\liminf_{n\to \infty} \mathrm{R}^n_2 \leq \liminf_{n\to \infty} \mathrm{R}^n_{2,a} + \liminf_{n\to \infty} \mathrm{R}^n_{2,b} \leq C \varepsilon.
\end{align*}

Passing with $\varepsilon \to 0$ verifies (<ref>).

Step <ref> – identifying the monotone operator:
Recall that $u \in L^\alpha_t V \cap C_t H \cap B^{1/2}_{2,\infty} H$. Let $\varepsilon >0$. Choosing $\phi = u + \varepsilon \xi$ in (<ref>), where $\xi  = \zeta v$ for $\zeta \in L^\infty_t$ and $v \in V$. Dividing by $\varepsilon$ and passing with $\varepsilon \to 0$ verify the following:

\begin{align*}
& \lim_{\varepsilon \to 0} \int_0^T \psi_t \bigg[  \int_0^t  \exp\left(-\int_0^s h_r + \eta(u_r + \varepsilon \xi_r) \dd r \right) \\
&\hspace{6em} 2 \langle \overline{A}_s -  A(s,u_s + \varepsilon \xi_s), \xi_s \rangle_{V^*,V} \dd s \bigg] \dd t  \leq 0.
\end{align*}

By repeating the argument for $\varepsilon <0$ we derive

\begin{align*}
& \lim_{\varepsilon \to 0} \int_0^T \psi_t \bigg[  \int_0^t  \exp\left(-\int_0^s h_r + \eta(u_r + \varepsilon \xi_r) \dd r \right) \\
&\hspace{6em} 2 \langle \overline{A}_s -  A(s,u_s + \varepsilon \xi_s), \xi_s \rangle_{V^*,V} \dd s \bigg] \dd t = 0.
\end{align*}

Since $\psi$ was arbitrary, we conclude for all $t \in [0,T]$

\begin{align*}
& \lim_{\varepsilon \to 0} \int_0^t  \exp\left(-\int_0^s h_r + \eta(u_r + \varepsilon \xi_r) \dd r \right) 2 \zeta_s \langle \overline{A}_s -  A(s,u_s + \varepsilon \xi_s), v \rangle_{V^*,V} \dd s = 0.
\end{align*}

Similarly, since $\zeta$ was arbitrary for almost all $s\in [0,T]$,

\begin{align*}
\lim_{\varepsilon \to 0}   \exp\left(-\int_0^s h_r + \eta(u_r + \varepsilon \xi_r) \dd r \right) \langle \overline{A}_s -  A(s,u_s + \varepsilon \xi_s), v \rangle_{V^*,V} = 0.
\end{align*}

Moreover, using the local boundedness of $\eta$,

\begin{align*}
\inf_{\abs{\varepsilon}\ll 1} \exp\left(-\int_0^s h_r + \eta(u_r + \varepsilon \xi_r) \dd r \right) \geq \exp\left(- [\norm{h}_{L^1_t} + \sup_{z \in B_V}\eta(z)] \right) > 0,
\end{align*}

where $B_V$ is a ball in $V$ such that $r \mapsto u_r \pm \norm{\zeta}_{L^\infty_t} v \in B_V$. This implies for all $v \in V$ and almost all $s \in [0,T]$

\begin{align} \label{eq:one-limit}
\lim_{\varepsilon \to 0}  \langle \overline{A}_s -  A(s,u_s + \varepsilon \xi_s), v\rangle_{V^*,V}  = 0.
\end{align}

The hemicontinuity <ref> establishes for all $v \in V$ and almost all $s \in [0,T]$

\begin{align} \label{eq:second-limit}
\lim_{\varepsilon \to 0} \langle A(s,u_s)  - A(s,u_s + \varepsilon \xi_s ), v\rangle_{V^*,V} = 0.
\end{align}

Combining (<ref>) and (<ref>) shows for all $v \in V$ and almost all $s \in [0,T]$

\begin{align*}
&\langle \overline{A}_s -  A(s,u_s) , v\rangle_{V^*,V} \\
&= \lim_{\varepsilon \to 0} \langle \overline{A}_s - A(s,u_s + \varepsilon \xi_s ), v\rangle_{V^*,V} + \lim_{\varepsilon \to 0} \langle A(s,u_s + \varepsilon \xi_s )-  A(s,u_s) , v\rangle_{V^*,V} =0.
\end{align*}

Since $v \in V$ was arbitrary, the identification is complete.

The estimate (<ref>) follows from (<ref>) using the weak lower semi-continuity of the norms.

§ PROOFS: INTEGRAL OPERATORS AND YOUNG REGIMES

In this section, we verify the theorems presented in Section <ref>.

§.§ Additive Young regime

Let $\gamma > 1/2$ and $Z \in C^\gamma_t H$. Define $I_t(u) \equiv Z_t - Z_0 $.

We omit the verification of Assumption <ref> for the operator $I$ since it is trivial. The existence of a weak solution follows by Theorem <ref>. Note that besides the locally monotone operator $A$, no non-linear expression of $u$ appears in the analysis, meaning we do not require the strong convergence (<ref>) along subsequences. Thus, we do not require the Aubin-Lions lemma and therefore we may drop Assumption <ref> in this setting.

Next, we discuss uniqueness: let $u$ and $v$ be solutions to (<ref>) started in $u_0$ and $v_0$, respectively. Subtracting (<ref>) for $u$ and $v$ yield for all $w \in V$ and $t \in [0,T]$

\begin{align*}
\left(u_t - v_t - u_0 - v_0, w \right)_H = \int_0^t \langle A(s,u_s) - A(s,v_s), w \rangle_{V^*,V} \dd s.
\end{align*}

Applying Theorem <ref> with $X_t = u_t-v_t$, $Y_t = A(t,u_t) - A(t,v_t)$ and $I_t = 0$ shows

\begin{align*}
\norm{u_t - v_t}^2_H &= \norm{u_0 - v_0}^2_H + \int_0^t 2  \langle A(s,u_s) - A(s,v_s), u_s - v_s \rangle_{V^*,V} \dd s \\
&\leq \norm{u_0 - v_0}^2_H + \int_0^t (h_s+\eta(u_s))\norm{u_s-v_s}_H^2 \dd s,
\end{align*}

where we used <ref> in the last step. The assertion follows by invoking Gronwall's lemma.

§.§ Abstract Young integrals

Let $\sigma: H\to H$ be such that

\begin{equation}
\norm{\sigma(u)}_H\leq C(1+\norm{u}_H), \qquad \norm{\sigma(u)-\sigma(v)}_H\leq L \norm{u-v}_H.
\label{sigma conditions}
\end{equation}

We want to construct the abstract Young integral

\[
I_t(u):=\int_0^t \sigma(u_s) \dd X_s
\]

for some given path $X$. Note that since locally, $I$ should behave as $\sigma(u_s)(X_t-X_s)$, we need to first ensure that $X$ is of such spatial regularity that $\sigma(u_s)(X_t-X_s)\in H$; in this context, we introduced in Definition <ref> the concept of multiplier. A multiplier is a sufficient condition for the spatial compatibility of abstract Young integration.

Let $H$ be a Hilbert space and $E$ a multiplier in $H$ in the sense of Definition <ref>. Suppose $\sigma:H\to H$ satisfies (<ref>). Let $X\in C^\gamma_t E$.

Then for any $q\in (2, \infty)$ such that $\gamma+1/q>1$, the mapping

\begin{equation}
\begin{split}
L^\infty_t H\cap B^{1/2}_{2, \infty}H&\to B^\gamma_{q, \infty}H\\
u&\to I(u):=\int_0^{(\cdot)}\sigma(u_s) \dd X_s
\end{split}
\end{equation}

is well-defined and satisfies

\begin{align} \label{eq:estimate-abstract-Young}
\seminorm{I(u)}_{B^\gamma_{q, \infty} H}\lesssim(C+L) \seminorm{X}_{C^\gamma_t E}(1+\norm{u}_{L^\infty_tH}+\seminorm{u}_{B^{1/2}_{2, \infty}H}).
\end{align}

In particular, for any sequence $(X^n)_n\subset C^1_tE$ such that $X^n\to X$ in $C^\gamma_t E$ the approximation $I^n(u):=\int_0^t \sigma(u_s) \dot{X}^n_s \dd s$ satisfies condition <ref> in Assumption <ref>.

Note that we start with an element $u$ on integrability scale $2$, yet the operator needs to enjoy integrability scale $q$, i.e., we need to gain integrability. This is possible, but at the price of imposing more regularity. Indeed, we use the embedding $B^{1/2}_{2, \infty}\hookrightarrow B^{1/q}_{q, \infty}$, which ensures such a trade in order to gain integrability. The price we pay in regularity becomes apparant in the condition $\gamma+1/q>1$, in that it becomes more restrictive in $\gamma$, the larger we intend to choose $q$.

Let $u \in L^\infty_t H\cap B^{1/2}_{2, \infty}H$ and $X \in C_t^\gamma E$.

We aim to apply the sewing lemma, Lemma <ref>, to the local approximation

\begin{align*}
G_{s,t}=\langle \sigma(u)\rangle_{s,t}(X_t-X_s),
\end{align*}

where $\mean{u}_{s,t} := \abs{t-s}^{-1} \int_s^t u_r \dd r$ is the average of $u$ on $[s,t]$. Ultimately, we define the abstract Young integral as the sewing of the local approximations, i.e., for all $t \in [0,T]$

\begin{align*}
I_t(u) := \mathscr{I}_t[G].
\end{align*}

It remains to verify the sew-ability of $G$; in other words, we need to check the assumptions of the sewing lemma.

