Proof.   Let us first note that, since \(b\) is symmetric, the definition of \(\beta\) actually makes sense due to Assumption 2. Note also that a priori we only have a supremum instead of a maximum in ?? ; we will later justify that it is a maximum. It follows from the definition of \(\beta\) that, for each \(\psi\in L^2(\Omega){\backslash \{0\}}\), \[k \min_{v_h \in V^h} \|{z_\psi}-v_h\|_{1,k} = k \min_{v_h \in V^h} \|(z_\psi/\|\psi\|_n) - v_h\|_{1,k} \|\psi\|_n = k \min_{v_h \in V^h} \|(z_{(\psi/\|\psi\|_n)} - v_h\|_{1,k} \|\psi\|_n \le \beta\|\psi\|_n.\] Hence, there exists \(v_h\in V^h\) such that \[\label{def95beta95reformulated} k\|{z_\psi} - v_h\|_{1,k} \le \beta\|\psi\|_n.\tag{13}\] Now let \(f\in L^2(\Omega)\) and let \(u\in H_0^1(\Omega)\) be the unique solution of 9 . Let us first assume that \(u_h \in V^h\) is any solution of 11 . By Assumption 2, there exists a unique \(\xi {= z_{k(u - u_h)}}\in H^1_0(\Omega)\) such that \[\label{existence95xi} b(w,\xi) = k(w,u-u_h)_{{n}} \qquad \text{for all} \;\; w \in H^1_0(\Omega).\tag{14}\] On one hand, (by 13 with \(\psi=k(u-u_h)\)), there exists \(\xi_h \in V^h\) such that \[\|\xi-\xi_h\|_{1,k} \le \beta \|u-u_h\|_{{n}}.\] On the other hand, taking the test function \(w = u-u_h\) in 14 we see that \[k\|u-u_h\|_{{n}}^2 = b(u-u_h,\xi) = b(u-u_h,\xi-\xi_h) \le \|u-u_h\|_{1,k}\|\xi-\xi_h\|_{1,k},\] where we used 10 and the fact that \(b(u-u_h,w_h)=0\) for \(w_h\in V^h\) since \(u\) solves 9 and \(u_h\) solves 11 . Combining these two estimates we obtain \[k\|u-u_h\|_{{n}} \le \beta\|u-u_h\|_{1,k},\] and hence \[\begin{align} (1-2\beta^2)\|u-u_h\|_{1,k}^2 &\le \|u-u_h\|_{1,k}^2-2k^2\|u-u_h\|_{{n}}^2 = b(u-u_h,u-u_h) \\[1ex] &= b(u-u_h,u-v_h) \le \|u-u_h\|_{1,k}\|u-v_h\|_{1,k} \end{align}\] for all \(v_h \in V^h\). Assuming that \(2\beta^2 < 1\) we obtain the first estimate in ?? . The second estimate follows from the definition of \(\beta\); see again 13 .

At that point, we have established ?? for any solution, but we do not know yet if such a solution exists. However, \(u_h\) is defined through a finite-dimensional linear system. In this setting, stability implies uniqueness, which in turns implies existence. This concludes the proof of ?? .

We now turn our attention to ?? . We first observe that, thanks to Fredholm alternative (applied in \(H^{-1}(\Omega)\)), Assumption 2 guarantees that the mapping \(S: H^{-1}(\Omega) \to H^1_0(\Omega)\) defined by \[b(S\psi,v) = \langle \psi,v \rangle\] for all \(\psi \in H^{-1}(\Omega)\) and \(v \in H^1_0(\Omega)\) is a well-defined continuous linear operator where \(\langle\cdot,\cdot\rangle\) denotes the \(H^{-1}(\Omega)\), \(H_0^1(\Omega)\)-pairing.

Let \(B_{L^2} = \{\phi \in L^2(\Omega) \; | \; \|{n^{-1}}\phi\|_{n} \le 1\}\) denote the closed unit ball in \(L^2(\Omega)\). Since the injection \(L^2(\Omega) \hookrightarrow H^{-1}(\Omega)\) is compact, we conclude that \(K \mathrel{\mathop{:}}=S(B_{L^2})\) is a compact set in \(H^1_0(\Omega)\). Note that for \(\phi \in L^2(\Omega)\) we have \(S\phi = z_{\{n^{-1} \phi\}}\) (with the notation introduced above). Hence, \[\label{beta95K} \max_{v \in K} \min_{v_h \in V^h} \|v-v_h\|_{1,k} = \max_{\substack{\phi \in L^2(\Omega) \\[0.2ex] {\|n^{-1}\phi\|_n \le 1}}}\; \min_{v_h \in V^h} \| z_{(n^{-1}\phi)}-v_h\|_{1,k} = \max_{\substack{{\psi} \in L^2(\Omega) \\[0.2ex] \|\psi\|_n \le 1}}\; \min_{v_h \in V^h} \| z_\psi-v_h\|_{1,k} = \frac{\beta}{k}\,.\tag{15}\] Let us fix \(\varepsilon > 0\). Since \(K\) is compact, there exist \(N\) functions \(\{v^j\}_{j=1}^N \subseteq K\) such that, for each \(v \in K\), there exists \(j \in \{1,\dots,N\}\) with \[k\|v-v^j\|_{1,k} \le \frac{\varepsilon}{3}.\] We then invoke the density of \(C_{\rm c}^\infty(\Omega)\) in \(H^1_0(\Omega)\), which guarantees that, for each \(j\in\{1,\ldots,N\}\), there exists \({\widetilde{v}^j} \in C_{\rm c}^\infty(\Omega)\) with \[k\|v^j-\widetilde{v}^j\|_{1,k} \le \frac{\varepsilon}{3}.\] Now we invoke a degree-explicit approximation result (e.g.Melenk and Sauter [30]) which implies that there exists a constant \(C\) (depending on \(C_{\mathrm{\textsf{SR}}}\)) such that \[\min_{v_h \in V^h} \| \widetilde{v}^j - v_h\|_{H^i(\Omega)} \le C \Bigl(\frac{h}{r}\Bigr)^{2-i}|\widetilde{v}^j|_{H^2(\Omega)} \qquad \text{for each} \;\; i = 0,1, \quad j = 1,\ldots , N.\] This estimate follows from [30] (given only on a unit element), by applying a standard scaling argument on each finite element and using the fact that the resulting global approximation operator maps continuous functions into \(V^h\); see, e.g.the steps in [30]. Hence there exist \(v^1_h,\ldots,v^N_h \in V^h\) such that \[\begin{align} k \|\widetilde{v}^j-v^j_h\|_{1,k} &\le k\Bigl(a_{\mathrm{\textsf{max}}}\|\widetilde{v}^j-v_h^j\|_{H^1(\Omega)}+k^2\|\widetilde{v}^j-v_h^j\|_{L^2(\Omega)}^2\Bigr)^{1/2} \nonumber\\[1ex] &\le C\frac{kh}{r}\biggl(a_{\mathrm{\textsf{max}}}+\Bigl(\frac{kh}{r}\Bigr)^2\biggr)^{1/2}|\widetilde{v}^j|_{H^2(\Omega)} \le C\frac{kh}{r}\biggl(a_{\mathrm{\textsf{max}}}+\Bigl(\frac{kh}{r}\Bigr)^2\biggr)^{1/2}M(\varepsilon) \label{est95interm95ml01} \end{align}\tag{16}\] for \(h/r\) small enough, where \(M(\varepsilon) \mathrel{\mathop{:}}=\max_{1 \le i \le N} |\widetilde{v}^i|_{H^2(\Omega)}.\) Putting the pieces together and using the triangle inequality we obtain \[k \min_{v_h \in V^h} \|v-v_h\|_{1,k} \le \frac{2\varepsilon}{3} + C\frac{kh}{r}\biggl(a_{\mathrm{\textsf{max}}}+\Bigl(\frac{kh}{r}\Bigr)^2\biggr)^{1/2}M(\varepsilon)\] for all \(v \in K\). Let us write \(\beta(h,r)\mathrel{\mathop{:}}=\beta\) to indicate the dependence on \(h\) and \(r\). We can assume, without loss of generality, that \(C \, \ge1\). It follows from 15 and 16 that \[\frac{\beta(h,r)}{C\,} \le \frac{2\varepsilon}{3} + C\frac{kh}{r}\biggl(a_{\mathrm{\textsf{max}}}+\Bigl(\frac{kh}{r}\Bigr)^2\biggr)^{1/2}M(\varepsilon).\] Clearly, there exists \(\delta>0\) such that \(\frac{\beta(h,r)}{C\,}\le\varepsilon\) if \(h/r\in(0,\delta)\). Now choose a decreasing sequence \(\varepsilon_\ell>0\) such that \(\lim_{\ell\to\infty}\varepsilon_\ell=0\). Then there exist \(\delta_\ell>0\) such that \((\delta_\ell)_{\ell=1}^\infty\) is strictly decreasing, \(\lim_{\ell\to\infty}\delta_\ell=0\) and \[\frac{\beta(h,r)}{C\,} \le \varepsilon_\ell \qquad\text{for} \;\; h,r>0 \quad \text{with}\;\;\; \frac{h}{r}<\delta_\ell.\] Defining \(\theta(\delta_\ell)\mathrel{\mathop{:}}=\varepsilon_\ell\) and interpolating linearly we obtain \(\theta\) satisfying ?? . ◻

