In this section, we prove Theorem 1.1. We split the proof into two main steps: The first concerns showing that discrete solutions converge to a Leray–Hopf weak solution, while the second proves the constructed solutions are, in fact, suitable.
Proof of Theorem 1.1
We first prove the convergence of the numerical sequence to a Leray–Hopf weak solution, mainly proving the correct balance of the global energy (2.2); then, we prove that the weak solution constructed is suitable, namely that it satisfies the local energy inequality (2.3).
Step 1: convergence toward a Leray–Hopf weak solution. We start by observing that from a simple density argument, the test functions considered in (2.1) can be chosen in the space \(L^{s}(0,T;H^{1}_{\operatorname{div}})\cap C^{1}(0,T;L^{2}_{ \operatorname{div}})\), with \(s\geq 4\). In particular, using (3.1) for any \(w\in L^{s}(0,T;H^{1}_{\operatorname{div}})\cap C^{1}(0,T;L^{2}_{ \operatorname{div}})\) such that \(w(T,x)=0\), we can find a sequence \(\{w_{h}\}_{h}\subset L^{s}(0,T;H^{1}_{\#})\cap C(0,T;L^{2}_{\#})\) such that
$$ \begin{aligned} &w_{h}\to w\quad \text{strongly in }L^{s}\bigl(0,T;H^{1}_{\#}\bigr) \text{ as }h\to 0, \\ &w_{h}(0)\to w(0)\quad \text{strongly in }L^{2}_{\#} \text{ as }h \to 0, \\ &\partial _{t}w_{h}\rightharpoonup \partial _{t}w\quad \text{weakly in }L^{2}\bigl(0,T;L^{2}_{ \#} \bigr) \text{ as }h\to 0. \end{aligned} $$
(5.1)
Let \(\{(v^{\Delta t}_{h}, v^{\Delta t}_{h}, p^{\Delta t}_{h})\}_{(\Delta t,h)}\), defined as in (3.12), be a family of solutions of (3.13). By Proposition 4.4a), we have that
$$\begin{aligned} & \bigl\{ v^{\Delta t}_{h} \bigr\} _{(\Delta t,h)}\subset L^{\infty}\bigl(0,T;L^{2}_{ \#}\bigr),\quad \text{with uniform bounds on the norms}. \end{aligned}$$
Then, by standard compactness arguments, there exists \(v\in L^{\infty}(0,T;L^{2}_{\#})\) such that (up to a sub-sequence)
$$ \begin{aligned} &v^{\Delta t}_{h} \rightharpoonup v\quad \text{weakly in }L^{2}\bigl(0,T;L^{2}_{ \#} \bigr)\text{ as }(\Delta t,h)\to (0,0). \end{aligned} $$
(5.2)
Again using Proposition 4.4 b), there exists \(u\in L^{\infty}(0,T;L^{2}_{\#})\) such that (up to a sub-sequence)
$$ \begin{aligned} &u^{\Delta t}_{h} \overset{*}{\rightharpoonup } u\quad \text{weakly* in }L^{ \infty} \bigl(0,T;L^{2}_{\#}\bigr) \text{ as }(\Delta t,h)\to (0,0), \\ &u^{\Delta t}_{h} \rightharpoonup u\quad \text{weakly in } L^{2}\bigl(0,T;H^{1}_{ \#}\bigr) \text{ as }( \Delta t,h)\to (0,0). \end{aligned} $$
(5.3)
Moreover, using (3.1), for any \(q\in L^{2}(0,T;L^{2}_{\#})\), we can find a sequence \(\{q_{h}\}_{h}\subset L^{2}(0,T;L^{2}_{\#})\) such that \(q_{h}\in L^{2}(0,T;M_{h})\) and