Estimating $G$: Let $t \in [0,T]$ and $h \in [0,T-t]$. Note that by (<ref>), we have

\[
\norm{G_{t, t+h}}_H\lesssim \norm{\langle\sigma(u)\rangle_{t,t+h}}_H\norm{X_{t+h}-X_t}_E\lesssim C(1+\norm{u}_{L^\infty_t H})\seminorm{X}_{C^\gamma_tE}h^\gamma,
\]

from which we conclude that

\[
\norm{G}_{\mathbb{B}^\gamma_{\infty, \infty}H}\lesssim C(1+\norm{u}_{L^\infty_t H})\seminorm{X}_{C^\gamma_tE}.
\]

Estimating $\delta G$: Let $s\leq v \leq t$. Moreover, we have

\begin{equation*}
\begin{split}
(\delta G)_{s, v, t}&=\left(\langle \sigma(u)\rangle_{s, t}- \langle \sigma(u)\rangle_{s, v}\right)(X_t-X_s)+\left(\langle \sigma(u)\rangle_{s, v}- \langle \sigma(u)\rangle_{v, t}\right)(X_t-X_v)\\
&=\left(\langle \sigma(u)\rangle_{s, t}- \sigma(u_s)\right)(X_t-X_s)-\left(\langle \sigma(u)\rangle_{s, v}- \sigma(u_s)\right)(X_t-X_s)\\
&+\left(\langle \sigma(u)\rangle_{s, v}- \sigma(u_v)\right)(X_t-X_v)-\left(\langle \sigma(u)\rangle_{v, t}- \sigma(u_v)\right)(X_t-X_v).
\end{split}
\end{equation*}

Let us further bound only the first of the above four expressions (the other ones follow by similar considerations). By Lemma <ref> we have

\begin{equation*}
\begin{split}
&\left( \int_0^{T-h}\norm{\left(\langle \sigma(u)\rangle_{s, s+h}- \sigma(u_s)\right)(X_{s+h}-X_s)}_{H}^q \dd s\right)^{1/q}\\
&\lesssim \seminorm{X}_{C^\gamma_t E}h^\gamma \left(\int_0^{T-h}\norm{\left(\langle \sigma(u)\rangle_{s, s+h}- \sigma(u_s)\right)}_H^{q} \dd s\right)^{1/q}\\
&\lesssim h^\gamma \seminorm{X}_{C^\gamma_tE}h^{1/q}L(2\norm{u}_{L^\infty_tH})^{1-2/q}\seminorm{u}_{B^{1/2}_{2, \infty}H}^{2/q}.
\end{split}
\end{equation*}

Bounding the remaining terms in a similar fashion, we obtain

\[
\bar{\Omega}_q(\delta G, h)\lesssim  h^{\gamma+1/q}\seminorm{X}_{C^\gamma_tE}L\seminorm{u}_{B^{1/2}_{2, \infty}H}.
\]

We are therefore able to apply the Besov Sewing Lemma <ref> with $\alpha = \gamma$, $p_1 = \infty$, $p_2= q$ and $\gamma = \gamma + 1/q$. Assertion (<ref>) follows from (<ref>); indeed,

\begin{align*}
\seminorm{I(u)}_{B^{\gamma}_{q,\infty} H} =  \seminorm{\mathscr{I}G }_{B^{\gamma}_{q,\infty} H} &\lesssim \norm{G}_{\mathbb{B}^\alpha_{\infty, \infty}H} + T^{1/q}\norm{\delta G}_{\bar{\mathbb{B}}^{\gamma+1/q}_{q, \infty}H} \\
&\lesssim  C(1+\norm{u}_{L^\infty_t H})\seminorm{X}_{C^\gamma_tE} + T^{1/q} \seminorm{X}_{C^\gamma_tE}L\seminorm{u}_{B^{1/2}_{2, \infty}H}.
\end{align*}

Suppose the setting of Lemma <ref> and let $(X^n)_n\subset C^1_tE$ be such that $X^n\to X$ in $C^\gamma_tE$. For $q\geq 2$ and $\gamma>1-1/q$, let $I(u)\in B^{\gamma}_{q, \infty}H$ be the abstract Young integral as defined in Lemma <ref> and $I^n(u)=\int_0^{\cdot}\sigma(u_s)\dot{X}^n_s \dd s$ be its approximate. Then the continuity property <ref> in Assumption <ref> holds.

Let us start by pointing out that in the notation of Assumption <ref>, we have $b^n(t, u)=\sigma(u)\dot{W}^n_t$. By (<ref>), it follows readily that

\begin{align*}
\norm{b^n(t, v)}_H&\lesssim C\norm{X^n}_{C^1_tE}(1+\norm{v}_H), \\
\norm{b^n(t, v)-b^n(t, u)}_H &\lesssim L\norm{X^n}_{C^1_tE}\norm{u-v}_H.
\end{align*}

We now take $v, v^n\in L^\infty_tH\cap B^{1/2}_{2, \infty}H$ such that $v^n\to v$ in $L^2_tH$. We need to show that $I^n(v_n)\to I(v)$ in $B^{\gamma}_{q, \infty}H$.
Note that due to the linearity of the integral in the integrator, we have

\[
\seminorm{I^n(v^n)-I(v^n)}_{B^\gamma_{q, \infty}}\lesssim \seminorm{X-X^n}_{C^\gamma_tE}(1+\norm{v^n}_{L^\infty_tH}+\seminorm{v^n}_{B^{1/2}_{2, \infty}H})\to 0.
\]

It thus remains to show $I(v^n)\to I(v)$ in $B^\gamma_{q, \infty}$, where we recall that $I$ is constructed as in Lemma <ref>. Towards this end, we employ Lemma <ref>, by which it suffices to show that for $G^n_{s,t}=\langle \sigma(v^n)\rangle_{s,t}(X_t-X_s)$ and $G_{s,t}=\langle \sigma(v)\rangle_{s,t}(X_t-X_s)$ we have $\norm{G^n-G}_{\mathbb{B}^\gamma_{q, \infty}}\to 0$. By Lemma <ref>, we have

\begin{align*}
&\left( \int_0^{T-h}\norm{G^n_{r, r+h}-G_{r, r+h}}_H^q \dd r\right)^{1/q}\\
&\lesssim h^\gamma \seminorm{W}_{C^\gamma_tH}\left(\int_0^{T-h}\norm{\langle \sigma(v^n)\rangle_{r,r+h}- \langle \sigma(v)\rangle_{r,r+h}}_H^q \dd r\right)^{1/q}\\
&\lesssim  h^\gamma \seminorm{W}_{C^\gamma_tH} (\norm{v^n}_{L^\infty_t H} + \norm{v}_{L^\infty_t H})^{1-2/q} L \norm{v^n-v}_{L^2_t H}^{2/q},
\end{align*}

and therefore $\norm{G^n-G}_{\mathbb{B}^\gamma_{q, \infty}}\to 0$. By Lemma <ref>, we conclude that $I(v^n)\to I(v)$ in $B^\gamma_{q, \infty}H$.

Let $I$, $I^n$ be as in Lemma <ref>.
By Lemma <ref>, the Young integral is a well defined element of $B^\gamma_{q, \infty}H$ provided $\gamma>1-1/q$. Note that Assumption <ref> and Theorem <ref> on the other hand require $\gamma>1/2+1/q$. Both constraints are satisfied for $q=4$ and $\gamma>3/4$. In this case, Lemma <ref> and <ref> assure that Assumption <ref> is satisfied. We may therefore conclude by Theorem <ref>.

§.§ Linear multiplicative Young regime

For existence, note that $E=\mathbb{R}$ is always a multiplicative ideal in $H$ in the sense of Definition <ref> and $\sigma(u)=u$ obviously satisfies (<ref>). Hence existence of weak solutions follows from Theorem <ref>.

Let $u$ and $v$ be two weak solutions to (<ref>) starting in $u_0$ and $v_0$, respectively. Using Theorem <ref> with $X_t = u_t -v_t$, $Y_t = A(t,u_t) - A(t,v_t)$ and $I_t = I_t(u-v)$ (notice that we used the linearity of $I$ at this point); and subtracting the resulting equations for $t > s$ yield

\begin{align*}
\norm{u_t-v_t}^2_H&=  \norm{u_s-v_s}^2_H+\int_s^t\langle A(r, u_r)-A(r, v_r), u_r-v_r\rangle_{V^*, V} \dd r\\
&+\mathscr{S}_{t}(u-v,  \dd I(u-v) ) - \mathscr{S}_{s}(u-v,  \dd I(u-v) ).
\end{align*}

Remark that $Z_t:=\norm{u_t-u_t}^2_H$ satisfies

\[
|Z_t-Z_s|\leq 2(\norm{u}_{C_tH}+\norm{v}_{C_tH})(\norm{u_t-u_s}_H+\norm{v_t-v_s}_H)
\]

by which we conclude that $Z\in B^{1/2}_{2, \infty}\mathbb{R}$. Hence, using the sewing Lemma <ref>, it can be easily verified that

\begin{align*}
\mathscr{S}_{t}(u-v,  \dd I(u-v) ) - \mathscr{S}_{s}(u-v,  \dd I(u-v) ) = \int_s^t \norm{u_r-v_r}_H^2 \dd \beta_r.
\end{align*}

Due to <ref> it holds moreover that $\int_s^t\langle A(r, u_r)-A(r, v_r), u_r-v_r\rangle_{V^*, V} \dd r \leq 0$. Thus, the above considerations lead to the inequality: for all $s\leq t$

\begin{equation}
Z_t\leq Z_s+\int_s^t Z_r \dd \beta_r.
\label{z inequality}
\end{equation}

Remark that by definition $Z\geq 0$. Let us further define

\[
B_t:=\int_0^t Z_r \dd \beta_r-Z_t.
\]

By (<ref>), $B$ is monotone increasing. Hence, the integral

\[
\int_0^t \exp{(-\beta_r)} \dd B_r\geq 0
\]

is well defined. Using Lemma <ref>, we obtain

\begin{align*}
0&\leq \int_0^t \exp{(-\beta_r)}\dd B_r=\int_0^t  \exp{(-\beta_r)} Z_r \dd \beta_r- \int_0^t\exp{(-\beta_r)}\dd Z_r\\
&=-\int_0^t \dd (Z_r\exp{(-\beta_r)}) = -Z_t\exp{(-\beta_t)}+Z_0\exp{(-\beta_0)},
\end{align*}

which finishes the proof.