Proof. The relation \(\widetilde{V}=\widehat{V}\oplus\ker\mathrm{\texttt{c}}\) is clear since \(\dim\widehat{V}+\dim\ker\mathrm{\texttt{c}}=\tilde{n}\) and \(\widehat{V}\cap\ker\mathrm{\texttt{c}}=\{0\}\). The form \(\mathrm{\texttt{c}}|_{\widehat{V}\times\widehat{V}}\) is symmetric and satisfies \(\mathrm{\texttt{c}}(v,v)>0\) for \(v\in\widehat{V}\setminus\{0\}\) by the assumption on \(\widehat{V}\). Hence \(\mathrm{\texttt{c}}|_{\widehat{V}\times\widehat{V}}\) is positive definite. ◻

Proof. Suppose \(s \ge 1\) and \(\mathrm{\texttt{b}}\) is positive definite on \(\ker \mathrm{\texttt{c}}\). Let us choose \(\widehat{V}\subseteq\widetilde{V}\) such that \(\widetilde{V}= \widehat{V}\oplus \ker \mathrm{\texttt{c}}\) (as in Lemma 3); then every \(p \in \widetilde{V}\) has a unique decomposition \(p = \widehat{p}+ p_{\mathrm{\texttt{c}}}\) with \(\widehat{p}\in \widehat{V}\) and \(p_{\mathrm{\texttt{c}}}\in \ker \mathrm{\texttt{c}}\). Thus the GEVP ?? is equivalent to seeking \(\lambda \in \mathbb{R}\), \(\widehat{p}\in \widehat{V}\) and \(p_{\mathrm{\texttt{c}}} \in \ker \mathrm{\texttt{c}}\) such that \(\widehat{p}+ p_{\mathrm{\texttt{c}}} \ne 0\) and \[\begin{align} \label{Ivan1a} \mathrm{\texttt{b}}(\widehat{p}+ p_{\mathrm{\texttt{c}}}, v ) = \lambda \mathrm{\texttt{c}}(\widehat{p}, v) \qquad \text{for all} \;\; v \in \widetilde{V}. \end{align}\tag{17}\] This, in turn, is equivalent to solving the coupled system \[\begin{align} {2} \mathrm{\texttt{b}}(\widehat{p}, v_\mathrm{\texttt{c}}) + \mathrm{\texttt{b}}(p_{\mathrm{\texttt{c}}}, v_{\mathrm{\texttt{c}}}) &= 0 \qquad && \text{for all} \;\; v_{\mathrm{\texttt{c}}} \in \ker \mathrm{\texttt{c}}, \tag{18}\\ \mathrm{\texttt{b}}(\widehat{p},\widehat{v}) + \mathrm{\texttt{b}}(p_{\mathrm{\texttt{c}}},\widehat{v}) &= \lambda \mathrm{\texttt{c}}(\widehat{p},\widehat{v}) \qquad\quad && \text{for all} \;\; \widehat{v}\in \widehat{V}, \tag{19} \end{align}\] for \(\lambda \in \mathbb{R}\), \(\widehat{p}\in \widehat{V}\) and \(p_{\mathrm{\texttt{c}}} \in \ker \mathrm{\texttt{c}}\) such that \(\widehat{p}+ p_{\mathrm{\texttt{c}}} \ne 0\).

Now, by assumption on \(\mathrm{\texttt{b}}\), for each \(\widehat{w}\in \widehat{V}\), there exists a unique \(w_\mathrm{\texttt{c}}\in \ker \mathrm{\texttt{c}}\) such that \[\begin{align} \label{Ivan1c1} \mathrm{\texttt{b}}(w_\mathrm{\texttt{c}}, v_\mathrm{\texttt{c}}) = - \mathrm{\texttt{b}}(\widehat{w}, v_\mathrm{\texttt{c}}) \qquad \text{for all} \;\; v_\mathrm{\texttt{c}}\in \ker \mathrm{\texttt{c}}. \end{align}\tag{20}\] Denoting the solution to 20 as \(w_\mathrm{\texttt{c}}= \mathcal{S}\widehat{w}\), this defines a linear map \(\mathcal{S}: \widehat{V}\rightarrow \ker \mathrm{\texttt{c}}\) with the property \[\begin{align} \label{Ivan1b1} \mathrm{\texttt{b}}(\mathcal{S}\widehat{w}, v_\mathrm{\texttt{c}}) = - \mathrm{\texttt{b}}(\widehat{w}, v_\mathrm{\texttt{c}}) \qquad \text{for all} \;\; \widehat{w}\in \widehat{V}\;\; \text{and} \;\; v_\mathrm{\texttt{c}}\in \ker \mathrm{\texttt{c}}, \end{align}\tag{21}\] or, equivalently, \[\begin{align} \label{Ivan1d} \mathrm{\texttt{b}}\bigl((I + \mathcal{S}) \widehat{w}, v_\mathrm{\texttt{c}}\bigr) = 0 \qquad \text{for all} \;\; \widehat{w}\in \widehat{V}\;\; \text{and} \;\; v_\mathrm{\texttt{c}}\in \ker \mathrm{\texttt{c}}. \end{align}\tag{22}\]

We now consider the GEVP: find \(\lambda \in \mathbb{C}\) and \(\widehat{p}\in \widehat{V}\backslash \{0\}\) such that \[\begin{align} \label{Ivan1e} \widetilde{\mathrm{\texttt{b}}} (\widehat{p}, \widehat{v}) \mathrel{\mathop{:}}=\mathrm{\texttt{b}}\bigl((I + \mathcal{S}) \widehat{p}, \widehat{v}\bigr) = \lambda \mathrm{\texttt{c}}(\widehat{p}, \widehat{v}) \qquad \text{for all} \;\; \widehat{v}\in \widehat{V}. \end{align}\tag{23}\] (This is the Schur complement of 18 , 19 .)

Note that \(\mathcal{S}\) is self-adjoint with respect to \(\mathrm{\texttt{b}}\). This is because, using the symmetry of \(\mathrm{\texttt{b}}\) (at the first and third steps) and the definition of \(\mathcal{S}\) (at the second and fourth steps), \[\mathrm{\texttt{b}}(\mathcal{S}\widehat{p}, \widehat{v}) = \mathrm{\texttt{b}}(\widehat{v}, \mathcal{S}\widehat{p}) = -\mathrm{\texttt{b}}(\mathcal{S}\widehat{v}, \mathcal{S}\widehat{p}) = -\mathrm{\texttt{b}}( \mathcal{S}\widehat{p},\mathcal{S}\widehat{v}) = \mathrm{\texttt{b}}(\widehat{p}, \mathcal{S}\widehat{v}).\] Thus the bilinear form \(\widetilde{\mathrm{\texttt{b}}}\) on the left-hand side of 23 is symmetric on \(\widehat{V}\). Now \(\mathrm{\texttt{c}}\) is symmetric and (by Lemma 3) positive definite on \(\widehat{V}\); so the GEVP 23 has \(\tilde{n}-s\) eigenpairs, which we denote \((\lambda_j, \widehat{p}_{j})\), \(j = 1, \ldots , \tilde{n}-s\), with real finite eigenvalues \(\lambda_j\), which can be ordered as in ?? . The \(\widehat{p}_{j} \in \widehat{V}\) can be chosen orthonormal with respect to \(\mathrm{\texttt{c}}\) and form a basis of \(\widehat{V}\). Now we define \[\begin{align} \label{eigdefj} p_j = (I + \mathcal{S})\widehat{p}_{j}, \qquad j = 1, \ldots, \tilde{n}-s. \end{align}\tag{24}\] By 24 , 23 and since \(\mathcal{S}\widehat{p}_{j} \in \ker \mathrm{\texttt{c}}\), we have \[\mathrm{\texttt{b}}(p_j, \widehat{v}) = {\widetilde{\mathrm{\texttt{b}}}(\widehat{p}_{j},\widehat{v})} = \lambda_j \mathrm{\texttt{c}}(\widehat{p}_{j}, \widehat{v}) = \lambda_j \mathrm{\texttt{c}}(p_{j}, \widehat{v}) \qquad \text{for all} \;\; \widehat{v}\in \widehat{V}.\] Also, by 24 and 22 we have \(\mathrm{\texttt{b}}(p_j, v_{\mathrm{\texttt{c}}}) = 0\) for all \(v_{\mathrm{\texttt{c}}} \in \ker \mathrm{\texttt{c}}\), and so (see 18 and 19 ), \((\lambda_j, p_j),\, j = 1, \ldots ,\tilde{n}-s\) are eigenpairs of the original GEVP ?? . Now, choosing an arbitrary basis, \(\{p_{\tilde{n}-s+1},\ldots,p_{\tilde{n}}\}\), of \(\ker\mathrm{\texttt{c}}\) we obtain a basis of \(\widetilde{V}\). For \(j \in \{1, \ldots , \tilde{n}-s\}\), relation ?? follows for \(j' \in \{ 1, \ldots, \tilde{n}-s\}\) because \(\mathcal{S}\widehat{p}_{j}, \mathcal{S}\widehat{p}_{j'} \in \ker \mathrm{\texttt{c}}\), and so \(\mathrm{\texttt{c}}(p_j,p_{j'}) = \mathrm{\texttt{c}}(\widehat{p}_{j},\widehat{p}_{j'}) = \delta_{j,j'}\) since the \(\{\widehat{p}_{j}: j = 1, \ldots, \tilde{n}-s\}\) were chosen orthonormal with respect to \(\mathrm{\texttt{c}}\). For \(j' \in \{\tilde{n}-s+1, \ldots, \tilde{n}\}\), ?? is trivial because then \(p_{j'} \in \ker \mathrm{\texttt{c}}\).