$$ q_{h}\to q\quad \text{strongly in }L^{2}\bigl(0,T;L^{2}_{\#} \bigr) \text{ as }h\to 0. $$
Then, using (5.3) and (3.13), we have that
$$ 0= \int _{0}^{T} \bigl(\operatorname{div}u^{\Delta t}_{h},q_{h} \bigr) \,dt\to \int _{0}^{T} (\operatorname{div}u,q ) \,dt\quad \text{as }(\Delta t,h)\to (0,0), $$
hence u is divergence-free, since it belongs to \(H^{1}_{\operatorname{div}}\). Let us consider (4.3), then
$$ \int _{0}^{T} \bigl\Vert v^{\Delta t}_{h}-u^{\Delta t}_{h} \bigr\Vert _{2}^{2} \,dt\leq \frac{\Delta t}{12} \sum _{m=1}^{N} \bigl\Vert u_{h}^{m}-u_{h}^{m-1} \bigr\Vert _{2}^{2} \leq C \biggl((\Delta t)^{2}+ \frac{\Delta t}{h^{1/2}} \biggr), $$
(5.4)
where in the last inequality we used again Proposition 4.4 and the estimate (4.2). Then, the integral \(\int _{0}^{T}\|v^{\Delta t}_{h}-u^{\Delta t}_{h}\|_{2}^{2} \,dt\) vanishes as \(\Delta t\to 0\) if \(\Delta t=o(h^{1/2})\), that is if (1.3) is satisfied. Then, from (5.2) and (5.3), it easily follows that \(v=u\).
The rest of the proof follows the same as in [5], so we just sketch out the proof, referring to [5] for complete details.
By Lemma 2.4 and the fact that \(u=v\), we get that \(u^{\Delta t}_{h} v^{\Delta t}_{h}\rightharpoonup |u|^{2}\) weakly in \(L^{1}((0,T)\times \mathbb{T}^{3})\) as \((\Delta t,h)\to (0,0)\). In particular, using (5.4), we have that
$$ \begin{aligned} &v^{\Delta t}_{h}, u^{\Delta t}_{h}\to u\quad \text{strongly in }L^{2} \bigl(0,T;L^{2}_{ \#}\bigr)\text{ as }(\Delta t,h)\to (0,0). \end{aligned} $$
(5.5)
Concerning the pressure term the uniform bound in Proposition 4.4 d) ensures the existence of \(p\in L^{{4}/{3}}(0,T;L^{2}_{\#})\) such that (up to a sub-sequence)
$$ p^{\Delta t}_{h}\rightharpoonup p\quad \text{weakly in }L^{{4}/{3}}\bigl(0,T;L^{2}_{ \#}\bigr) \text{ as }( \Delta t,h)\to (0,0). $$
(5.6)
Then, using (5.1) and (5.2), we have that
$$ \begin{aligned} \lim_{(\Delta t,h)\to (0,0)} \int _{0}^{T}\bigl(\partial _{t}v^{\Delta t}_{h},w_{h} \bigr) \,dt& =- \int _{0}^{T}(u,\partial _{t}w) \,dt- \bigl(u_{0}, w(0)\bigr), \end{aligned} $$
Next, using (5.3), (5.1), and (5.6), we also get
$$ \begin{aligned} &\lim_{(\Delta t,h)\to (0,0)} \int _{0}^{T}\bigl(\nabla u^{\Delta t}_{h}, \nabla w_{h}\bigr) \,dt= \int _{0}^{T}(\nabla u, \nabla w) \,dt, \\ &\int _{0}^{T}\bigl(p^{\Delta t}_{h}, \operatorname{div}w_{h}\bigr) \,dt\to 0 \quad \text{as }(\Delta t,h)\to (0,0). \end{aligned} $$
Concerning the non-linear term, let \(s\geq 4\), with a standard compactness argument