§ EXAMPLES

In this section we present three equations that fall in the general framework of Theorem <ref>: the $p$-Laplace equation, the porous medium equation and an equation that models shear-thickening fluids. We derive the existence of weak solutions for additive drivers and Young integrals. Moreover, we show that our framework captures regularization by noise effects for the $p$-Laplace equation.

§.§ The p-Laplace equation

Let $\Lambda\subset \mathbb{R}^d$ be a bounded Lipschitz domain and $p\in (\frac{2d}{d+2}, \infty)$.

§.§.§ The Gelfand triple
We set $V=W^{1, p}_0(\Lambda)$ and $H=L^2(\Lambda)$.

Since $p>\frac{2d}{d+2}$, the Rellich-Kondrachov theorem ensures that the embedding $W^{1, p}_0(\Lambda)\hookrightarrow L^2(\Lambda)$ is compact. Thus, Assumption <ref> is satisfied.

§.§.§ The monotone operator

The $p$-Laplace operator $\Delta_p: V\to V^*$ is defined by

\begin{equation}
\begin{split}
\langle \Delta_p u, v\rangle_{V^*,V}:=-\int_\Lambda |\nabla u|^{p-2} \nabla u\cdot \nabla v \dd x.
\label{p-laplace}
\end{split}
\end{equation}

It is now well-known that $\Delta_p$ satisfies <ref>; <ref>; <ref> with $c_1=1, c_2=0, f=0$; and <ref> with $c_3=1, g=0$ (see <cit.>). Thus, Assumption <ref> holds.

Let $T > 0$ and $u_0 \in L^2(\Lambda)$. Let $I$ be given by one of the following:

* (Additive driver)  Let $\gamma > 1/2$ and $Z \in C^\gamma(0,T;L^2(\Lambda))$.

Define $I_t(u) := Z_t - Z_0$;

* (Young integral) Let $\gamma > 3/4$, $X\in C^\gamma(0,T; L^\infty(\Lambda))$ and $\sigma:\mathbb{R}\to \mathbb{R}$ satisfy

\[
|\sigma(x)|\leq C(1+|x|), \qquad |\sigma(x)-\sigma(y)|\leq L|x-y|.
\]

Define $I_t(u) := \int_0^t \sigma(u_s) \dd X_s$ as the Young integral given by Lemma <ref>;

Then there exists a (in Case <ref> unique) weak solution

\begin{align*}
u \in C([0,T]; L^2(\Lambda)) \cap B^{1/2}_{2,\infty}(0,T; L^2(\Lambda)) \cap L^p(0,T; W^{1,p}_{0}(\Lambda))
\end{align*}

\begin{align*}
\dd u - \Delta_p u \dd t = \dd I_t(u), \qquad u(0) = u_0.
\end{align*}

The first part of the statement follows directly from Theorem <ref>.

For the second part of the statement, note that $\sigma$ gives rise to a Nemytskii operator in $H=L^2(\Lambda)$ satisfying (<ref>). Moreover, $E=L^\infty(\Lambda)$ is a multiplicative ideal in $H=L^2(\Lambda)$ in the sense of Definition <ref>. Hence, we may conclude by Theorem <ref>.

§.§.§ Regularization by noise
It is well-known that randomness can lead to regularization; an effect nowadays called regularization by noise. In particular, the translation with respect to a rough process $w$ is regularizing. One main tool behind the regularization effect is a formula that relates classical integration of differences and the sewing of a convolution with respect to the local time $L^w$ of $w$ (for brief overview of its basic definition and some properties, refer to the corresponding section in the Appendix <ref>) evaluated at a time-averaged function value $\mean{u}$ of $u$:

\begin{align} \label{eq:main-formula-reg-by-noise}
\int_0^t b(u_s - w_s) \dd s = \mathscr{I}_t \big( b * L^w (\mean{u}) \big).
\end{align}

Let us briefly motivate (<ref>). Since we assume $w$ to be a highly oscillating noise, we would expect that on small time scales, i.e. $|t-s|\ll 1$, we have

\[
\int_s^tb(u_r-w_r)\dd r\simeq \int_s^tb(\langle u\rangle_{s, t}-w_r) \dd r=(b*L^w_{s,t})(\langle u\rangle_{s,t})
\]

where in the last step, we used the occupation times formula. Sewing together these small scale approximation, one can recover (<ref>), provided $L^w$ is sufficiently regular with respect to $u$ and $b$.
Note in particular that the right-hand-side can in principle still makes sense for irregular $b$; in other words, $w$ can regularizes (<ref>) through the regularity of its local time.

Our abstract framework captures this regularization effect by choosing

\begin{align*}
I_t(u) := \mathscr{I}_t \big( b * L^w (\mean{u}) \big).
\end{align*}

It generalizes our results on regularization by noise for the $p$-Laplace equation [11] to a broader class of singularities $b$.

The next lemma justifies (<ref>).

Let $q \in [2,\infty]$, $\gamma > 1-1/q$, $b \in \mathscr{S}'(\mathbb{R})$ be a Schwartz distribution, and $w:[0,T] \to \mathbb{R}$ such that its local time $L^w$ satisfies $b * L^w \in C^\gamma(0,T;C^{0,1}(\mathbb{R}))$.

Then the generalized integral

\begin{equation}
\begin{split}
L^\infty(0,T; L^2(\Lambda))\cap B^{1/2}_{2, \infty}(0,T;L^2(\Lambda)) &\to B^\gamma_{q, \infty}(0,T;L^2(\Lambda))\\
u&\to I(u)=  \mathscr{I} \big( b * L^w (\mean{u}) \big),
\end{split}
\end{equation}

where $\mathscr{I}_t \big( b * L^w (\mean{u}) \big)$ is the sewing of the local approximation $G_{s,t} = b* L_{s,t}^w( \mean{u}_{s,t})$ defined in Lemma <ref>, is well-defined and satisfies

\begin{align} \label{eq:Integral-reg-by-noise}
\begin{aligned}
&\seminorm{I(u)}_{B^\gamma_{q, \infty}(0,T;L^2(\Lambda))}\\
&\hspace{2em} \lesssim \norm{b*L^w}_{C^\gamma(0,T;C^{0,1}(\mathbb{R}))}(1+\norm{u}_{L^\infty(0,T;L^2(\Lambda))}+\seminorm{u}_{B^{1/2}_{2, \infty}(0,T;L^2(\Lambda))}).
\end{aligned}
\end{align}

If additionally $b$ is more regular, i.e., it satisfies

\begin{equation}
\label{b nice}
|b(x)|\leq C(1+|x|), \qquad |b(x)-b(y)|\leq L |x-y|,
\end{equation}

then the generalized integral has the alternative representation: for all $t \in [0,T]$

\begin{align*}
I_t(u) = \int_0^t b(u_s - w_s) \dd s.
\end{align*}

To prove the first part, we need to show that the germ $G_{s,t}=(b*L^w_{s,t})(\langle u\rangle_{s,t})$ admits a sewing in $L^2(\Lambda)$.

