divided into several logical steps. We recall the following auxiliary technical lemma from [@Mishura2012mixed]. The proof can be obtained by [@Mishura2012mixed Lemma 2.1, Proposition 2.1], thus, we omit it.

\[smooth\] Let $\varpi\in(0,1]$ and $\hbar:[0,T]\rightarrow \mathbb{R}$ be a $\varpi$- Hölder continuous function. Define for $\epsilon>0,$ $$\hbar^\epsilon(t)=\epsilon^{-1}\int_{0  \vee (t-\epsilon)}^{t}\hbar(s)ds.$$ Then, for $\alpha\in(1-\varpi,1)$, there exists a constant $C>0$ such that $$\|\hbar-\hbar^\epsilon\|_{\alpha,0,T}\leq C K_{\varpi}(\hbar)\epsilon^{\varpi+\alpha-1},$$ where $K_{\varpi}(\hbar)=\sup_{0\leq s <t \leq T}\frac{|\hbar(t)-\hbar(s)|}{(t-s)^{\varpi}}$ is the $\varpi$-Hölder seminorm of $\hbar$.

Fix $N \geq 1$, we define the following stopping time $\tau_N$, $$\begin{aligned}
\label{stoptime}
\tau_N :=\inf \{t\geq 0:\Lambda^{0,t}_{\alpha,B^H} \geq N \} \wedge T,\end{aligned}$$ where $\Lambda_{\alpha,B^H}^{0,t}:=\sum_{i=1}^\infty\sqrt{\lambda_i} \Lambda_\alpha^{0,t}(\beta_i^H)$.

Put $\beta_i^{H, N}(t)=\beta^H_{i}(t\wedge \tau_N), t\in [0,T],i \in \mathbb{N}$ and taking $\epsilon=\frac{1}{n}$ in Lemma \[smooth\], then, for each $n,i \in \mathbb{N}$, define an approximation of $\beta_i^{H, N, n}$ by $$\begin{aligned}
\label{betaHn}
\beta_i^{H, N, n}(t)=n\int_{(t-\frac{1}{n})\vee 0}^{t}\beta_i^{H,N}(s)\mathrm{d}s.\end{aligned}$$

Similarly, denote $\Lambda_{\alpha,B^{H,N, n}}^{0,T} :=\sum_{i=1}^\infty\sqrt{\lambda_i} \Lambda_\alpha^{0,T}(\beta_i^{H,N,n})$ which will be used in next step.

\[betaHn1\] For any $ i\in \mathbb{N}$ and $N$, we have $$\begin{aligned}
    \|\beta_{i}^{H,N}\|_{\alpha,0,T} \leq C \|\beta^H_i\|_{\alpha,0,\tau_N},
    \end{aligned}$$ almost surely, where $C$ is a constant which is independent 