$$ nl_{h} \bigl(u^{\Delta t}_{h},u^{\Delta t}_{h} \bigr) \rightharpoonup u\cdot \nabla u, \quad \text{in } L^{s'} \bigl(0,T;H^{-1}\bigr)\text{ as }(\Delta t,h)\to (0,0). $$
Then, also from (5.1), it follows that
$$ \int _{0}^{T} b_{h}\bigl(u^{\Delta t}_{h},u^{\Delta t}_{h},w_{h} \bigr) \,dt \to \int _{0}^{T} \bigl((u\cdot \nabla ) u,w \bigr) \,dt \quad \text{as }(\Delta t,h)\to (0,0). $$
(5.7)
Finally, the energy inequality follows by Lemma 4.1, using the lower semi-continuity of the \(L^{2}\)-norm with respect to the weak convergence, since the estimate (4.1) can be rewritten as
$$ \frac{1}{2} \bigl\Vert v^{\Delta t}_{h}(T) \bigr\Vert ^{2}_{2}+\nu \int _{0}^{T} \bigl\Vert \nabla u^{ \Delta t}_{h}(t) \bigr\Vert ^{2}_{2} \,dt \leq \frac{1}{2} \Vert u_{0} \Vert ^{2}_{2}. $$
The treatment of the Case 2 and Case 3 can be done with minor changes, concerning the tri-linear term, just using the estimate already proved in the previous section. The other terms are unchanged, and the energy estimate remains the same.
Remark 5.1
The results for Case 2 and Case 3 can be easily adapted also to cover the θ-scheme for \(\theta >1/2\), hence completing the results in [5], which were focusing only on the treatment of Case 1.
Note that the tri-linear term based on the rotational formulation from Case 2 and Case 3 \((\nabla \times u^{\Delta t}_{h})\times u^{\Delta t}_{h}\) converges exactly as in the previous step, since
$$ \bigl(\nabla \times u^{\Delta t}_{h}\bigr)\times u^{\Delta t}_{h} \rightharpoonup (\nabla \times u)\times u, \text{ in } L^{s'}\bigl(0,T;H^{-1}\bigr) \quad \text{as }(\Delta t,h)\to (0,0), $$
which implies (5.7), ending the proof in Case 2.
In Case 3, the term, which needs some care, is the projected Bernoulli pressure. In this case, note that for \(\frac{1}{s^{*}}+\frac{1}{2}=\frac{1}{s'}\),
$$ \begin{aligned} \int _{0}^{T} \bigl\Vert \mathcal{K}_{h} \bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2} \bigr)-\mathcal{K}_{h}\bigl( \vert u \vert ^{2}\bigr) \bigr\Vert ^{s'}_{2} \,dt &\leq \int _{0}^{T} \bigl\Vert \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}- \vert u \vert ^{2} \bigr\Vert ^{s'}_{2} \,dt \\ & \leq \int _{0}^{T} \bigl\Vert \bigl\vert u^{\Delta t}_{h}-u \bigr\vert \bigl\vert u^{\Delta t}_{h}+u \bigr\vert \bigr\Vert ^{s'}_{2} \,dt \\ &\leq \bigl\Vert u^{\Delta t}_{h}-u \bigr\Vert _{L^{s^{*}}(L^{3})}\bigl( \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{L^{2}(L^{6})}+ \Vert u \Vert _{L^{2}(L^{6})}\bigr). \end{aligned} $$
This shows that
$$ \mathcal{K}_{h}\bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}\bigr)\to \mathcal{K}_{h}\bigl( \vert u \vert ^{2}\bigr) \quad \text{in }L^{s'}\bigl(0,T;L^{2} \bigl(\mathbb{T}^{3}\bigr)\bigr). $$