Firstly, for $r \in [0,T]$ and $h \in [0,T-r]$,

\begin{align*}
\norm{G_{r, r+h}}_{L^2(\Lambda)}&=\norm{(b*L^w_{r,r+h})(\langle u\rangle_{r,r+h})}_{L^2(\Lambda)} \\
&\lesssim \norm{b *L^w_{r,r+h}}_{L^\infty(\mathbb{R})}\leq \norm{b*L^w}_{C^\gamma(0,T; C^{0,1}(\mathbb{R}))} h^\gamma,
\end{align*}

from which we conclude that

\[
\norm{G}_{\mathbb{B}^\gamma_{\infty, \infty}(0,T;L^2(\Lambda))}\lesssim  \norm{b*L^w}_{C^\gamma(0,T; C^{0,1}(\mathbb{R}))}.
\]

Secondly, we have

\begin{align*}
(\delta G)_{s,v,t}&=(b*L^w_{s,v})(\langle u\rangle_{s,t})-(b*L^w_{s,v})(\langle u\rangle_{s,v})+(b*L^w_{v,t})(\langle u\rangle_{s,t})-(b*L^w_{v,t})(\langle u\rangle_{v,t})\\
&=(b*L^w_{s,v})(\langle u\rangle_{s,t})-(b*L^w_{s,v})(u_s)+(b*L^w_{s,v})(u_s)-(b*L^w_{s,v})(\langle u\rangle_{s,v})\\
&+(b*L^w_{v,t})(\langle u\rangle_{s,t})-(b*L^w_{v,t})(u_t)+(b*L^w_{v,t})(u_t)-(b*L^w_{v,t})(\langle u\rangle_{v,t})
\end{align*}

For the first difference term, we employ Lemma <ref> for $H=L^2(\Lambda)$ and $\sigma(x)=(b*L^w_{r, r+\theta h})(x)$ with $\theta\in [0, 1]$ to obtain

\begin{align*}
&\left( \int_0^{T-h} \norm{(b^n*L_{r,r+\theta h})(\langle u\rangle_{r, r+h})-(b^n*L_{r, r+\theta h})(u_r)}^q_{L^2_x}dr\right)^{1/q}\\
&\leq \norm{(b^n*L_{r, r+\theta h})}_{Lip_x} h^{1/q}(2\norm{u}_{L^\infty_t L^2_x})^{1-2/q}\seminorm{u}^{2/q}_{B^{1/2}_{2, \infty}L^2_x}\\
&\leq h^{\gamma+1/q}\norm{b^n*L}_{C^\gamma_tLip_x}(2\norm{u}_{L^\infty_t L^2_x})^{1-2/q}\seminorm{u}^{2/q}_{B^{1/2}_{2, \infty}L^2_x}
\end{align*}

Bounding the remaining terms in a similar fashion, we obtain

\[
\bar{\Omega}_q(\delta G, h)\lesssim h^{\gamma+1/q}\norm{b^n*L}_{C^\gamma_tLip_x}(2\norm{u}_{L^\infty_t L^2_x})^{1-2/q}\seminorm{u}^{2/q}_{B^{1/2}_{2, \infty}L^2_x}
\]

meaning that for $\gamma+1/q>1$, by the Sewing Lemma <ref> $G$ admits a sewing. By the a priori bound that comes with the Sewing Lemma <ref>, this concludes the claim.

It can then be shown that this sewing coincides with the Bochner integral $I^n(u)$ in the statement by means of Lemma <ref>.

We only need to check Assumption <ref>, since Assumptions <ref> and <ref> have already been verified. The assertion then follows by Theorem <ref>.

The approximate operator:
We define the Nemytskii operator $ \sigma^n: [0,T] \times L^2(\Lambda) \to L^2(\Lambda)$ by

\begin{align*}
\sigma^n(t,u) := b^n*L^w_t(u).
\end{align*}

Clearly, $\sigma^n$ is $\mathcal{B}(0,T) \otimes \mathcal{B}(L^2(\Lambda))$-measurable.

Next, we show that $\sigma^n$ satisfies the linear growth and Lipschitz condition. By the occupation times formula (<ref>) it holds

\begin{align*}
b^n*L^w_t(u) = \int_{\mathbb{R}} b^n(u -z) L^w_t(z) \dd z = \int_0^t b^n(u -w_s) \dd s.
\end{align*}

Now, Hölder's inequality implies

\begin{align*}
\norm{\sigma^n(t,u)}_{L^2(\Lambda)}^2 &= \int_{\Lambda} \abs{\int_0^t b^n(u(x) -w_s) \dd s }^2 \dd x \\
&\leq t \int_{\Lambda} \int_{0}^t \abs{b^n(u(x) -w_s)}^2 \dd s \dd x \\
&\leq 2t^2 C_n^2 \left( (1 + \norm{w}_{C([0,t];\mathbb{R})})^2 \abs{\Lambda}+ \norm{u}_{L^2(\Lambda)}^2   \right).
\end{align*}

\begin{align*}
\norm{\sigma^n(t,u) - \sigma^n(t,v)}_{L^2(\Lambda)}^2 &\leq t \int_{\Lambda} \int_0^t \abs{b^n(u(x) -w_s) - b^n(v(x) -w_s) }^2 \dd s \dd x \\
&\leq t^2 C_n^2 \norm{u -v}_{L^2(\Lambda)}^2.
\end{align*}

Define $I^n_t(u) = \int_0^t \sigma^n(s, u_s) \dd s$.

Ad <ref>:
An application of Lemma <ref> with $q=4$ and $\gamma > 3/4$ implies

\begin{align*}
I^n_t(u) = \int_0^t \sigma^n(s, u_s) \dd s = \int_0^t b^n(u_s - w_s) \dd s.
\end{align*}

Following the derivation of (<ref>) (replacing $(0,T)$ by $(s,t)$ for arbitrary $s<t \in [0,T]$) it holds

\begin{align*}
&\seminorm{I^n(u)}_{B^\gamma_{4, \infty}(s,t;L^2(\Lambda))}\\
&\hspace{2em} \lesssim \norm{b^n*L^w}_{C^\gamma([s,t];C^{0,1}(\mathbb{R}))}(1+\norm{u}_{L^\infty(s,t;L^2(\Lambda))}+\seminorm{u}_{B^{1/2}_{2, \infty}(s,t;L^2(\Lambda))}).
\end{align*}

Thus, <ref> holds with $\lambda(\abs{t-s}):= \sup_{n \in \mathbb{N}}  \norm{b^n*L^w}_{C^\gamma([s,t];C^{0,1}(\mathbb{R}))}$ and some constant $c_4$ independent of $n$.

Ad <ref>: Let us take $(u^n)_{n \in \mathbb{N}}, u \in  L^\infty(0,T; L^2(\Lambda))\cap B^{1/2}_{2, \infty}(0,T;L^2(\Lambda))$ such that

\begin{align*}
\sup_{n \in \mathbb{N}} \norm{u^n}_{ L^\infty(0,T; L^2(\Lambda))} + \norm{u^n}_{B^{1/2}_{2, \infty}(0,T;L^2(\Lambda))} < \infty,
\end{align*}

and $u^n \to u \in L^2(0,T;L^2(\Lambda))$. We need to show $I^n(u^n)\to I(u)$ in $B^\gamma_{q, \infty} (0,T;L^2(\Lambda))$.

Note that by construction, we have

\begin{align*}
&\seminorm{I^n(u^n)-I(u^n)}_{B^\gamma_{q, \infty}(0,T;L^2(\Lambda))}\lesssim \norm{(b^n*L)-(b*L)}_{C^\gamma(0;T;C^{0,1}(\Lambda))}\\
&\hspace{13em}(1+\norm{u^n}_{ L^\infty(0,T; L^2(\Lambda))} + \norm{u^n}_{B^{1/2}_{2, \infty}(0,T;L^2(\Lambda))})\to 0.
\end{align*}

An application of Lemma <ref> with $q=4$ and $\gamma > 3/4$ shows $I(u^n)\to I(u)$ in $B^\gamma_{q, \infty}(0,T;L^2(\Lambda))$.

At this point, we have verified that Assumption <ref> holds for $q=4$ and $\gamma > 3/4$. Therefore the claim follows by Theorem <ref>.

Let $q \in [2,\infty)$, $\gamma > 1-1/q$, $b \in \mathscr{S}'(\mathbb{R})$ be a Schwartz distribution, and $w:[0,T] \to \mathbb{R}$ such that its local time $L^w$ satisfies $b * L^w \in C^\gamma(0,T;C^{0,1}(\mathbb{R}))$.

Moreover, let $(u^n)_{n \in \mathbb{N}}, u \in  L^\infty(0,T; L^2(\Lambda))\cap B^{1/2}_{2, \infty}(0,T;L^2(\Lambda))$ such that

\begin{align*}
\sup_{n \in \mathbb{N}} \norm{u^n}_{ L^\infty(0,T; L^2(\Lambda))} + \norm{u^n}_{B^{1/2}_{2, \infty}(0,T;L^2(\Lambda))} < \infty,
\end{align*}

and $u^n \to u \in L^2(0,T;L^2(\Lambda))$.

Then $I(u^n) \rightarrow I(u) \in B^{\gamma}_{q,\infty}(0,T;L^2(\Lambda))$.

The assertion follows from Lemma <ref> provided we show that for $G^n_{s,t}=(b*L^w_{s,t})(\langle u^n\rangle_{s,t})$ and $G_{s,t}=(b*L^w_{s,t})(\langle u \rangle_{s,t})$ it holds $\norm{G^n-G}_{\mathbb{B}^\gamma_{q, \infty}(0,T;(L^2(\Lambda))}\to 0$.

Using Lemma <ref> we have

\begin{align*}
&\left( \int_0^{T-h}\norm{G^n_{r, r+h}-G_{r, r+h}}_{L^2(\Lambda)}^q \dd r\right)^{1/q}\\
&\leq \norm{b*L^w_{r, r+h}}_{C^{0,1}(\mathbb{R})}(\norm{u^n}_{L^\infty(0,T;L^2(\Lambda))}+\norm{u}_{L^\infty(0,T;L^2(\Lambda))})^{1-2/q}\norm{u^n-u}^{2/q}_{L^2(0,T;L^2(\Lambda))}\\
&\lesssim h^\gamma \norm{b*L^w}_{C^\gamma(0,T;C^{0,1}(\mathbb{R})}\norm{v^n-v}^{2/q}_{L^2_tL^2_x},
\end{align*}

from which we conclude that indeed $\norm{G^n-G}_{\mathbb{B}^\gamma_{q, \infty}(0,T;L^2(\Lambda))}\to 0$.