Relation ?? then follows directly from ?? because, for \(j \in \{1, \ldots, \tilde{n}-s\}\) and any \(j' {\in \{1,\dots,\tilde{n}\}}\), \(\mathrm{\texttt{b}}(p_j, p_{j'}) = \lambda_j\mathrm{\texttt{c}}(p_j, p_{j'})\). This completes the proof when \(s\ge1\), while the case \(s = 0\) is trivial since ?? can then be reduced to a standard eigenvalue problem. ◻

Proof.   Let \(v \in {\widetilde{V}}\) be expressed as in ?? and set \(\alpha_j\mathrel{\mathop{:}}=\mathrm{\texttt{c}}(v,p_j)\), \(j\in\{1,\ldots,\tilde{n}-s\}\). We use the properties from Corollary 1 and Theorem 3 to obtain, \[\begin{align} \mathrm{\texttt{b}}(v-\Pi v,v-\Pi v) &= \mathrm{\texttt{b}}\Biggl(v_0 + \sum_{j=m+1}^{\tilde{n}-s} \alpha_j p_j\, ,\, v_0 + \sum_{j=m+1}^{\tilde{n}-s} \alpha_j p_j \Biggr) \nonumber\\[1ex] &= \mathrm{\texttt{b}}(v_0,v_0) + 2 \sum_{j = m+1}^{\tilde{n}-s} \alpha_j \mathrm{\texttt{b}}( p_j, v_0) + \sum_{j=m+1}^{\tilde{n}-s} \lambda_j \alpha_j^2 \nonumber \\ &= \mathrm{\texttt{b}}(v_0,v_0) + 2 \sum_{j = m+1}^{\tilde{n}-s} \alpha_j \lambda_j \mathrm{\texttt{c}}(p_j, v_0) + \sum_{j=m+1}^{\tilde{n}-s} \lambda_j \alpha_j^2. \end{align}\] Then we use the the fact that \(v_0 \in \ker \mathrm{\texttt{c}}\) and the positive definiteness of \(\mathrm{\texttt{b}}\) on \(\ker \mathrm{\texttt{c}}\) to obtain \[\label{Ivan281} \mathrm{\texttt{b}}(v-\Pi v,v-\Pi v) = \mathrm{\texttt{b}}(v_0,v_0) + \sum_{j=m+1}^{\tilde{n}-s} \lambda_j \alpha_j^2 \ge \sum_{j=m+1}^{\tilde{n}-s} \lambda_j \alpha_j^2 \ge \lambda_{m+1} \sum_{j=m+1}^{\tilde{n}-s} \alpha_j^2.\tag{25}\] Using an almost identical argument and the results from Theorem 3 we also obtain \[\mathrm{\texttt{c}}(v-\Pi v,v-\Pi v) = \sum_{j=m+1}^{\tilde{n}-s} \alpha_j^2. \label{Ivan282}\tag{26}\] Now ?? follows from 25 and 26 . ◻

Proof.   Due to the assumption that we are working with Lagrange elements, for any \(T \in \mathcal{T}_h\), there is an affine map \(F_T\) mapping the unit simplex \(\widehat{T}\) to \(T\). Then (see, e.g.[34]), for any \(w \in V^h\), setting \(\widehat{w} = w \circ F_T\), we have \[\begin{align} \label{eqnorm} \Vert w \Vert_T^2 \;\sim\; h_T^{d} \Vert \widehat{w}\Vert_{\widehat{T}}^2 \;\sim \;h_T^{d} \sum_{j \in \overline{\mathop{\mathrm{dof}}}(T)} w_j^2, \end{align}\tag{39}\] where \(w_j\) is the value of \(w\) at \(j \in \mathop{\mathrm{dof}}(T)\) and the hidden constants are independent of the mesh and \(w\). (The second relation in 39 uses (\(r\)-dependent) equivalence of norms on the space of polynomials of degree \(r\).) Hence, by Definition 2 of \(\Xi_i^{{c}}(v)\) and since freedoms of \(\Xi_i^c(v)\) are interior to \(\Omega_i^c\), we have for \(v \in V_{\Omega_i^c}\), \[\begin{align} \|\Xi_i^{{c}}(v)\|_{\Omega_i^c}^2 &= \sum_{T \subseteq \overline{\Omega_i^c}}\|\Xi_i^{c}(v)\|_{T}^2 \sim \sum_{T \subseteq \overline{\Omega_i^c}} h_T^d \sum_{j \in \overline{\mathop{\mathrm{dof}}}(T) \cap \mathop{\mathrm{dof}}(\Omega_i^c)} (\Xi_i^{{c}}(v))_j^2 = \sum_{T \subseteq \overline{\Omega_i^c}} h_T^d \sum_{j \in \overline{\mathop{\mathrm{dof}}}(T) \cap \mathop{\mathrm{dof}}(\Omega_i^c)} \mu_j^{-2} v_j^2 \nonumber\\[0.5ex] &\ge \Lambda_c^{-2} \sum_{T \subseteq \overline{\Omega_i^c}} h_T^d \sum_ {j \in \overline{\mathop{\mathrm{dof}}}(T) \cap \mathop{\mathrm{dof}}(\Omega_i^c)} v_j^2 \sim \Lambda_c^{-2}\sum_{T \subseteq \overline{\Omega_i^c}}\|v\|_{T}^2 = \Lambda_c^{-2} \|v\|_{\Omega_i^c}^2, \end{align}\] proving ?? . The proof of ?? is analogous but simpler. ◻

Proof. Assume first that ?? is satisfied. Since \(\Lambda_\ell\ge1\), \(\Lambda_c\ge1\), \(a_{\mathrm{\textsf{max}}}\ge1\), \(n_{\mathrm{\textsf{min}}}\le 1\) and \(C_{\ast}\ge 1\), we have \[2C_2 k \Theta^{\frac{1}{2}} (1+C_{\mathrm{\textsf{stab}}}) + 3k H_\ell < \frac{1}{2C_3},\] and hence \[k H_\ell < \frac{1}{6C_3} \qquad\text{and}\qquad (1+C_{\mathrm{\textsf{stab}}})^2 k^{2}\Theta < \frac{1}{16C_2^2C_3^2},\] which yields ?? .

Conversely, assume that ?? is satisfied with \(C>0\) such that ?? holds. Since, by assumption, \({k\ge k_0}\), we have \[\Theta \le \frac{C}{(1+C_{\mathrm{\textsf{stab}}})^2 k^{2}} \le C{k_0^{-2}}.\] Hence, using this with ?? and ?? , we have \[\begin{align} s \le 2C_3 \bigl(1 + \Lambda_\ell \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}C k_0^{-2}\bigr) \bigl(2 C_2\Lambda_c(a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}C)^{1/2} +3\Lambda_\ell C\bigr) < 1, \label{ests} \end{align}\tag{41}\] i.e.@eq:eq:324953 is satisfied.

Moreover, combining ?? with 41 , we obtain a \(k\)-independent lower bound for \(\gamma\). ◻

Proof.   The hidden constants in this proof may depend on \(C_{\mathrm{\textsf{SR}}}\). To obtain ?? , note that, from Definition 2, if \(v \in \widetilde{V}_i\) with \(v_l = 0\) for all \(l \in \mathop{\mathrm{dof}}\Omega_i^c\), then \(\Xi_i^c(v) \equiv 0\) and so \(v \in \ker c_i\). Conversely, if \(v \in \ker c_i\), then, by Assumption 1, \[0 = c_i(v,v) = \int_{\Omega_i^c} \left({\nabla} {\Xi_i^c}(v)\cdot A \nabla {\Xi_i^c}(v) + k^2 {n} \,{(\Xi_i^c(v))^2} \right) \ge {\Vert \nabla \Xi_i^c(v)\Vert_{a,\Omega_i^c}^2 + k^2 n_{\mathrm{\textsf{min}}}\Vert \Xi_i^c(v) \Vert_{\Omega_i^c}^2},\] which implies \(\Xi_i^c(v)\equiv 0\), and thus \(v_l = 0\) for all \(l \in \mathop{\mathrm{dof}}\Omega_i^c\), hence proving ?? .

To obtain ?? , note first that, for any \(v \in \widetilde{V}_i\) (since \(a_{\mathrm{\textsf{min}}}=1 = n_{\mathrm{\textsf{max}}}\)), \[b_{i}(v,v) = \int_{\Omega_i^c} \bigl(\nabla v \cdot (A\nabla v) - k^2 {n} v^2\bigr) \ge \|\nabla v\|_{\Omega_i^c}^2 - k^2 \|v\|_{\Omega_i^c}^2. \label{iggeq1}\tag{42}\] Now let \(v \in \ker c_i\) and note that \(v\) is supported only on elements which touch \(\partial \Omega_i\). For any degree of freedom \(l \in \partial \Omega_i^c\) let \(\mathcal{T}_\ell^0\) denote the set of elements containing the freedom \(l\) and let \(\mathcal{T}_l\) denote the union of \(\mathcal{T}_\ell^0\) with all elements which touch the elements in \(\mathcal{T}_\ell^0\). Then let \(h_\ell = \mathrm{diam}(\mathcal{T}_\ell)\), and let \(B(l,h_l)\) denote the open ball of radius \(h_l\) centred at \(l\). Then it is easy to see that \(\mathop{\mathrm{supp}}v \subseteq \bigcup_{l\in \partial \Omega_i^c} B(l,h_l)\). Moreover, for each \(l\), \(v\) vanishes on a portion of the boundary of \(B(l,h_l)\) having \(d-1\) dimensional measure \(\gtrsim h_l\). So, by the generalised Friedrichs’ inequality (see, e.g.[16]) and since \(h_l \lesssim h\), we have \[\label{Fried} \Vert \nabla v \Vert_{B(l,2h_l)}^2 \gtrsim {h_{l}}^{-2} \Vert v \Vert_{B(l,2h_l)}^2 \gtrsim {h}^{-2} \Vert v \Vert_{B(l,2h_l)}^2.\tag{43}\] Also, by shape regularity, each \(T \in \mathop{\mathrm{supp}}v\) can only be overlapped by a bounded number of balls \(B(l,h_l)\) as \(h \rightarrow 0\); so we have \[\begin{align} \Vert \nabla v \Vert_{\Omega_i^c}^2 &= \sum_{T\in \mathcal{T}_h(v)} \Vert \nabla v \Vert_{T}^2 \sim \sum_{T\in \mathcal{T}_h(v)} \sum_{l \in \partial \Omega_i^c}\Vert \nabla v \Vert_{T\cap B(l,h_l)}^2 = \sum_{l \in \partial \Omega_i^c} \sum_{T\in \mathcal{T}_h(v)} \Vert \nabla v \Vert_{T\cap B(l,h_l)}^2 = \sum_{l \in \partial \Omega_i^c} \Vert \nabla v \Vert_{B(l,h_l)}^2. \end{align}\] Combining this with 43 we obtain \(\Vert \nabla v \Vert_{\Omega_i^c}^2 \gtrsim h^{-2} \Vert v \Vert_{\Omega_i^c}^2\).