Moreover, since \(\mathcal{K}_{h}(|w|^{2})\to |w|^{2}\) in \(L^{2}(\mathbb{T}^{3})\) for a.e. \(t\in (0,T)\) and \(\|\mathcal{K}_{h}(|w|^{2})\|_{2}\leq \|w\|^{2}_{2}\), by Lebesgue dominated convergence, we have \(\mathcal{K}_{h}(|w|^{2})\to |w|^{2}\) in \(L^{s'}(0,T;L^{2}(\mathbb{T}^{3}))\), finally showing that
$$ \int _{0}^{T}\bigl(\mathcal{K}_{h}\bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}\bigr), \operatorname{div}v\bigr) \,dt\to \int _{0}^{T}\bigl( \vert u \vert ^{2},\operatorname{div}v\bigr) \,dt. $$
This proves, after integration by parts, that
$$ \int _{0}^{T}b_{h}\bigl(u^{\Delta t}_{h}, u^{\Delta t}_{h},w\bigr) \,dt \to \int _{0}^{T}(u\cdot \nabla ) u,w) \,dt. $$
Note also that since in all cases, it holds that
$$ \frac{1}{2} \bigl\Vert v^{\Delta t}_{h}(T) \bigr\Vert ^{2}_{2}+\nu \int _{0}^{T} \bigl\Vert \nabla u^{ \Delta t}_{h}(t) \bigr\Vert ^{2}_{2} \,dt \leq \frac{1}{2} \Vert u_{0} \Vert ^{2}_{2}, $$
(5.8)
a standard lower semi-continuity argument is enough to infer that the weak solution
$$ u=v=\lim_{(h,\Delta t)\to (0,0)} u^{\Delta t}_{h}=\lim _{(h,\Delta t) \to (0,0)} v^{\Delta t}_{h} $$
also satisfies the global energy inequality (2.2).
Step 2: proof of the local energy inequality. In order to conclude the proof of Theorem 1.1, we need to prove that the Leray–Hopf weak solution constructed in Step 1 is suitable. According to Definition 2.2, this requires just to prove the local energy inequality. To this end, let us consider a smooth, periodic in the space variable function \(\phi \geq 0\), vanishing for \(t=0,T\); we use \(P_{h}(u^{\Delta t}_{h}\phi )\) as test function in the momentum equation in (3.13).
The term involving the time-derivative is treated as in [5].
$$\begin{aligned} & \int _{0}^{T} \bigl(\partial _{t} v^{\Delta t}_{h},P_{h}\bigl(u^{\Delta t}_{h} \phi \bigr) \bigr) \,dt \\ &\quad = \int _{0}^{T} \bigl(\partial _{t} v^{\Delta t}_{h},u^{ \Delta t}_{h}\phi \bigr) \,dt + \int _{0}^{T} \bigl(\partial _{t} v^{ \Delta t}_{h},P_{h}\bigl(u^{\Delta t}_{h} \phi \bigr)-u^{\Delta t}_{h}\phi \bigr) \,dt=:I_{1}+I_{2}. \end{aligned}$$
Concerning the term \(I_{1}\), we have that
$$ \begin{aligned} &= \int _{0}^{T}\bigl(\partial _{t} v^{\Delta t}_{h},v^{\Delta t}_{h} \phi \bigr)\,dt+ \int _{0}^{T}\bigl(\partial _{t} v^{N}_{h},\bigl(u^{\Delta t}_{h}-v^{ \Delta t}_{h} \bigr) \phi \bigr) \,dt =:I_{11}+I_{12}. \end{aligned} $$
Let us first consider \(I_{11}\). By splitting the integral over \([0,T]\) as the sum of integrals over \([t_{m-1},t_{m}]\) and by integration by parts, we get