§.§ Porous medium equation

Let $\Lambda\subset \mathbb{R}^d$ be a bounded Lipschitz domain and $p\in (\frac{2d}{d+2}, \infty)$.

§.§.§ The Gelfand triple
We set $V=L^p(\Lambda)$ and $H=W^{-1, 2}(\Lambda)$.

By duality and the Rellich-Kondrachov theorem $p>\frac{2d}{d+2}$ ensures again that $L^p(\Lambda)\hookrightarrow W^{-1, 2}(\Lambda)$ is compact. Thus, Assumption <ref> is satisfied.

§.§.§ The monotone operator

Let $\Psi: \mathbb{R}\to \mathbb{R}$ be a function satisfying:

($\Psi$1) $\Psi$ is continuous;

($\Psi$2) $\Psi$ is monotone in the sense that $(\Psi(t)-\Psi(s))\cdot (t-s)\geq 0$ for all $t, s\in \mathbb{R}$;

($\Psi$3) $\Psi$ is coercive in the sense that there exist $a\in (0, \infty)$ and $c\in [0, \infty)$ such that for all $s\in \mathbb{R}$

\[
s\cdot \Psi(s)\geq a|s|^{p}-c;
\]

There exist $c_3, c_4$ such that for all $s\in \mathbb{R}$

\[
|\Psi(s)|\leq c_4+c_3|s|^{p-1}.
\]

It can then be shown that for any $\Psi$ satisfying <ref>-<ref>, the operator

\begin{equation}
\begin{split}
A: V&\to V^*\\
u&\to \Delta \Psi(u)
\label{porous medium operator}
\end{split}
\end{equation}

is well-defined and satisfies the Conditions <ref>-<ref>, refer to <cit.>. Thus, Assumption <ref> holds.

Let $T > 0$ and $u_0 \in W^{-1,2}(\Lambda)$. Let $I$ be given by one of the following:

* (Additive driver)  Let $\gamma > 1/2$ and $Z \in C^\gamma(0,T;W^{-1,2}(\Lambda))$.

Define $I_t(u) := Z_t - Z_0$;

* (Young integral) Let $\gamma > 3/4$, $\varepsilon > 0$, $X\in C^\gamma(0,T; C^{1+\varepsilon}(\Lambda))$ and $\sigma: W^{-1, 2}(\Lambda)\to W^{-1, 2}(\Lambda)$ satisfy

\[
\norm{\sigma(u)}_{W^{-1, 2}(\Lambda)}\leq C(1+\norm{u}_{W^{-1, 2}(\Lambda)}), \quad \norm{\sigma(u)-\sigma(v)}_{W^{-1, 2}(\Lambda)}\leq L\norm{u-v}_{W^{-1, 2}(\Lambda)}.
\]

Define $I_t(u) := \int_0^t \sigma(u_s) \dd X_s$ as the Young integral given by Lemma <ref>;

Then there exists a (in Case <ref> unique) weak solution

\begin{align*}
u \in C([0,T]; W^{-1, 2}(\Lambda)) \cap B^{1/2}_{2,\infty}(0,T; W^{-1, 2}(\Lambda)) \cap L^p(0,T; L^p(\Lambda))
\end{align*}

\begin{align*}
\dd u - \Delta \Psi(u) \dd t = \dd I_t(u), \qquad u(0) = u_0.
\end{align*}

The first part of the statement follows again directly from Theorem <ref>. Using Theorem <ref>, we note that

\[
\norm{u\cdot v}_{W^{-1, 2}(\Lambda)}\leq C\norm{u}_{W^{-1, 2}(\Lambda)}\norm{v}_{C^{1+\epsilon}(\Lambda)}
\]

for $\epsilon>0$. Thus, $E=C^{1+\epsilon}(\Lambda)$ is a multiplier in $H=W^{-1, 2}(\Lambda)$ in the sense of Definition <ref>. Now, the second claim follows from Theorem <ref>.

§.§ Shear-thickening fluids

Power-law fluids are non-Newtonian fluids, whose rheology – the interaction of shear-stress and strain-rate – is described by a power-law behaviour. Here, we assume that this rheology is given by

\begin{align*}
S(B) := \mu(1 + \abs{B})^{p-2} B \in \mathbb{R}^{d \times d },
\end{align*}

where $p\in (1, \infty)$ and $\mu > 0$. More details can be found in e.g. [7].

The evolution equation for velocity field $u$ and pressure $\pi$ of an unforced incompressible power-law fluid reads

\begin{align} \label{eq:Power-law-fluids}
\begin{cases}
\partial_t u - \Div \,S(\varepsilon u) + (u\cdot \nabla)u + \nabla \pi &= 0,\\
\hfill \Div \, u &= 0.
\end{cases}
\end{align}

Here $\varepsilon u \in \mathbb{R}^{d\times d}$ denotes the symmetric gradient.

We show that if the fluid is shear-thickening for a sufficiently large power-law index, our general framework provides the existence of weak solutions for the forced evolution equation (<ref>). We only discuss the existence of velocity. The pressure can be reconstructed once the existence of velocity is established.

From now on, let $d \in \mathbb{N}$, $p\in (d,\infty) \cap [3, \infty)$ and $\Lambda\subset \mathbb{R}^d$ be a bounded Lipschitz domain.

§.§.§ The Gelfand triple

We set

\begin{align*}
V &=W^{1, p}_{0,\Div}(\Lambda) = \overline{\{ u\in  C^\infty_c(\Lambda):\, \Div \, u =0 \}}^{\norm{\cdot}_{W^{1,p}(\Lambda)}}, \\
H &= L_{\Div}^2(\Lambda) =  \overline{\{ u\in  C^\infty_c(\Lambda):\, \Div \, u =0 \}}^{\norm{\cdot}_{L^{2}(\Lambda)}}.
\end{align*}

§.§.§ The locally monotone operator

We define the operator $A: V\to V^*$ by

\begin{align} \label{eq:power-law-operator}
\forall v \in V: \quad   \langle A (u), v\rangle_{V^*,V}:=-\int_\Lambda S(\varepsilon u) : \varepsilon v + (u \cdot \nabla)u \cdot v \dd x.
\end{align}

The following result establishes the existence and conditional uniqueness of the velocity field for shear-thickening fluids.

Let $T > 0$, $u_0 \in L^2_{\Div}(\Lambda)$, and $I$ satisfy Assumption <ref>.

Then there exists a weak solution

\begin{align*}
u \in C([0,T];L^2_{\Div}(\Lambda)) \cap B^{1/2}(0,T;L^2_{\Div}(\Lambda)) \cap L^{p}(0,T; W^{1,p}_{0,\Div}(\Lambda))
\end{align*}

such that for all $t \in [0,T]$ and $v \in W^{1,p}_{0,\Div}(\Lambda)$ it holds

\begin{align*}
\int_{\Lambda} (u_t - u_0) \cdot v \dd x + \int_0^t\int_\Lambda S(\varepsilon u) : \varepsilon v + (u \cdot \nabla)u \cdot v \dd x \dd s = \int_{\Lambda} I_t(u)\cdot v \dd x.
\end{align*}

In particular, if $I_t(u) =Z_t - Z_0$ for $Z \in C^\gamma(0,T;L^2_{\Div}(\Lambda))$ with $\gamma > 1/2$, then this weak solution is unique.

Notice that $Z = 0$ is a valid choice in Theorem <ref>, recovering existence and uniqueness for the velocity field of unforced shear-thickening fluids.

Existence follows from Theorem <ref>, while uniqueness is derived by Theorem <ref>. It remains to verify their assumptions.

Ad Assumption <ref>: By the Rellich-Kondrachov theorem, it follows that the embedding $W^{1, p}_{0,\Div}(\Lambda)\hookrightarrow L^2_{\Div}(\Lambda)$ is compact.

Closely related to the tensor $S$ is the tensor

\begin{align*}
F(B) := \sqrt{\mu} (1 + \abs{B})^{(p-2)/2} B.
\end{align*}

The tensor $F$ quantifies the monotonicity of the tensor $S$ and is a convenient tool for the verification of Assumption <ref>. More details on the relation between $F$ and $S$ can be found in e.g. [13, 3, 51].

Keep in mind that by Korn's and Poincaré's inequalities the following norms on $W^{1,p}_{0,\Div}(\Lambda)$ are equivalent:

\begin{align*}
\norm{\varepsilon u}_{L^p(\Lambda)} \eqsim \norm{\nabla u}_{L^p(\Lambda)} \eqsim \norm{ u}_{W^{1,p}(\Lambda)}.
\end{align*}

Next, we verify Assumption <ref>.

Ad <ref>: Let $u,v,w \in W^{1,p}_{0,\Div}(\Lambda)$ and $\lambda \in \mathbb{R}$. We split

\begin{align*}
&\langle A (u + \lambda v) ,w\rangle_{V^*,V}  \\
&\hspace{2em} = -\int_\Lambda S(\varepsilon u + \lambda \varepsilon v) : \varepsilon w \dd x - \int_{\Lambda} \big( (u + \lambda v) \cdot \nabla\big)(u + \lambda v) \cdot w \dd x =: \mathrm{K}_1 + \mathrm{K}_2.
\end{align*}

We show that $\mathrm{K}_1$ and $\mathrm{K}_2$ are continuous functions in $\lambda$. Let $\lambda_1, \lambda_2 \in \mathbb{R}$.