Hence, from 42 , there exists a constant \(C'>0\) (which may depend on \(C_{\mathrm{\textsf{SR}}}\)) such that \[b_i (v,v) \ge (C' h^{-2} - k^2) \|v\|_{\Omega_i^c}^2 = (C' - (hk)^2) h^{-2} \|v\|_{\Omega_i^c}^2\] for all \(h\) and \(v \in \ker c_i\). Hence, if \((hk)^2 \le C_1\mathrel{\mathop{:}}=C'/2\), we have \(b_i (v,v) \ge C_1 h^{-2} \|v\|_{\Omega_i^c}^2\). ◻

Proof of Theorem 13. Let \(i\in\{1,\ldots,N\}\). Note that if \(l \in \mathop{\mathrm{dof}}(\Omega_i^c)\), then \(\phi_l^{{i}} \in V_i\). Denoting the dofs in the assumption by \(l_1,\ldots,l_{s_i+1}\) we use Corollary 1 to write \[\begin{align} \label{each95side} \phi_{{l_j}}^{{i}} = v_0^{i,{j}} + \sum_{m=1}^{n_i-s_i}\alpha_m^{i,{j}} p_m^i, \qquad j\in\{1,\ldots,s_i+1\}, \end{align}\tag{44}\] with \(v_0^{i,j}\in\ker c_i\) and \(\alpha_m^{i,j}\in\mathbb{R}\). Since \(s_i=\dim\ker c_i\), there exist \(\beta_1^i,\ldots,\beta_{s_i+1}^i\in\mathbb{R}\), not all vanishing, such that \(\sum_{j=1}^{s_i+1}\beta_j^i v_0^{i,j}=0\). Set \(\psi^i\mathrel{\mathop{:}}=\sum_{{j=1}}^{s_i+1}\beta_{{j}}^i\phi_{{l_j}}^i\). Then multiplying each side of 44 by \(\beta_j^i\) and summing over \(j\) we obtain \[\label{ml02} \psi^i = \sum_{m=1}^{n_i-s_i}\widetilde{\alpha}_m^i p_m^i \qquad \text{with} \;\; \widetilde{\alpha}_m^i = \sum_{j=1}^{s_i+1} \beta_j \alpha_m^{i,j}.\tag{45}\] Further, set \(d_l^i \mathrel{\mathop{:}}=\mathop{\mathrm{diam}}\mathop{\mathrm{supp}}\phi_l^i\). For \(j\in\{1,\ldots,s_i+1\}\), it follows from Friedrichs’ inequality ?? and the assumptions \(a_{\mathrm{\textsf{min}}}= n_{\mathrm{\textsf{max}}}= 1\) that \[\begin{align} b_{i}(\phi_{l_j}^{{i}},\phi_{l_j}^{{i}}) \ge\|\phi_{l_j}^i\|_{a,\Omega_i^c}^2 - k^2\|\phi_{l_j}^i\|_{\Omega_i^c}^2 \ge \Bigl(\frac{2}{(d_{l_j}^i)^2}-k^2\Bigr)\|\phi_{l_j}^i\|_{\Omega_i^c}^2 = \frac{1}{(h_{l_j}^i)^{2}}\biggl(2\Bigl(\frac{h_{l_j}^i}{d_{l_j}^i}\Bigr)^2-(kh_{l_j}^i )^2\biggr)\|\phi_l^i\|_{\Omega_i^c}^2. \label{ml01} \end{align}\tag{46}\] where \(h_{l_j}^i \mathrel{\mathop{:}}=\max\{ h_T: T \subseteq \mathop{\mathrm{supp}}\phi_{l_j}^i\}\). The assumed shape-regularity of the mesh implies that \({h_{l_j}^i/d_{l_j}^i}\) is bounded below by a positive constant, and we have \(k h_{l_j}^i \le kh\); so, for small enough \(hk\), the right-hand side of 46 is strictly positive for all \(j\in\{1,\ldots,s_i+1\}\). Since \(\mathop{\mathrm{supp}}\phi_{l_j}^i\cap\mathop{\mathrm{supp}}\phi_{l_{j'}}^i\) has measure zero for \(j\ne j'\), we also have \(b_{i}(\phi_{l_j}^i,\phi_{l_{j'}}^i)=0\) for \(j\ne j'\) and hence \(b_{i}(\psi^i,\psi^i) = \sum_{j=1}^{s_i+1} (\beta_j^i)^2 b_i(\phi_{l_j}^i,\phi^i_{l_j}) > 0\).

Now, if \(\lambda_m^i\le0\) for all \(m=1,\ldots,n_i-s_i\), then, by 45 and ?? , \[b_{i}(\psi^i,\psi^i) = \sum_{m=1}^{n_i-s_i}\lambda_m^i (\widetilde{\alpha}_m^i)^2 \le 0,\] which is a contradiction. ◻