$$ \begin{aligned} & \int _{0}^{T}\bigl(\partial _{t} v^{\Delta t}_{h},v^{\Delta t}_{h}\phi \bigr) \,dt\\ &\quad= \sum_{m=1}^{N} \int _{t_{m-1}}^{t_{m}}\bigl(\partial _{t} v^{\Delta t}_{h},v^{ \Delta t}_{h}\phi \bigr) \,dt= \sum_{m=1}^{N} \int _{t_{m-1}}^{t_{m}}\biggl( \frac{1}{2}\partial _{t} \bigl\vert v^{\Delta t}_{h} \bigr\vert ^{2},\phi \biggr) \,dt \\ &\quad=\frac{1}{2}\sum_{m=1}^{N}\bigl( \bigl\vert u_{h}^{m} \bigr\vert ^{2},\phi (t_{m},x)\bigr)- \bigl( \bigl\vert u_{h}^{m-1} \bigr\vert ^{2}, \phi (t_{m-1},x)\bigr)-\sum _{m=1}^{N} \int _{t_{m-1}}^{t_{m}}\biggl(\frac{1}{2} \bigl\vert v^{ \Delta t}_{h} \bigr\vert ^{2},\partial _{t}\phi \biggr) \,dt, \end{aligned} $$
where we used that \(\partial _{t} v^{\Delta t}_{h}(t)= \frac{u_{h}^{m}-u_{h}^{m-1}}{\Delta t}\), for \(t\in [t_{m-1},t_{m}[\). Next, since the sum telescopes and ϕ is with compact support in \((0,T)\), we get
$$ \int _{0}^{T}\bigl(\partial _{t} v^{\Delta t}_{h},v^{\Delta t}_{h}\phi \bigr) \,dt =- \int _{0}^{T} \biggl(\frac{1}{2} \bigl\vert v^{\Delta t}_{h} \bigr\vert ^{2},\partial _{t} \phi \biggr) \,dt. $$
By the strong convergence of \(v^{\Delta t}_{h}\rightarrow u\) in \(L^{2}(0,T;L^{2}_{\#})\), we can conclude that
$$ \lim_{(\Delta t,h)\to (0,0)} \int _{0}^{T}\bigl(\partial _{t} v^{\Delta t}_{h},v^{ \Delta t}_{h}\phi \bigr) \,dt= - \int _{0}^{T} \biggl(\frac{1}{2} \vert u \vert ^{2}, \partial _{t}\phi \biggr) \,dt. $$
Then, we consider the term \(I_{12}\). Since \(u^{\Delta t}_{h}\) is constant on the interval \([t_{m-1}, t_{m}[\), we can write
$$ \begin{aligned} \int _{0}^{T}\bigl(\partial _{t} v^{\Delta t}_{h},\bigl(u^{\Delta t}_{h}-v^{ \Delta t}_{h} \bigr) \phi \bigr) \,dt&= -\sum_{m=1}^{N} \int _{t_{m-1}}^{t_{m}}\bigl( \partial _{t} \bigl(v^{\Delta t}_{h}-u^{\Delta t}_{h}\bigr), \bigl(v^{\Delta t}_{h}-u^{ \Delta t}_{h}\bigr) \phi \bigr) \,dt \\ &=\sum_{m=1}^{N} \int _{t_{m-1}}^{t_{m}} \biggl( \frac{ \vert v^{\Delta t}_{h}-u^{\Delta t}_{h} \vert ^{2}}{2}, \partial _{t}\phi \biggr) \,dt, \end{aligned} $$
since the sum telescopes. Hence, we have that \(u^{\Delta t}_{h}-v^{\Delta t}_{h}\) vanishes (strongly) in \(L^{2}(0,T;L^{2}_{\#})\), provided that \(\Delta t=o(h^{1/2})\). Then, \(I_{12}\rightarrow 0\) as \((\Delta t,h)\rightarrow (0,0)\).
Remark 5.2
It is at this point that the coupling between h and Δt plays a role. For the convergence of the other terms, the discrete commutator property is needed. We are skipping some details from the other proofs because they are very close to that in the cited references.