Continuity of $\mathrm{K}_1$: Due to continuity of $S$ it holds

\begin{align*}
\big( S(\varepsilon u + \lambda_1 \varepsilon v) -  S(\varepsilon u + \lambda_2 \varepsilon v) \big) : \varepsilon w \rightarrow 0 \quad \text{ for } \quad \lambda_1 \to \lambda_2.
\end{align*}

Moreover, using Hölder's inequality and $\abs{S(B)}^{p'} \lesssim \abs{B}^p + 1$,

\begin{align*}
&\int_\Lambda \big( S(\varepsilon u + \lambda_1 \varepsilon v) - S(\varepsilon u + \lambda_2 \varepsilon v) \big): \varepsilon w \dd x \\
&\hspace{2em} \leq \big(\norm{S(\varepsilon u + \lambda_1 \varepsilon v)}_{L^{p'}(\Lambda)} +\norm{S(\varepsilon u + \lambda_2 \varepsilon v)}_{L^{p'}(\Lambda)} \big)  \norm{\varepsilon w}_{L^p(\Lambda)} \\
&\hspace{2em} \lesssim \big( \norm{\varepsilon u + \lambda_1 \varepsilon v}_{L^p(\Lambda)}^p +\norm{\varepsilon u + \lambda_2 \varepsilon v}_{L^p(\Lambda)}^p + \abs{\Lambda} \big)^{1/p'}  \norm{\varepsilon w}_{L^p(\Lambda)} < \infty.
\end{align*}

Therefore, dominated convergence ensures that

\begin{align*}
\lambda \mapsto \int_\Lambda S(\varepsilon u + \lambda \varepsilon v) : \varepsilon w \dd x
\end{align*}

is continuous.

Continuity of $\mathrm{K}_2$: Hölder's inequality implies

\begin{align*}
&\int_{\Lambda} \big[ \big( (u + \lambda_1 v) \cdot \nabla \big)(u + \lambda_1 v) - \big( (u + \lambda_2 v) \cdot \nabla \big)(u + \lambda_2 v) \big] \cdot w \dd x \\
&\hspace{2em} = (\lambda_1 - \lambda_2) \int_{\Lambda} \big( v \cdot \nabla)(u + \lambda_1 v) + \big( (u + \lambda_2 v) \cdot \nabla) v \cdot w \dd x \\
&\hspace{2em} \leq  \abs{\lambda_1 - \lambda_2} \big( \norm{\nabla u + \lambda_1 \nabla v}_{L^{p}(\Lambda)} \norm{v \otimes w }_{L^{p'}(\Lambda)} \\
&\hspace{9em} + \norm{\nabla v}_{L^{p}(\Lambda)} \norm{(u + \lambda_2 v) \otimes w}_{L^{p'}(\Lambda)} \big).
\end{align*}

Notice that, since $p \geq 3d/(d+2)$ it holds $W^{1,p}_{0}(\Lambda) \hookrightarrow L^{2p'}(\Lambda)$,

\begin{align*}
\norm{v \otimes w }_{L^{p'}(\Lambda)} \leq \norm{v}_{L^{2p'}(\Lambda)} \norm{w}_{L^{2p'} (\Lambda)} \lesssim \norm{\nabla v}_{L^p(\Lambda)}  \norm{\nabla w}_{L^p(\Lambda)}.
\end{align*}

\begin{align*}
&\norm{\nabla u + \lambda_1 \nabla v}_{L^{p}(\Lambda)} \norm{v \otimes w }_{L^{p'}(\Lambda)}  + \norm{\nabla v}_{L^{p}(\Lambda)} \norm{(u + \lambda_2 v) \otimes w}_{L^{p'}(\Lambda)} \\
&\hspace{2em}  \lesssim \norm{\nabla u + \lambda_1 \nabla v}_{L^{p}(\Lambda)} \norm{\nabla v}_{L^p(\Lambda)}  \norm{\nabla w}_{L^p(\Lambda)} \\
&\hspace{4em} + \norm{\nabla v}_{L^{p}(\Lambda)} \norm{\nabla u + \lambda_2 \nabla v}_{L^p(\Lambda)}  \norm{\nabla w}_{L^p(\Lambda)}  < \infty.
\end{align*}

This implies the continuity of

\begin{align*}
\lambda \to \int_{\Lambda} \big( (u + \lambda v) \cdot \nabla\big)(u + \lambda v) \cdot w \dd x.
\end{align*}

Thus, Condition <ref> holds.

Ad <ref>: Let $u,v \in W^{1,p}_{0,\Div}(\Lambda)$. We split

\begin{align*}
&2\langle A (u) - A(v),u- v\rangle_{V^*,V} = -2\int_\Lambda \big(S(\varepsilon u) - S(\varepsilon v)\big) : \big(\varepsilon u- \varepsilon v \big) \dd x \\
&\hspace{4em} -2 \int_{\Lambda} \big( (u \cdot \nabla)u - (v \cdot \nabla)v \big) \cdot (u-v) \dd x =: \mathrm{J}_1 + \mathrm{J}_2,
\end{align*}

into strongly monotone and lower-order parts.

Using <cit.>, we find

\begin{align} \label{eq:strong-dissipation}
\begin{aligned}
2\int_\Lambda \big(S(\varepsilon u) - S(\varepsilon v)\big) : \big(\varepsilon u- \varepsilon v \big) \dd x  &\eqsim \norm{F( \varepsilon u) - F(\varepsilon v)}_{L^2(\Lambda)}^2 \\
&\geq c \mu \norm{\varepsilon (u - v)}_{L^2(\Lambda)}^2.
\end{aligned}
\end{align}

Controlling $\mathrm{J}_2$: We write

\begin{align*}
& \int_{\Lambda} \big( (u \cdot \nabla)u - (v \cdot \nabla)v \big) \cdot (u-v) \dd x \\
&\hspace{2em} =  \int_{\Lambda} \big( (u \cdot \nabla)(u-v) + ((u-v) \cdot \nabla)v \big) \cdot (u-v) \dd x.
\end{align*}

Notice that, using integration by parts and the divergence-free condition,

\begin{align} \label{eq:Vanish}
\int_{\Lambda} (u \cdot \nabla)(u-v)  \cdot (u-v) \dd x = - \frac{1}{2} \int_{\Lambda} \Div\, u \abs{u-v}^2 \dd x = 0.
\end{align}

Similarly, together with Hölder's inquality and $W^{1,p}_0(\Lambda) \hookrightarrow L^\infty(\Lambda)$,

\begin{align*}
\int_{\Lambda} ((u-v) \cdot \nabla)v  \cdot (u-v) \dd x &= -   \int_{\Lambda} ((u-v) \cdot \nabla)(u-v)  \cdot v \dd x \\
&\leq \norm{\nabla (u-v)}_{L^2(\Lambda)} \norm{ u-v}_{L^{2}(\Lambda)} \norm{v}_{L^\infty(\Lambda)} \\
&\lesssim \norm{\nabla (u-v)}_{L^2(\Lambda)} \norm{ u-v}_{L^{2}(\Lambda)} \norm{\nabla v}_{L^p(\Lambda)}.
\end{align*}

Korn's inequality and weighted Young's inequality imply

\begin{align} \label{eq:bound-lower-order}
\begin{aligned}
\mathrm{J}_2 &\leq \delta  \norm{\varepsilon (u-v)}_{L^2(\Lambda)}^2 + c_\delta \norm{ u-v}_{L^2(\Lambda)}^2  \norm{\nabla v}_{L^{p}(\Lambda)}^{2}.
\end{aligned}
\end{align}

Combining (<ref>) and (<ref>), and choosing $\delta >0$ sufficiently small, yield

\begin{align*}
2\langle A (u) - A(v),u- v\rangle_{V^*,V} &\leq (\delta-c \mu)  \norm{\varepsilon (u-v)}_{L^2(\Lambda)}^2  +c_\delta \norm{ u-v}_{L^2(\Lambda)}^2  \norm{\nabla v}_{L^{p}(\Lambda)}^{2} \\
&\leq c_\delta \norm{ u-v}_{L^2(\Lambda)}^2  \norm{\nabla v}_{L^{p}(\Lambda)}^{2}.
\end{align*}

Thus, Condition <ref> holds with $h = 0$ and $\eta(v) = c_\delta  \norm{\nabla v}_{L^{p}(\Lambda)}^{2}$.

Ad <ref>: Using $\abs{F(B)}^2 = \mu(1+ \abs{B})^{p-2} \abs{B}^2 \geq \mu \abs{B}^p $, and an argument similar to (<ref>),

\begin{align*}
\langle A (u), u\rangle_{V^*,V} &=-\int_\Lambda S(\varepsilon u) : \varepsilon u + (u\cdot \nabla)u \cdot u \dd x \\
&= - \norm{F(\varepsilon u)}_{L^2(\Lambda)}^2 \leq - \mu \norm{\varepsilon u}_{L^p(\Lambda)}^p.
\end{align*}

Thus, Condition <ref> holds with $\alpha = p$, $c_1 = \mu$, $c_2 = 0$ and $f(t) = 0$.