Proof.   We recall from equation 23 that \(\lambda_1^i\) is the minimum eigenvalue of the problem: find \(\lambda^i \in \mathbb{R}\) and \(\widehat{p}^i \in \widehat{V}_i\backslash \{0\}\) such that \[\label{Ivan3195old} \widetilde{b}_i (\widehat{p}^i, \widehat{v}^i) \mathrel{\mathop{:}}=b_i\bigl((I + \mathcal{S}_i ) \widehat{p}^i, \widehat{v}^i\bigr) = \lambda^i c_i(\widehat{p}^i, \widehat{v}^i) \qquad \text{for all} \;\; \widehat{v}^i\in \widehat{V}_i,\tag{49}\] where \(\mathcal{S}_i: \widehat{V}_i \rightarrow \ker c_i\) is the operator defined by \[\label{Ivan31aa} b_i(\mathcal{S}_i \widehat{v}^i, {w_{c_i}}) = - b_i(\widehat{v}^i, {w_{c_i}}) \qquad \text{for all} \;\; {w_{c_i}} \in \ker c_i.\tag{50}\] Using 50 and the fact that \(\mathcal{S}_i\) maps into \(\ker c_i\), we have that 49 is equivalent to \[\label{Ivan31} b_i\bigl((I + \mathcal{S}_i ) \widehat{p}^i, (I + \mathcal{S}_i) \widehat{v}^i\bigr) = \lambda^i c_i(\widehat{p}^i, \widehat{v}^i) \qquad \text{for all} \;\; \widehat{v}^i\in \widehat{V}_i.\tag{51}\] Since, by Lemma 3, \(c_i\) is positive definite on \(\widehat{V}_i\), we obtain from Rayleigh’s principle that \[\label{Rayleigh} \lambda_1^i = \min_{0 \ne {\widehat{v}^i} \in \widehat{V}_i}\frac{{b_i\bigl((I + \mathcal{S}_i ) \widehat{v}^i, (I + \mathcal{S}_i) \widehat{v}^i\bigr)}}{c_{i} (\widehat{v}^i,\widehat{v}^i)}.\tag{52}\] To estimate 52 from below, we first derive an estimate from above for \(\Vert \mathcal{S}_i\widehat{v}^i \Vert_{\Omega_i^c}\) for any \(\widehat{v}^i \in \widehat{V}_i\). Using ?? , the definition 50 of \(\mathcal{S}_i\), the fact that \(n_{\mathrm{\textsf{max}}}= 1\) and ?? , we have, for any \(\widehat{v}^i \in \widehat{V}_i\), \[\begin{align} C_1 h^{-2} \Vert \mathcal{S}_i \widehat{v}^i \Vert_{\Omega_i^c}^2 &\le b_i(\mathcal{S}_i \widehat{v}^i ,\mathcal{S}_i \widehat{v}^i ) = -b_i(\widehat{v}^i ,\mathcal{S}_i \widehat{v}^i ) \nonumber \\[0.5ex] &\le \Vert \nabla(\mathcal{S}_i \widehat{v}^i )\Vert_{\Omega_i^c} \Vert \nabla \widehat{v}^i \Vert_{\Omega_i^c} + k^2 \Vert \mathcal{S}_i \widehat{v}^i \Vert_{\Omega_i^c} \Vert \widehat{v}^i \Vert_{\Omega_i^c} \nonumber \\[0.5ex] & \le (a_{\mathrm{\textsf{max}}}C^2_{\mathrm{\textsf{inv}}}+ 1)k^2 \Vert \mathcal{S}_i \widehat{v}^i \Vert_{\Omega_i^c} \Vert \widehat{v}^i \Vert_{\Omega_i^c}, \label{Ivan31a} \end{align}\tag{53}\] and thus \[\label{Ivan32} \|\mathcal{S}_i\widehat{v}^i\|_{\Omega_i^c} \le (a_{\mathrm{\textsf{max}}}C^2_{\mathrm{\textsf{inv}}}+ 1) C_1^{-1} (hk)^2 \Vert \widehat{v}^i \Vert_{\Omega_i^c} \le (a_{\mathrm{\textsf{max}}}C^2_{\mathrm{\textsf{inv}}}+ 1)\Vert \widehat{v}^i \Vert_{\Omega_i^c}.\tag{54}\] As a result, for the numerator in 52 , we have \[\label{Ivan34} b_i\bigl((I+\mathcal{S}_i)\widehat{v}^i , (I+\mathcal{S}_i)\widehat{v}^i\bigr) \ge -k^2 \|(I+\mathcal{S}_i)\widehat{v}^i\|_{\Omega_i^c}^2 \ge -k^2 (a_{\mathrm{\textsf{max}}}C^2_{\mathrm{\textsf{inv}}}+ 2)^2 \|\widehat{v}^i\|_{\Omega_i^c}^2.\tag{55}\] For the denominator note that \(\widehat{V}_i=V_i\) by 47 ; hence Lemma 4 implies that \[c_i(\widehat{v}^i, \widehat{v}^i) \ge k^2 n_{\mathrm{\textsf{min}}}\Vert \Xi_i^c( \widehat{v}^i )\Vert_{\Omega_i^c}^2 \gtrsim k^2 n_{\mathrm{\textsf{min}}}\Lambda_c^{-2} \Vert \widehat{v}^i \Vert_{\Omega_i^c}^2. \label{Ivan35}\tag{56}\] For the rest of this proof (only) let \(\mathbf{min}\) denote the minimum of any quantity over all \(0 \ne {\widehat{w}^i} \in \widehat{V}_i\). Recall also that \(c_i\) is positive definite on \(\widehat{V}_i\). If \(\mathbf{min}\,\widetilde{b}_i(\widehat{w}^i,\widehat{w}^i)\ge0\), then \(\lambda_1^i\ge0\) and ?? is trivially true. Now, assume that \(\mathbf{min}\,\widetilde{b}_i(\widehat{w}^i,\widehat{w}^i) < 0\). Then, for all \(0 \ne \widehat{v}^i \in \widehat{V}_i\), \[\frac{\widetilde{b}_i(\widehat{v}^i, \widehat{v}^i)}{c_i(\widehat{v}^i, \widehat{v}^i)} \ge \frac{\mathbf{min}\,\widetilde{b}_i(\widehat{w}^i, \widehat{w}^i)}{c_i(\widehat{v}^i, \widehat{v}^i)} \ge \frac{\mathbf{min}\,\widetilde{b}_i(\widehat{w}^i, \widehat{w}^i)}{\mathbf{min}\,{c_i(\widehat{w}^i, \widehat{w}^i)}} = \frac{\mathbf{min}\,{b}_i((I + \mathcal{S}_i)\widehat{w}^i, (I + \mathcal{S}_i)\widehat{w}^i)}{\mathbf{min}\,{c_i(\widehat{w}^i, \widehat{w}^i)}},\] and so, by 55 , 56 , for all \(0 \ne \widehat{v}^i \in \widehat{V}_i\), \[\frac{ \widetilde{b}_i( \widehat{v}^i, \widehat{v}^i )}{c_i(\widehat{v}^i, \widehat{v}^i)} \gtrsim - \Lambda_c^{2}n_{\mathrm{\textsf{min}}}^{-1}(a_{\mathrm{\textsf{max}}}C^2_{\mathrm{\textsf{inv}}}+2)^2,\] which proves ?? also in this case. ◻

Proof.   The inequality in ?? follows directly from ?? of Proposition 5. Set \(\alpha_m^i \mathrel{\mathop{:}}=c_{i}(v,p_m^i)\). Then \(w = v_0 + \sum_{m = m_i+1}^{n_i-s_i} \alpha_m^i p_m^i\) with \(v_0 \in \mathop{\mathrm{span}}\{ p_{n_i-s_1+1}\ldots , p_{n_i}\} = \ker c_i\). Hence using property ?? , then Lemma 5 and the fact that \(\lambda_m^i > 0\) for all \(m \in \{m_i +1, \ldots n_i-s_i\}\), we obtain \[\begin{align} b_{i} (w,w) &= b_{i} (v_0,v_0) + 2 \sum_{m= m_i+1}^{n_i-s_i} \alpha_m^i b_{i} (p_m^i,v_0) + \sum_{m,m'= m_i+1}^{n_i-s_i} \alpha_m^i \alpha_{m'}^i b_{i} (p_m^i,p_{m'}^i) \\ &= b_{i} (v_0,v_0) + \sum_{m = m_i+1}^{n_i-s_i} (\alpha_m^i)^2 \lambda_m^i \ge 0, \end{align}\] i.e.the left-hand inequality in ?? . To obtain the right-hand inequality in ?? , note that \(b_{i}(w,\Pi_{m_i}v) = 0\), and hence \[\label{Lem341} b_{i}\!\left( w , w \right) = b_{i}(v,v) - b_{i}( \Pi_{m_i} v,\Pi_{m_i} v) = a_{\Omega_i^c}(v,v) - k^2 \|v\|_{n,\Omega_i^c}^2 - b_{i}( \Pi_{m_i} v, \Pi_{m_i} v).\tag{57}\] To estimate the last term in 57 we use ?? , ?? , ?? and then ?? to obtain \[\begin{align} -b_{i}\!\left( \Pi_{m_i} v, \Pi_{m_i} v \right) &= - \sum_{m=1}^{m_i} (\alpha_m^i)^2 \, \lambda_m^i \le -\lambda_1^i \sum_{m=1}^{m_i}(\alpha_m^i)^2 \nonumber\\ &\lesssim C_{\mathrm{\textsf{eig}}}\sum_{m=1}^{m_i} (\alpha_m^i)^2 \le C_{\mathrm{\textsf{eig}}}\sum_{m=1}^{n_i-s_i} (\alpha_m^i)^2 = C_{\mathrm{\textsf{eig}}}c_{i}(v,v). \label{Lem342} \end{align}\tag{58}\] Hence combining 57 and 58 we obtain \[\label{Lem343} b_i(w,w) \lesssim a_{\Omega_i^c}(v,v) + C_{\mathrm{\textsf{eig}}}\, c_i(v,v).\tag{59}\] To conclude, we now use ?? , the fact that \(a_{\mathrm{\textsf{max}}}\ge 1\) and \(C_{\mathrm{\textsf{inv}}}\ge 1\), and ?? to obtain \[\begin{align} c_i(v,v) &\le a_{\mathrm{\textsf{max}}}\|\nabla \Xi_i^c(v)\|_{\Omega_i^c}^2 + k^2 \|\Xi_i^c(v)\|_{\Omega_i^c}^2 \nonumber\\ &\le (a_{\mathrm{\textsf{max}}}C^2_{\mathrm{\textsf{inv}}}+ 1)k^2 \Vert \Xi_i^c(v)\Vert_{\Omega_i^c}^2 \lesssim (a_{\mathrm{\textsf{max}}}C^2_{\mathrm{\textsf{inv}}}+ 1 ) k^2\Vert v \Vert_{\Omega_i^c}^2 \lesssim a_{\mathrm{\textsf{max}}}C^2_{\mathrm{\textsf{inv}}}k^2 \Vert v \Vert_{\Omega_i^c}^2, \label{hidden} \end{align}\tag{60}\] and the proof follows on combining 60 with 59 , using \(C_{\mathrm{\textsf{eig}}}\ge1\) and recalling Notation 1. ◻

Proof.   Since \(\Pi_{m_i}^c v|_{\Omega_i^c}\in\mathop{\mathrm{span}}\{p_1^i,\ldots,p_{m_i}^i\}\), it is clear that \(z_0 \in V_0\), which also proves the first inequality in ?? . Using 38 , 31 , ?? and ?? we obtain \[\begin{align} \|v-z_0\|_{1,k}^2 &= \Bigg\| \sum_{i=1}^N E_i^c (\Xi_i^c (v|_{\Omega_i^c}))-\sum_{i=1}^N E^c_i \Xi^c_i \bigl(\Pi_{m_i} {v|_{\Omega_i^c}}\bigr) \Bigg\|_{1,k}^2 \\ &\le \Lambda_c \sum_{i=1}^N \left \|\Xi_i^c \left(v|_{\Omega_i^c}- \Pi_{m_i} {v|_{\Omega_i^c}} \right) \right\|_{1,k,{\Omega_i^c}}^2 \nonumber = \Lambda_c \sum_{i=1}^N c_{i}\Bigl( \bigl({v|_{\Omega_i^c}}-\Pi_{m_i}{v|_{\Omega_i^c}}\bigr), \bigl({v|_{\Omega_i^c}}-\Pi_{m_i}{v|_{\Omega_i^c}}\bigr)\Bigr) \nonumber \\ &\le \Lambda_c \sum_{i=1}^N \frac{1}{\lambda_{m_i+1}^i} b_{i}\Bigl(\bigl({v|_{\Omega_i^c}}-\Pi_{m_i}{v|_{\Omega_i^c}}\bigr),\bigl({v|_{\Omega_i^c}}-\Pi_{m_i}{v|_{\Omega_i^c}}\bigr)\Bigr) \nonumber \\ &\le C_2^2 \Lambda_c a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1}C_{\ast}\sum_{i=1}^N\frac{1}{\lambda_{m_i+1}^i}\big\|v|_{\Omega_i^c}\big\|_{1,k,\Omega_i^c}^2 \le C_2^2 \Lambda_c a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1}C_{\ast}\Theta\sum_{i=1}^N\big\|v|_{\Omega_i^c}\big\|_{1,k,\Omega_i^c}^2, \label{proof95inequ95sum95z95i} \end{align}\tag{61}\] and the result follows from 28 . ◻