We have that the \(I_{2}\to 0\) as \((\Delta t,h)\rightarrow (0,0)\). Indeed, by the discrete commutator property (3.4), Proposition 4.4, and the inverse inequality (3.3), we can infer
$$\begin{aligned} \vert I_{2} \vert &\leq \int _{0}^{T} \bigl\Vert \partial _{t} v^{\Delta t}_{h} \bigr\Vert _{H^{-1}} \bigl\Vert P_{h}\bigl(u^{ \Delta t}_{h}\phi \bigr)-u^{\Delta t}_{h} \phi \bigr\Vert _{H^{1}}\,dt \\ & \leq ch^{{1}/{2}} \bigl\Vert \partial _{t} v^{\Delta t}_{h} \bigr\Vert _{L^{ {4}/{3}}(H^{-1})} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert ^{{1}/{2}}_{L^{\infty}(L^{2})} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert ^{{1}/{2}}_{L^{2}(H^{1})} \leq c h^{ {1}/{2}}. \end{aligned}$$
Hence, this term also vanishes as \(h\to 0\), ending the analysis of the term involving the time-derivative.
Concerning the viscous term, we write
$$ \begin{aligned} \bigl(\nabla u^{\Delta t}_{h},\nabla P_{h}\bigl(u^{\Delta t}_{h} \phi \bigr)\bigr) & =\bigl( \bigl\vert \nabla u^{\Delta t}_{h} \bigr\vert ^{2}, \phi \bigr)-\biggl(\frac{1}{2} \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}, \Delta \phi \biggr)+R_{visc}, \end{aligned} $$
with the “viscous remainder” \(R_{visc}:= (\nabla u^{\Delta t}_{h},\nabla [P_{h}(u^{\Delta t}_{h} \phi )-u^{\Delta t}_{h}\phi ] )\). Since \(u^{\Delta t}_{h}\) converges to u weakly in \(L^{2}(0,T;H_{\#}^{1})\) and strongly in \(L^{2}(0,T;L^{2}_{\#})\),
$$ \begin{aligned} &\liminf_{(\Delta t,h)\to (0,0)} \int _{0}^{T} \bigl( \bigl\vert \nabla u^{\Delta t}_{h} \bigr\vert ^{2}, \phi \bigr) \,dt\geq \int _{0}^{T} \bigl( \vert \nabla u \vert ^{2},\phi \bigr) \,dt, \\ & \frac{1}{2} \int _{0}^{T} \bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2},\Delta \phi \bigr) \,dt \to \frac{1}{2} \int _{0}^{T} \bigl( \vert u \vert ^{2},\Delta \phi \bigr) \,dt. \end{aligned} $$
For the remainder \(R_{visc}\), using again the discrete commutator property from Definition 3.1, we have that
$$ \biggl\vert \int _{0}^{T} R_{visc} \,dt \biggr\vert \leq c h \int _{0}^{T} \bigl\Vert \nabla u^{\Delta t}_{h} \bigr\Vert ^{2}_{2} \,dt\to 0 \quad \text{as }( \Delta t,h)\to (0,0). $$
We consider now the nonlinear term \(b_{h}\). We have
$$ \begin{aligned} b_{h}\bigl(u^{\Delta t}_{h},u^{\Delta t}_{h},P_{h} \bigl(u^{\Delta t}_{h}\phi \bigr)\bigr) & =b_{h} \bigl(u^{\Delta t}_{h},u^{\Delta t}_{h},u^{\Delta t}_{h} \phi \bigr)+R_{nl}. \end{aligned} $$
(5.9)