Ad <ref>: Using $W^{1,p}_0(\Lambda) \hookrightarrow L^\infty(\Lambda)$, $\abs{S(B)}^{p'} \leq \mu^{p'}(1+\abs{B})^p$, Hölder's inequality and Korn's inequality,

\begin{align*}
\norm{A(u)}_{V^*} &\leq \norm{S(\varepsilon u)}_{L^{p'}(\Lambda)} + C\norm{(u\cdot \nabla)u}_{L^1(\Lambda)} \\
&\leq \mu \norm{1+\varepsilon u}_{L^p(\Lambda)}^{p-1} + C \abs{\Lambda}^{1/p'}\norm{u}_{L^{\infty}(\Lambda)} \norm{\nabla u}_{L^{p}(\Lambda)} \\
&\leq \mu(\abs{\Lambda}+ \norm{\varepsilon u}_{L^p(\Lambda)})^{p-1} + C^2 C_{\mathrm{Korn}}\abs{\Lambda}^{1/p'} \norm{\varepsilon u}_{L^{p}(\Lambda)}^2
\end{align*}

Since $p \geq 3$,

\begin{align*}
\norm{\varepsilon u}_{L^{p}(\Lambda)}^2 \leq (1+\norm{\varepsilon u}_{L^{p}(\Lambda)})^2 \leq  (1+\norm{\varepsilon u}_{L^{p}(\Lambda)})^{p-1} \leq 2^{p-2}(1 +\norm{\varepsilon u}_{L^{p}(\Lambda)}^{p-1}).
\end{align*}

\begin{align*}
\norm{A(u)}_{V^*} &\leq 2^{p-2} (\mu \abs{\Lambda}^{p-1} +  C^2 C_{\mathrm{Korn}}\abs{\Lambda}^{1/p'}) +  \mu  2^{p-2}(\mu + C^2 C_{\mathrm{Korn}}\abs{\Lambda}^{1/p'}) \norm{\varepsilon u}_{L^p(\Lambda)}^{p-1}.
\end{align*}

Thus, Condition <ref> holds with $g(t) = 2^{p-2} (\mu \abs{\Lambda}^{p-1} +  C^2 C_{\mathrm{Korn}}\abs{\Lambda}^{1/p'}) $ and $c_3 = \mu  2^{p-2}(\mu + C^2 C_{\mathrm{Korn}}\abs{\Lambda}^{1/p'})$.

By assumption, Assumption <ref> is satisfied.

We have verified all assumptions of Theorems <ref> and <ref>, and the proof is complete.

§ APPENDIX

§.§ Local time and occupation times formula

We recall for the reader the basic concepts of occupation measures, local times and the occupation times formula. A comprehensive review paper on these topics is [23].

Let $w:[0,T]\to \R^N$ be a measurable path. Then the occupation measure at time $t\in [0,T]$, written $\mu^w_t$ is the Borel measure on $\R^d$ defined by

\[
\mu^w_t(A):=\lambda(\{ s\in [0,t]:\ w_s\in A\}), \quad A\in \mathcal{B}(\R^N),
\]

where $\lambda$ denotes the standard Lebesgue measure.

The occupation measure thus measures how much time the process $w$ spends in certain Borel sets. Provided for any $t\in [0,T]$, the measure is absolutely continuous with respect to the Lebesgue measure on $\R^N$, we call the corresponding Radon-Nikodym derivative local time of the process $w$:

Let $w:[0,T]\to \R^N$ be a measurable path. Assume that there exists a measurable function $L^w:[0,T]\times \R^N\to \R_+$ such that

\[
\mu^w_t(A)=\int_A L^w_t(z) \dd z,
\]

for any $A\in \mathcal{B}(\R^N)$ and  $t\in [0,T]$. Then we call $L^w$ local time of $w$.

Note that by the definition of the occupation  measure, we have for any bounded measurable function $f:\R^N\to \R$ that

\begin{equation}
\int_0^tf(w_s)ds=\int_{\R^N} f(z)\mu^w_t(\dd z).
\label{occupation times formula}
\end{equation}

The above equation (<ref>) is called occupation times formula. Remark that in particular, provided $w$ admits a local time, we also have for any $u\in \R^N$

\begin{equation}
\int_0^tf(u-w_s)\dd s=\int_{\R^N} f(u-z)\mu^w_t(\dd z)=\int_{\R^N}f(u-z)L^w_t(z) \dd z=(f*L^w_t)(u).
\end{equation}

§.§ Some helpful Lemmata

Suppose $\sigma:H\to H$ is Lipschitz-continuous, i.e., there exists a constant $L>0$ such that for all $x,y \in H$ it holds

\begin{align} \label{eq:Lipschitz-ass}
\norm{\sigma(x)-\sigma(y)}_H\leq L\norm{x-y}_H.
\end{align}

Moreover, let $u \in B^{1/2}_{2,\infty} H\cap L^\infty_tH$. Then we have for $q\geq 2$

\[
\left( \int_0^{T-h} \norm{\langle \sigma(u)\rangle_{r, r+h}-\sigma(u_r)}_H^qdr\right)^{1/q}\leq h^{1/q} (2\norm{u}_{L^\infty_t H})^{1-2/q} L \seminorm{u}_{B^{1/2}_{2, \infty}H}^{2/q}.
\]

\[
\left( \int_0^{T-h} \norm{\sigma(\langle u\rangle_{r, r+h})-\sigma(u_r)}_H^qdr\right)^{1/q}\leq h^{1/q} (2\norm{u}_{L^\infty_t H})^{1-2/q} L \seminorm{u}_{B^{1/2}_{2, \infty}H}^{2/q}.
\]

Using Jensen's inequality, (<ref>), and Hölder's inequalities,

\begin{align*}
&\int_0^{T-h}\norm{ \langle \sigma(u) \rangle_{r,r+h}-\sigma(u_r)}_{H}^q \dd r \\
&\hspace{2em} \leq\frac{1}{h^q}\int_0^{T-h}\left(\int_{r}^{r+h}\norm{\sigma(u_z)-\sigma(u_r)}_H \dd z\right)^q \dd r \\
&\hspace{2em} \leq \frac{L^q}{h^q} (2\norm{u}_{L^\infty_t H})^{q-2} \int_0^{T-h}\left(\int_{r}^{r+h}\norm{u_z-u_r}^{2/q}_H \dd z \right)^q \dd r \\
&\hspace{2em} \leq \frac{L^q}{h} (2\norm{u}_{L^\infty_t H})^{q-2} \int_0^{T-h} \int_{r}^{r+h}\norm{u_z-u_r}^{2}_H \dd z \dd r.
\end{align*}

A substitution $z = r+s$ and Fubini's theorem show

\begin{align*}
\int_0^{T-h} \int_{r}^{r+h}\norm{u_z-u_r}^{2}_H \dd z \dd r &= \int_0^h \int_0^{T-h}\norm{u_{r+s}-u_r}_H^2 \dd r \dd s \\
&\leq \int_0^h s \dd s \seminorm{u}_{B^{1/2}_{2,\infty}H}^2 \leq h^2 \seminorm{u}_{B^{1/2}_{2,\infty}H}^2.
\end{align*}

The first assertion follows by combining the estimates. The second assertion is proven analogously.

Let $\sigma:H\to H$ be Lipschitz-continuous, i.e.

\[
\norm{\sigma(x)-\sigma(y)}_H\leq L\norm{u-v}.
\]

Let $u,v \in L^\infty_t H$. Then we have for $q\geq 2$

\begin{align}
\begin{aligned}
&\left(\int_0^{T-h}\norm{\langle \sigma(u)\rangle_{r, r+h}-\langle \sigma(v)\rangle_{r, r+h}}_H^q \dd r \right)^{1/q} \\
&\hspace{2em} \leq (\norm{u}_{L^\infty_t H} + \norm{v}_{L^\infty_t H})^{1-2/q} L \norm{u-v}_{L^2_t H}^{2/q}.
\end{aligned}
\end{align}

as well as

\begin{align}
\begin{aligned}
&\left(\int_0^{T-h}\norm{ \sigma(\langle u\rangle_{r, r+h})- \sigma(\langle v\rangle_{r, r+h})}_H^q \dd r \right)^{1/q} \\
&\hspace{2em} \leq (\norm{u}_{L^\infty_t H} + \norm{v}_{L^\infty_t H})^{1-2/q} L \norm{u-v}_{L^2_t H}^{2/q}.
\end{aligned}
\end{align}

The proof proceeds analogously to the one of Lemma <ref>.

Let $\alpha, \beta\in \mathbb{R}\backslash \{0\}$ with $\alpha<\beta$ and $\alpha+\beta>0$. Then there exists a $C>0$ such that for all $p_1, p_2, p, q_1, q_2, r\in [1, \infty]$ satisfying

\[
\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2},\qquad r\leq q_1
\]

we have

\[
\norm{u\cdot v}_{B^\alpha_{p, r}}\leq C \norm{u}_{B^\alpha_{p_1, q_1}}\norm{v}_{B^\beta_{p_2, q_2}}
\]

Let $q \in (1,\infty]$, $\gamma \in (1/q,1]$ and $E$ be a Banach space. Then for all finite intervals $J$ and $u \in B^{\gamma}_{q,\infty}(J;E)$ it holds

\begin{align} \label{eq:Shift-bound}
\sup_{s\in J}  \left(\int_J \norm{u_t - u_s}_E^q \dd t \right)^{1/q} \leq  C_{\gamma,q} \abs{J}^{\gamma} \seminorm{u}_{B^{\gamma}_{q,\infty}(I;E)},
\end{align}

where $C_{\gamma,q} = \frac{3^{2-(\gamma - 1/q)}}{\gamma - 1/q}$.