Proof.   We choose a partition of unity consisting of real-valued Lipschitz functions \(\{\Phi_j^{\ell}\}_{j=1}^Q\) on \(\Omega\) satisfying \(0 \le \Phi_j^{\ell} \le 1\), \(\mathop{\mathrm{supp}}(\Phi_j^{\ell}) \subseteq \overline{\Omega_j^\ell}\) for each \(j\), and \(\sum_{j=1}^Q \Phi_j^{\ell} \equiv 1\) on \(\Omega\); see, e.g.[16]. Then \(\|\nabla \Phi_j^{\ell}\|_{L^\infty(\Omega)} \le C_{\mathrm{\textsf{POU}}}\delta_\ell^{-1}\) for all \(j\), where \(\delta_\ell\) is the overlap parameter from Assumption 4. By [32] and then Assumption 4, for any \(v \in H^1_0(\Omega)\), \[\label{multPOU} \sum_{j=1}^Q \| \Phi_j^{\ell} v \|_{1,k,\Omega_j^\ell}^2 \lesssim \Lambda_\ell \left(1 + \frac{1}{(k{\delta_\ell})^2} \right) \|v\|_{1,k,\Omega}^2 \lesssim \Lambda_\ell \|v\|_{1,k,\Omega}^2,\tag{62}\]

Now define \(z_j^\ell \mathrel{\mathop{:}}=\Phi_j^\ell(v - z_0) \in V_j^\ell\). Then \[\sum_{j=1}^Q E_j^\ell z_j^\ell = \sum_{j=1}^Q E_j^\ell \Phi_j^\ell (v - z_0) = (v - z_0)\sum_{j=1}^Q {\Phi_j^\ell} = v - z_0.\] Moreover, from Lemma 6 we have \[\|z_0\|_{1,k}^2 \le 2\left(\|z_0 - v\|_{1,k}^2 + \|v\|_{1,k}^2\right) \lesssim (1 + \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}\Theta) \|v\|_{1,k}^2.\] Also, using 62 , we obtain \[\sum_{j=1}^Q \|z_j^\ell\|_{1,k,\Omega_j^\ell}^2 = \sum_{j=1}^Q \| \Phi_j^\ell (v - z_0)\|_{1,k,\Omega_j^\ell}^2 \lesssim \Lambda_\ell \|v - z_0\|_{1,k}^2.\] Combining this with Lemma 6 and the bound for \(z_0\) we obtain the result. ◻

Proof.   Using Lemma 7, the definitions of \(P_j^\ell\) and \(P_0\) and then Cauchy–Schwarz, we get \[\begin{align} \|v\|_{1,k}^2 &= \Biggl(v, z_0 + \sum_{j=1}^{Q} {E_j^\ell} {z_j^\ell}\Biggr)_{1,k} = (v, z_0)_{1,k} + \Biggl(v,\sum_{j=1}^{Q} {E_j^\ell} {z_j^\ell}\Biggr)_{1,k} \\ &= (P_0v, z_0)_{1,k} + \sum_{j=1}^{Q}( {P_j^\ell} v, {z_j^\ell})_{1,k, {\Omega^\ell_j}} \le \| P_0v\|_{1,k} \|z_0\|_{1,k} + \sum_{j=1}^{Q}\| {P_j^\ell} v \|_{1,k, {\Omega^\ell_j}} \|{z_j^\ell}\|_{1,k, {\Omega^\ell_j}}{.} \end{align}\] We now apply Cauchy–Schwarz and Lemma 7 again to obtain \[\begin{align} \|v\|_{1,k}^2 &\le \left(\| P_0v\|_{1,k}^2 + \sum_{j=1}^{Q}\| {P_j^\ell} v \|_{1,k, {\Omega^\ell_j}}^2 \right)^{1/2} C_3^{1/2}\bigl(1 + \Lambda_\ell \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}\Theta \bigr)^{1/2} \|v\|_{1,k} \\[1ex] &= \left(( E_0 P_0v, v)_{1,k} + \sum_{j=1}^{Q} ( {E_j^\ell} {P_j^\ell} v, v)_{1,k} \right)^{1/2} C_3^{1/2}\bigl(1 + \Lambda_\ell \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}\Theta \bigr)^{1/2}\|v\|_{1,k} \\[1ex] &= \left( E_0 P_0 v + \sum_{j=1}^{Q}{E_j^\ell} {P_j^\ell} v, v \right)_{1,k}^{1/2} C_3^{1/2}\bigl(1 + \Lambda_\ell \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}\Theta \bigr)^{1/2} \|v\|_{1,k} \\ &= (P v,v)_{1,k}^{1/2} \, C_3^{1/2}\bigl(1 + \Lambda_\ell \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}\Theta \bigr)^{1/2}\|v\|_{1,k}, \end{align}\] and the inequality ?? follows. ◻

Proof. Using Friedrichs’ inequality ?? and the assumption that \(a_{\mathrm{\textsf{min}}}= 1 = n_{\mathrm{\textsf{max}}}\), we obtain, for \(u\in V_{\Omega_j^\ell}\) \[b_{{\Omega^\ell_j}}(u,u) = a_{{\Omega^\ell_j}}(u,u) - k^2({n}u,u)_{{\Omega^\ell_j}} \ge \frac{2}{{{(H_\ell)}}^2}\|u\|_{{\Omega^\ell_j}}^2 - k^2\|u\|_{{\Omega^\ell_j}}^2 = \frac{2-{{(H_\ell)}}^2k^2}{{{(H_\ell)}}^2}\|u\|_{{\Omega^\ell_j}}^2,\] and hence \(b_{{\Omega^\ell_j}}(\cdot,\cdot)\) is positive definite when \(H_\ell k < \sqrt{2}\). The square system 35 defining \(T_j^\ell v\) is then positive definite and the result follows. ◻

Proof.   Using 35 , 10 , the fact that \(n_{\mathrm{\textsf{max}}}= 1 = a_{\mathrm{\textsf{min}}}\), and Friedrichs’ inequality ?? , we obtain \[\begin{align} \|{T_j^\ell} v\|^2_{1,k,{\Omega^\ell_j}} &= b_{{\Omega^\ell_j}}({T_j^\ell} v,{T_j^\ell} v) + 2k^2\|T_j^\ell v\|_{n,\Omega_i^\ell}^2 = b_{\Omega_i^\ell}\bigl(v|_{\Omega^\ell_j},{T_j^\ell} v\bigr) + 2k^2\|T_j^\ell v\|_{n,\Omega^\ell_j}^2 \\ &\le \big\|v|_{\Omega^\ell_j}\big\|_{1,k,\Omega^\ell_j}\|T_j^\ell v\|_{1,k,\Omega^\ell_j} + 2k^2\|T_j^\ell v\|_{\Omega^\ell_j}^2 \\ &\le \big\|v|_{\Omega^\ell_j}\big\|_{1,k,\Omega^\ell_j}\|{T_j^\ell} v\|_{1,k,\Omega^\ell_j} + k^2(H_\ell)^2\|T_j^\ell v\|^2_{1,k,\Omega^\ell_j}, \end{align}\] which implies \(\bigl(1-k^2 (H_\ell)^2\bigr)\|T_j^\ell v\|_{1,k,\Omega^\ell_j} \le \big\|v|_{\Omega^\ell_j}\big\|_{1,k,\Omega^\ell_j}\), and the result follows. ◻