The “nonlinear remainder” \(R_{nl}:=b_{h}(u^{\Delta t}_{h},u^{\Delta t}_{h},P_{h}(u^{\Delta t}_{h} \phi )-u^{\Delta t}_{h}\phi )\) can be estimated using the discrete commutator property, (3.3), and (3.8), (3.10), (3.11) for the choices of the nonlinear term approximation in Case 1, Case 2, and Case 3, respectively. Indeed, we have
$$ \begin{aligned} \vert R_{nl} \vert & \leq \bigl\Vert nl_{h}\bigl(u^{\Delta t}_{h},u^{\Delta t}_{h} \bigr) \bigr\Vert _{H^{-1}} \bigl\Vert P_{h} \bigl(u^{\Delta t}_{h}\phi \bigr)-u^{\Delta t}_{h} \phi \bigr\Vert _{H^{1}} \leq c h^{1/2} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{2} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{H^{1}}^{2}, \end{aligned} $$
(5.10)
hence, by integrating over time
$$ \int _{0}^{T}R_{nl} \,dt\to 0 \quad \text{as }(\Delta t,h)\to (0,0). $$
The last term we consider is that involving the pressure. By integrating by parts, we have
$$ \begin{aligned} \bigl(p^{\Delta t}_{h}, \operatorname{div}P_{h}\bigl(u^{\Delta t}_{h}\phi \bigr) \bigr) =\bigl(p_{h}^{ \Delta t} u^{\Delta t}_{h}, \nabla \phi \bigr)+R_{p1}+R_{p2}. \end{aligned} $$
where the two “pressure remainders” are defined as follows
$$ R_{p1}:= \bigl(p^{\Delta t}_{h},\operatorname{div} \bigl(P_{h}\bigl(u^{\Delta t}_{h} \phi \bigr)-u^{\Delta t}_{h}\phi \bigr) \bigr)\quad \text{and}\quad R_{p2}:= \bigl(\phi p^{\Delta t}_{h}, \operatorname{div}u^{\Delta t}_{h} \bigr). $$
Using again the discrete commutator property (3.5) and (3.3), we easily get
$$ \begin{aligned} \vert R_{p1} \vert &\leq c h \bigl\Vert p^{\Delta t}_{h} \bigr\Vert _{2} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{H^{1}} \end{aligned} $$
which implies
$$ \int _{0}^{T} R_{p1} \,dt\to 0\quad \text{as }(\Delta t,h)\to (0,0). $$
The term \(R_{p2}\) can be treated in the same way but now using the discrete commutator property for the projector over \(Q_{h}\)
$$ \begin{aligned} \vert R_{p2} \vert &\leq c \bigl\Vert Q_{h}\bigl(p^{\Delta t}_{h}\phi \bigr)-p^{\Delta t}_{h}\phi \bigr\Vert _{2} \bigl\Vert u^{\Delta t}_{h}\phi \bigr\Vert _{H^{1}} \leq c h^{{1}/{2}} \bigl\Vert p^{ \Delta t}_{h} \bigr\Vert _{L^{{4}/{3}}(L^{2})} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{L^{2}(H^{1})}^{ {1}/{2}} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{L^{\infty}(L^{2})}^{{1}/{2}}, \end{aligned} $$
and finally this implies that
$$ \int _{0}^{T} R_{p2} \,dt\to 0\quad \text{as }(\Delta t,h)\to (0,0). $$
The convergence
$$ \begin{aligned} \int _{0}^{T}\bigl(p_{h}^{\Delta t} u^{\Delta t}_{h},\nabla \phi \bigr)\to \int _{0}^{T}(p u,\nabla \phi ) \end{aligned} $$
is an easy consequence of (5.5), (5.6) and Proposition 4.4 b). These steps are common to the three cases.