Let $\nu \in J$. Using <cit.> with $s = \gamma$, $r = \gamma - 1/q$, $p = q$ and $q = \infty$ shows

\begin{align*}
\left(\int_J \norm{u_t - u_\nu}_E^q \dd t \right)^{1/q} &\leq \seminorm{u}_{C^{\gamma - 1/q}(J;E)} \left(\int_J \abs{t-\nu}^{\gamma q-1} \dd t \right)^{1/q} \\
&\leq C_{\gamma,q} \seminorm{u}_{B^{\gamma}_{q,\infty}(J;E)}\abs{J}^\gamma.
\end{align*}

Since $\nu \in J$ was arbitrary, (<ref>) follows by taking the supremum.

Let $J = [a,b]$, $a<b$ be an interval, $\mathcal{P} = \{ [t_{k-1}, t_k]: \, t_0 = a < t_1  < \ldots < t_N = b\}$ be a finite partition of $J$, and $E$ be a Banach space. Then

\begin{align*}
u \in  \bigcap_{k=1}^N B^{1/2}_{2,\infty}([t_{k-1},t_k];E) \cap L^\infty(J;E) \quad \Rightarrow \quad u \in B^{1/2}_{2,\infty}(J;E).
\end{align*}

\begin{align} \label{eq:Local-Nikolskii}
\seminorm{u}_{B^{1/2}_{2,\infty}(J;E)}^2 \leq \sum_{k=1}^N  \seminorm{u}_{B^{1/2}_{2,\infty}([t_{k-1},t_k];E)}^2 + \norm{u}_{L^\infty(J;E)}^2 2(N-1+h_{\mathrm{min}}^{-1}\abs{J} ),
\end{align}

where $h_{\mathrm{min}} = \min_{k=1,\ldots,N}\abs{t_k - t_{k-1}}$.

We distinguish two cases: differences for small and large times. Here small and large times should be understood with respect to the minimal partition mesh-size $h_{\mathrm{min}}$.

$h \leq h_{\mathrm{min}}$: For short times we can use the control provided by the local Nikolskii norm and an $L^\infty$-estimate. Notice that each $[t_{k-1},t_k]$ splits into $[t_{k-1}, t_{k} - h] \cup [t_{k} - h,t_{k}]$. Therefore,

\begin{align*}
\int_a^{b-h} \norm{u_{t+h} - u_t}_E^2 \dd t &= \sum_{k=1}^N \int_{t_{k-1}}^{t_k-h} \norm{u_{t+h} - u_t}_E^2 \dd t +  \sum_{k=1}^{N-1} \int_{t_{k}-h}^{t_k} \norm{u_{t+h} - u_t}_E^2 \dd t \\
&\leq h \left( \sum_{k=1}^N  \seminorm{u}_{B^{1/2}_{2,\infty}([t_{k-1},t_k];E)}^2 + \norm{u}_{L^\infty(J;E)}^2 2(N-1) \right) .
\end{align*}

It follows

\begin{align} \label{eq:small-h}
\sup_{h \leq h_{\mathrm{min}}} h^{-1} \int_a^{b-h} \norm{u_{t+h} - u_t}_E^2 \dd t \leq  \sum_{k=1}^N  \seminorm{u}_{B^{1/2}_{2,\infty}([t_{k-1},t_k];E)}^2 + \norm{u}_{L^\infty(J;E)}^2 2(N-1).
\end{align}

$h > h_{\mathrm{min}}$: For large times we no longer need to guarantee a decay of increments. Thus, it is sufficient to estimate

\begin{align*}
h^{-1} \int_a^{b-h} \norm{u_{t+h} - u_t}_E^2 \dd t \leq  2h_{\mathrm{min}}^{-1} \norm{u}_{L^\infty(J;E)}^2 \abs{J},
\end{align*}

which implies

\begin{align} \label{eq:large-h}
\sup_{h > h_{\mathrm{min}}} h^{-1} \int_a^{b-h} \norm{u_{t+h} - u_t}_E^2 \dd t \leq  2h_{\mathrm{min}}^{-1} \norm{u}_{L^\infty(J;E)}^2 \abs{J}.
\end{align}

Combining (<ref>) and (<ref>) yields (<ref>).

Let $Z\in B^{1/2}_{2, \infty}\mathbb{R}$ and $\beta\in C^\gamma_{t}\mathbb{R}$ with $\gamma>3/4$. Then

\[
B_t:=\int_0^t Z_r \dd \beta_r-Z_t
\]

is a well-defined object in $B^{1/2}_{2, \infty}$ and moreover, we have

\[
\int_0^t\exp{(-\beta_r)}\dd B_r=-Z_t\exp{(-\beta_t)}+Z_0\exp{(-\beta_0)}.
\]

The first part of the statement is an immediate consequence of the Sewing Lemma <ref>. Note that if $\beta^n, Z^n$ are smooth, we have for the corresponding $B^n$

\begin{align*}
& \int_0^t \exp{(-\beta^n_r)} \dd B^n_r=\int_0^t  \exp{(-\beta^n_r)}Z^n_r \dd \beta^n_r- \int_0^t\exp{(-\beta^n_r)}\dd Z^n_r\\
&=-\int_0^t \dd (Z^n_r\exp{(-\beta^n_r)}) = -Z^n_t\exp{(-\beta^n_t)}+Z^n_0\exp{(-\beta^n_0)}.
\end{align*}

Using the stability of sewings under mollification as expressed in Lemma <ref>, we are able to pass to the limit in the above, yielding the claim.

§.§ An Aubin Lions type Lemma

In an abstract sense, upon establishing suitable a priori bounds on approximate problems, one is confronted with the problem of the passage to the limit. However, as only weak or weak-$*$ convergences are available from a priori bounds, the passage to the limit in non-linear terms typically poses an issue as this requires some form of strongly convergent subsequences. Such strong convergence is classically provided by the Rellich-Kondrachov theorem in combination with the Aubin-Lions Lemma. Remark however that due to the irregularity of $I(u)$, no $L^q([0, T], X)$ estimates on $\partial_t u^n$ will be available. We therefore establish an Aubin-Lions type Lemma suited for our context.

Let $p\in [1, \infty)$ and let $T\in (0, \infty)$. Let $X$ be a Banach space. A set $F\subset L^p([0, T], X)$ is relatively compact in $L^p([0, T], X)$ if and only if

* $\sup_{f\in F} \norm{f}_{L^p([0, T], X)}<\infty$

* $\lim_{h\to 0}\int_0^{T-h} \norm{f_{r+h}-f_r}_X^pdr=0$ uniformly in $f\in F$

* For every $\epsilon>0$, there exists a compact set $Q_\epsilon\subset X$ such that for every $f\in F$ there exists a set $A_{f, \epsilon}\subset [0, T]$ with $\lambda([0, T]\backslash A_{f, \epsilon})\leq \epsilon$ and

\[
f_t\in Q_\epsilon, \quad \forall t\in A_{f, \epsilon}
\]

Given the above result, we may conclude the following.

Let $X_0, X$ be Banach spaces, such that the embedding $X_0\hookrightarrow \hookrightarrow X$ is compact. Let $p, q\in [1, \infty)$ and $\gamma>0$. Then the space

\[
\mathcal{Y}:=L^q_tX_0\cap B^{\gamma}_{p, \infty}X
\]

embeds compactly into $L^p_tX$.

We apply Theorem <ref>. Points $(1)$ and $(2)$ thereof follow immediately as

\[
\int_0^{T-h} \norm{f_{r+h}-f_r}_X^pdr\leq h^{\gamma p} \norm{f}_{B^\gamma_{p, \infty}X}\to 0.
\]

Let $(f^n)_n$ be a bounded sequence in $\mathcal{Y}$. Given $\epsilon>0$, we choose $R(\epsilon)$ sufficiently large such that

\[
\frac{1}{R(\epsilon)^q}\sup_n \norm{f^n}_{L^q_tX_0}^q<\epsilon
\]

Further set $A_{n, \epsilon}:=\{ t\in [0, T]\ |\ \norm{f_n}_{X_0}\geq R(\epsilon)\}$ and for the canonical imbedding $i:X_0\to X$, set $Q_\epsilon:=\overline{i(B_{X_0}(0, R(\epsilon)))}$, which is a compact set in $X$ such that $f^n_t\in Q_\epsilon$ for all $t\in A_{n, \epsilon}$. Finally, by Markov's inequality we have

\[
\lambda([0, T]\backslash A_{n, \epsilon})\leq \frac{1}{R(\epsilon)^q}\int_0^T \norm{f^n_r}_{X_0}^qdr<\epsilon.
\]

This establishes $(3)$, thus allowing us to apply Theorem <ref> concluding the proof.

§ ACKNOWLEDGEMENT
The first authors warmly thanks the organizers and participants of the "GEN-Y research workshop in stochastic analysis" at the University of Münster for feedback on a preliminary version of this work.  The first author acknowledges support from the European Research Council (ERC) under the European Union’s Horizon 2020 research
and innovation programme (grant agreement No. 949981). The second author acknowledges support from the Australian Government through the Australian Research Council’s Discovery Projects funding scheme (grant number DP220100937).

All the authors declare that they have no conflicts of interest. Data availability statement is not applicable in the context to the present article since all the results here are theoretical in nature and do not involve any data.

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