Proof.   Let \(C\) and \(\Theta\) be as in Lemma 1. Then there exists \(h_1>0\) such that \[C\,\Theta\Bigl(\frac{h}{r}\Bigr) \le \min\biggl\{\frac{1}{2},\frac{k_0n_{\mathrm{\textsf{min}}}^{1/2}}{2}\biggr\} \qquad \text{for all} \;\; h\in(0,h_1).\] Let \(h\in(0,h_1)\) and assume (for a contradiction) that there exists a \(w_0 \in V_0\backslash \lbrace 0 \rbrace {\subseteq V^h}\) such that \[\label{eq:w95095sol} b(w_0,z) = 0 \qquad \text{for all} \;\; z \in V_0.\tag{66}\] Let \(w \in H^1_0(\Omega)\) be the solution of \(b(w,v)=(w_0,v)\) for all \(v \in H^1_0(\Omega)\), which exists and is unique by Assumption 2. It follows from Lemma 1 that \(\beta\le\min\bigl\{1/2,k_0n_{\mathrm{\textsf{min}}}^{1/2}/2\bigr\}<1/\sqrt{2}\) (where \(\beta\) is defined as in Lemma 1) and there exists \(w_h \in V^h\) such that \[b(w_h, v) = (w_0,v) \quad \text{for all} \;\; v \in V^h, \quad \text{and} \quad \|w - w_h\|_{1,k} \le \frac{2 \beta}{k n_{\mathrm{\textsf{min}}}^{1/2}} \|w_0\| \le \|w_0\|. \label{ScottWang}\tag{67}\] Let \(z\in V_0\) be arbitrary. Inserting \(v=w_0\) in 67 and combining with 66 and then 10 , we obtain \[\|w_0\|^2 = b(w_h, w_0) = b(w_h - z, w_0) \le \|w_h - z\|_{1,k} \|w_0\|_{1,k} \qquad\text{for all} \;\; z \in V_0.\] Since this is true for all \(z \in V_0\), Lemma 6 yields \[\label{eq:32395595f95sq} \|w_0\|^2 \le \inf_{z \in V_0}\|w_h-z\|_{1,k} \|w_0\|_{1,k} \le C_2\, \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} \Theta^{\frac{1}{2}}\|w_h\|_{1,k} \, \|w_0\|_{1,k}.\tag{68}\] Also, equation 66 with \(z=w_0\) shows that \(0 = b(w_0,w_0) = \|w_0\|_{1,k}^2-2k^2\|w_0\|_n^2\), which, together with 68 , implies \[\begin{align} \|w_0\|^2 &\le C_2 \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} \Theta^{\frac{1}{2}} \|w_h\|_{1,k}\sqrt{2}\, k \|w_0\|_n \\ &\le \sqrt{2}\, C_2 \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} k \Theta^{\frac{1}{2}} \|w_h\|_{1,k} \|w_0\|, \end{align}\] and hence \[\label{eq:323955951} \|w_0\| \le \sqrt{2}\, C_2 \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} k \Theta^{\frac{1}{2}} \|w_h\|_{1,k}.\tag{69}\] Then we use the triangle inequality, 67 and ?? to obtain \[\|w_h\|_{1,k} \le \|w-w_h\|_{1,k} + \|w\|_{1,k} \le (1 + C_{\mathrm{\textsf{stab}}})\|w_0\|. \label{eq:32w95h95comb}\tag{70}\] With 70 used in 69 , we arrive at \[\|w_0\| \le \sqrt{2}\, C_2 \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} k \Theta^{\frac{1}{2}} (1+C_{\mathrm{\textsf{stab}}})\|w_0\|.\] Together with ?? this leads to a contradiction. ◻

Proof.   Let \(h_1>0\) be as in Lemma 10. Further, let \(u \in V^h\) and consider the auxiliary problem: \[\label{eq323957954} \text{find } w_h \in V^h \;\; \text{such that} \quad b(w_h, v) = \bigl(T_0 u - u,v\bigr) \quad \text{for all} \;\; v \in V^h.\tag{71}\] Analogously to 70 , problem 71 has a unique solution \(w_h\) for \(h\in(0,h_1)\), and \[\label{estwh} \|w_h\|_{1,k} \le (1+C_{\mathrm{\textsf{stab}}}) \|T_0 u - u\|.\tag{72}\] Now, by the definition 64 of \(T_0\), we have \(b(T_0 u-u,z)=0\) for all \(z \in V_0\). Inserting \(v = T_0 u - u \in V^h\) into 71 and using 10 , we obtain, for every \(z\in V_0\), \[\|T_0 u - u\|^2 = b(w_h, T_0 u - u) = b(w_h - z, T_0 u - u) \le \|w_h-z\|_{1,k}\,\|T_0 u - u\|_{1,k}.\] Combining this with Lemma 6 we obtain \[\label{ineq:T0u-u} \|T_0 u - u\|^2 \le \|T_0 u - u\|_{1,k}\inf_{z \in V_0}\|w_h-z\|_{1,k} \le \|T_0 u - u\|_{1,k} {C_2}\, \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} \Theta^{1/2} \|w_h\|_{1,k}.\tag{73}\] Together with 72 , this proves ?? .

Let us now show ?? . Inserting \(P_0 u - T_0 u \in V_0\) in the definition 36 of \(T_0\) we obtain \[b(u-T_0 u, P_0 u - T_0 u) = 0.\] Using this together with the link between the bilinear forms \((\cdot,\cdot)_{1,k}\) and \(b\), and the Cauchy–Schwarz inequality we obtain \[\begin{align} \|u - T_0 u\|_{1,k}^2 &= (u - T_0 u, u - T_0 u)_{1,k} = b(u - T_0 u, u - T_0 u) + 2k^2(u - T_0 u, u - T_0 u)_n \\[0.5ex] &= b(u - T_0 u, u - T_0 u) - b(u - T_0 u, P_0 u - T_0 u) + 2 k^2 (u - T_0 u, u - T_0 u)_n \\[0.5ex] &= b(u - T_0 u, u - P_0 u) + 2k^2(u - T_0 u, u - T_0 u)_n \\[0.5ex] &= (u - T_0 u, u - P_0 u)_{1,k} - 2k^2(u - T_0 u, u - P_0 u)_n + 2k^2(u - T_0 u, u - T_0 u)_n \\[0.5ex] &= (u - T_0 u, u - P_0 u)_{1,k} + 2k^2(u - T_0 u, P_0 u - T_0 u)_n \\[0.5ex] &\le \|u - T_0 u\|_{1,k}\|u - P_0 u\|_{1,k} + 2k^2\|u - T_0 u\|_n \|T_0 u - P_0 u\|_n \\[0.5ex] &\le \|u - T_0 u\|_{1,k}\|u - P_0 u\|_{1,k} + 2k\|u - T_0 u\|_n \|T_0 u - P_0 u\|_{1,k} \\[0.5ex] &= \|u - T_0 u\|_{1,k}\|u - P_0 u\|_{1,k} + 2k\|u - T_0 u\|_n \|P_0(T_0 u - u)\|_{1,k} \\[0.5ex] &\le \|u - T_0 u\|_{1,k}\|u\|_{1,k} + 2k\|u - T_0 u\|_n \|T_0 u - u\|_{1,k}, \end{align}\] where in the last step we used the fact that \(P_0\) is the orthogonal projection onto \(V_0\) with respect to \((\cdot,\cdot)_{1,k}\). Dividing both sides by \(\|u - T_0 u\|_{1,k}\) and then using ?? , \(n_{\mathrm{\textsf{max}}}= 1\) and the assumption ?? , we arrive at \[\begin{align} \|u - T_0 u\|_{1,k} &\le \|u\|_{1,k} + 2k\|u - T_0 u\|_n \le \|u\|_{1,k} + 2{C_2}\, \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} k \Theta^{1/2} (1 + C_{\mathrm{\textsf{stab}}})\|u - T_0 u\|_{1,k} \\[0.5ex] &\le \|u\|_{1,k} + \frac{1}{2}\|u - T_0 u\|_{1,k}, \end{align}\] which proves ?? . ◻