We now treat the inertial term. In Case 1, the definition of \(nl_{h}\) in (3.6) allows us to handle the first term on the right-hand side in (5.9) with some integration by parts as follows:
$$\begin{aligned} b_{h}\bigl(u^{\Delta t}_{h},u^{\Delta t}_{h},u^{\Delta t}_{h} \phi \bigr) & = - \biggl(u^{\Delta t}_{h} \frac{1}{2} \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2},\nabla \phi \biggr). \end{aligned}$$
By arguing as in [5], it can be proved that
$$ u^{\Delta t}_{h} \frac{1}{2} \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}\to u \frac{1}{2} \vert u \vert ^{2} \quad \text{strongly in } L^{1}\bigl(0,T;L^{1}\bigr), \text{ as }( \Delta t,h)\to (0,0), $$
and one shows that
$$ \int _{0}^{T} b_{h}\bigl(u^{\Delta t}_{h},u^{\Delta t}_{h},u^{\Delta t}_{h} \phi \bigr) \,dt \to - \int _{0}^{T} \biggl(u \frac{1}{2} \vert u \vert ^{2},\nabla \phi \biggr) \,dt\quad \text{as } (\Delta t,h)\to (0,0). $$
In Case 2, the result is much simpler since by direct computations, one shows that for smooth enough w, we have (by a point-wise equality, where \(\epsilon _{ijk}\) the totally anti-symmetric tensor)
$$ \begin{aligned} \bigl[ ( \nabla \times w)\times w \bigr]\cdot (\phi w)&=\sum_{i,j,k,l}( \epsilon _{jki}-\epsilon _{jlm}) \partial _{l}w_{m}w_{k}w_{i} \phi \\ &=\phi \sum_{i,k}w_{k}\partial _{k}w_{i}w_{i}-w_{i}\partial _{i}w_{k}w_{k}=0. \end{aligned} $$
Hence, we get
$$ b_{h}\bigl(u^{\Delta t}_{h},u^{\Delta t}_{h},u^{\Delta t}_{h} \phi \bigr)=0, $$
and there are no terms to be estimated.
In Case 3, we get instead (cf. [16, Lemma 4.1])
$$ \begin{aligned} b_{h}\bigl(u^{\Delta t}_{h},u^{\Delta t}_{h},u^{\Delta t}_{h} \phi \bigr)&=- \frac{1}{2} \bigl(\mathcal{K}_{h}\bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}\bigr), \operatorname{div}\bigl(\phi u^{\Delta t}_{h}\bigr) \bigr) \\ &=-\frac{1}{2} \bigl(u^{\Delta t}_{h} \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2},\nabla \phi \bigr)+R_{1}+R_{2} \end{aligned} $$
with
$$ R_{1}:=-\frac{1}{2} \bigl(u^{\Delta t}_{h} \mathcal{K}_{h}\bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}\bigr)-u^{ \Delta t}_{h} \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2},\nabla \phi \bigr) \quad \text{and}\quad R_{2}:=-\frac{1}{2} \bigl(\phi \mathcal{K}_{h}\bigl( \bigl\vert u^{ \Delta t}_{h} \bigr\vert ^{2}\bigr),\operatorname{div}u^{\Delta t}_{h} \bigr), $$
The strong \(L^{s'}(0,T;L^{2})\)-convergence of \(\mathcal{K}_{h}(|u^{\Delta t}_{h}|^{2}\) implies that \(\int _{0}^{T}|R_{1}|\,dt\to 0\). While using the discrete commutator property for \(R_{2}\), we estimate
$$ \begin{aligned} \vert R_{2} \vert &= \frac{1}{2} \bigl(\phi \mathcal{K}_{h}\bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}\bigr)-Q_{h}\bigl( \phi \mathcal{K}_{h}\bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}\bigr)\bigr) ,\operatorname{div}u^{ \Delta t}_{h} \bigr)| \\ &\leq c h \bigl\Vert \mathcal{K}_{h}\bigl( \bigl\vert u^{\Delta t}_{h} \bigr\vert ^{2}\bigr) \bigr\Vert _{2} \bigl\Vert u^{ \Delta t}_{h} \bigr\Vert _{H^{1}}\leq c h \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{4}^{2} \bigl\Vert u^{ \Delta t}_{h} \bigr\Vert _{H^{1}} \\ &\leq c h^{1/2} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{2}^{2} \bigl\Vert u^{\Delta t}_{h} \bigr\Vert _{H^{1}}^{2}, \end{aligned} $$
which shows that \(\int _{0}^{T}|R_{2}|\,dt\to 0\). □