Proof.   Let \(u \in V^h\). We proceed in several steps.
Step 1 (Relation between \((Tu,u)_{1,k}\) and \((Pu,u)_{1,k}\)). Using 65 and the definition of \(b\), we obtain \[\begin{align} (Pu,u)_{1,k} & {= (u, Pu)_{1,k}} = \biggl(u,E_0 P_0u + \sum^{Q}_{j=1} E^{\ell}_j P^{\ell}_j u\biggr)_{1,k} \\ &= b(u,E_0 P_0u) + 2k^2(u,E_0 P_0u)_n + \sum^{Q}_{j=1} \left[ b\bigl(u, E^{\ell}_j P^{\ell}_j u\bigr) + 2k^2\bigl(u,E^{\ell}_j P^{\ell}_j u\bigr)_n \right]. \end{align}\] Then, using 35 and 36 , we obtain \[\begin{align} (Pu,u)_{1,k} &= b(T_0 u,P_0u) + 2k^2(u,E_0 P_0u)_n + \sum^{Q}_{j=1} \left[ b_{\Omega^{\ell}_j}\bigl(T^{\ell}_j u, P^{\ell}_j u\bigr) + 2k^2\bigl(u,E^{\ell}_j P^{\ell}_j u\bigr)_n \right] \\[0.5ex] &= (T_0 u,P_0u)_{1,k} - 2k^2(T_0 u,P_0u)_n + 2k^2(u,E_0 P_0u)_n \\[0.5ex] &\quad + \sum^{Q}_{j=1} \left[ \bigl(T^{\ell}_j u, P^{\ell}_j u\bigr)_{1,k,\Omega^{\ell}_j} - 2k^2\bigl(T^{\ell}_j u, {P^{\ell}_j} u\bigr)_{n,\Omega^{\ell}_j} + 2k^2\bigl(u,E^{\ell}_j P^{\ell}_j u\bigr)_n \right]. \end{align}\] Now using the definitions 63 and 64 of \(P_j^\ell\) and \(P_0\), we have \[\begin{align} (Pu,u)_{1,k} &= (E_0 T_0 u,u)_{1,k} - 2k^2(E_0 T_0 u,E_0 P_0u)_n + 2k^2(u,E_0 P_0u)_n \\[0.5ex] &\quad + \sum^{Q}_{j=1} \left[ (E^{\ell}_j T^{\ell}_j u, u)_{1,k} - 2k^2\bigl(E^{\ell}_j T^{\ell}_j u, E^{\ell}_j P^{\ell}_j u\bigr)_n + 2k^2\bigl(u,E^{\ell}_j P^{\ell}_j u\bigr)_n \right]. \end{align}\] We now rearrange the right-hand side to obtain \[\begin{align} (Pu,u)_{1,k} &= (E_0 T_0 u,u)_{1,k} + \sum^{Q}_{j=1} \bigl(E^{\ell}_j T^{\ell}_j u, u\bigr)_{1,k} - 2k^2 \biggl(\bigl(E_0 T_0 u - u,E_0 P_0u\bigr)_n + \sum^{Q}_{j=1} \bigl(E^{\ell}_j T^{\ell}_j u - u, E^{\ell}_j P^{\ell}_j u\bigr)_n \biggr) \nonumber\\[0.5ex] &= (T u, u)_{1,k} - R, \label{relation} \end{align}\tag{74}\] with \(T\) defined in 37 and \[\begin{align} \label{defR} R \mathrel{\mathop{:}}=2k^2 \biggl( \bigl(E_0 T_0 u - u, E_0 P_0u\bigr)_n + \sum^{Q}_{j=1} \bigl(E^{\ell}_j T^{\ell}_j u - u, E^{\ell}_j P^{\ell}_j u\bigr)_n \biggr). \end{align}\tag{75}\] Step 2 (Bounding the first term in \(R\)). Recalling that \(E_0 w_0 = w_0\) for all \(w_0\in V_0\), and noting that the assumption in ?? implies that ?? is satisfied, we can use ?? , ?? and the fact that \(P_0\) is the orthogonal projection onto \(V_0\) with respect to \((\cdot, \cdot)_{1,k}\), to obtain \[\begin{align} k^2\bigl(E_0 T_0 u - u, E_0 P_0 u\bigr)_n &\le k^2 \|E_0 T_0 u - u\|_n \|E_0 P_0 u\|_n = k^2\| T_0 u -u\|_n \| P_0 u\|_n \nonumber\\[0.5ex] &\le k\|T_0 u -u\|\, \|P_0 u\|_{1,k} \nonumber\\[0.5ex] &\le C_2\, \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} k \Theta^{1/2} (1+C_{\mathrm{\textsf{stab}}})\|T_0 u - u\|_{1,k}\|P_0 u\|_{1,k} \nonumber\\[0.5ex] &\le 2C_2\, \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} k \Theta^{1/2} (1+C_{\mathrm{\textsf{stab}}})\|u\|_{1,k}\|P_0 u\|_{1,k} \nonumber\\[0.5ex] &\le 2C_2\, \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} k \Theta^{1/2} (1+C_{\mathrm{\textsf{stab}}})\|u\|_{1,k}^2. \label{eq:10950} \end{align}\tag{76}\] Step 3 (Bounding the second term in \(R\)). For each \(j\), we use Friedrichs’ inequality ?? to arrive at \[\begin{align} k^2\bigl(E^{\ell}_j T^{\ell}_j u-u,E^{\ell}_j P^{\ell}_j u\bigr)_n &= k^2\bigl({T^{\ell}_j} u-u|_{\Omega^{\ell}_j}, P^{\ell}_j u\bigr)_{n,\Omega^{\ell}_j} \le k^2 \big\|T^{\ell}_j u-u|_{\Omega^{\ell}_j}\big\|_{n,\Omega^{\ell}_j}\big\|P^{\ell}_j u\big\|_{n,\Omega^{\ell}_j} \nonumber\\[0.5ex] &\le k\bigl\|T^{\ell}_j u-u|_{\Omega^{\ell}_j}\big\|_{1,k,\Omega^{\ell}_j}\big\|{P^{\ell}_j} u\big\|_{n,\Omega^{\ell}_j} \le k\bigl\|{T^{\ell}_j} u-u|_{\Omega^{\ell}_j}\big\|_{1,k,\Omega^{\ell}_j} \frac{H_{\ell}}{\sqrt{2}\,}\big\|{P^{\ell}_j} u\big\|_{1,k,\Omega^{\ell}_j} \nonumber\\[0.5ex] &\le \frac{k H_{\ell}}{\sqrt{2}\,} \Bigl(\|{T^{\ell}_j} u\|_{1,k,\Omega^{\ell}_j}+\big\|u|_{\Omega^{\ell}_j}\big\|_{1,k,{\Omega^{\ell}_j}}\Bigr) \big\|P^{\ell}_j u\big\|_{1,k,\Omega^{\ell}_j}. \label{lemma411} \end{align}\tag{77}\] Now, note that the hypothesis ?? implies that \(\sqrt{2} k {H_{\ell}} < \sqrt{2}/(3 \Lambda_\ell) < 1\); so we can apply Lemma 9 to obtain from 77 that \[k^2\bigl(E^{\ell}_j T^{\ell}_j u-u, E^{\ell}_j P^{\ell}_j u\bigr)_n \le \frac{3kH_{\ell}}{\sqrt{2}\,}\big\|u|_{\Omega^{\ell}_j}\big\|_{1,k,\Omega^{\ell}_j}\big\|P^{\ell}_j u\big\|_{1,k,\Omega^{\ell}_j} < 3kH_{\ell}\big\|u|_{\Omega^{\ell}_j}\big\|_{1,k,\Omega^{\ell}_j}\big\|P^{\ell}_j u\big\|_{1,k,\Omega^{\ell}_j}.\] Summing over \(j\), applying Cauchy–Schwarz and 28 , we obtain \[\begin{align} & k^2 \sum_{j=1}^{Q} \bigl(E^{\ell}_j T^{\ell}_j u-u, E^{\ell}_j P^{\ell}_j u\bigr)_n \le 3kH_{\ell} \sum_{j=1}^{Q} \big\|u|_{\Omega^{\ell}_j}\big\|_{1,k,\Omega^{\ell}_j}\|P^{\ell}_j u\|_{1,k,\Omega^{\ell}_j} \nonumber\\[0.5ex] &\le 3kH_{\ell} \Biggl(\sum_{j=1}^{Q} \big\|u|_{\Omega^{\ell}_j}\big\|_{1,k,\Omega^{\ell}_j}^2\Biggr)^{1/2} \Biggl( \sum_{j=1}^{Q} \|P^{\ell}_j u\|_{1,k,\Omega^{\ell}_j}^2\Biggr)^{1/2} \le 3kH_{\ell} \Lambda_{\ell}^{1/2} \|u\|_{1,k}\Biggl( \sum_{j=1}^{Q} \|P^{\ell}_j u\|_{1,k,\Omega^{\ell}_j}^2\Biggr)^{1/2}. \label{eq:10951} \end{align}\tag{78}\] To estimate the sum on the right-hand side of 78 , note that \[\|E_j^\ell P_j^\ell u \|_{1,k}^2 = (P_j^\ell u,P_j^\ell u)_{1,k,\Omega_j^\ell} = (P_j^\ell u, u)_{1,k,\Omega_j^\ell} = (E_j^\ell P_j^\ell u, u)_{1,k},\] and so, using 30 , we have \[\begin{align} \Bigg\|\sum_{j=1}^{Q} E^{\ell}_j P^{\ell}_j u\Bigg\|_{1,k}^2 &\le \Lambda_{\ell} \sum_{j=1}^{Q} \|E^{\ell}_j P^{\ell}_j u\|_{1,k}^2 = \Lambda_{\ell} \sum_{j=1}^{Q} (E^{\ell}_j P^{\ell}_j u,u)_{1,k} \\[0.5ex] &= \Lambda_{\ell} \Biggl(\sum_{j=1}^{Q} E^{\ell}_j P^{\ell}_j u,\,u\Biggr)_{1,k} \le \Lambda_{\ell} \,\Bigg\|\sum_{j=1}^{Q} E^{\ell}_j P^{\ell}_j u\Bigg\|_{1,k} \|u\|_{1,k}. \end{align}\] Thus \[\begin{align} \sum_{j=1}^{Q} \|P^{\ell}_j u\|_{1,k,\Omega^{\ell}_j}^2 &= \sum_{j=1}^{Q} (E^{\ell}_j P^{\ell}_j u,u)_{1,k} = \Biggl(\sum_{j=1}^{Q} E^{\ell}_j P^{\ell}_j u,\,u\Biggr)_{1,k} \le \Bigg\|\sum_{j=1}^{Q} E^{\ell}_j P^{\ell}_j u\Bigg\|_{1,k}\|u\|_{1,k} \le \Lambda_{\ell} \|u\|_{1,k}^2. \end{align}\] This, together with 78 , leads to \[\label{eq:10952} k^2 \sum_{j=1}^{Q} \bigl(E^{\ell}_j T^{\ell}_j u-u, E^{\ell}_j P^{\ell}_j u\bigr)_n \le 3kH_{\ell} \Lambda_{\ell} \|u\|_{1,k}^2.\tag{79}\] Step 4 (Obtaining ?? ). We can now combine 75 , 76 and 79 to obtain \[\begin{align} |R| &\le 2 \Bigl(2 C_2\, \Lambda_c (a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast})^{1/2} k \Theta^{1/2} (1+C_{\mathrm{\textsf{stab}}}) + 3 k \Lambda_\ell H_\ell\Bigr) \|u\|_{1,k}^2 \nonumber\\[0.5ex] &= \frac{s}{C_3 \bigl(1+\Lambda_\ell \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}\Theta \bigr)} \|u\|_{1,k}^2, \end{align}\] with \(s\) and \(C_3\) as defined in Theorem 10. Hence, using this, 74 and Proposition 17, we obtain \[\begin{align} (Tu,u)_{1,k} &= (Pu,u)_{1,k} + R \ge C_3^{-1} \bigl(1 + \Lambda_\ell \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}\Theta \bigr)^{-1} \|u\|_{1,k}^2 - |R| \nonumber\\[0.5ex] &\ge C_3^{-1} \bigl(1 + \Lambda_\ell \Lambda_c^2 a_{\mathrm{\textsf{max}}}n_{\mathrm{\textsf{min}}}^{-1} C_{\ast}\Theta \bigr)^{-1} (1-s) \|u\|_{1,k}^2, \end{align}\] as required.
Step 5 (Obtaining ?? ). Since \(E_0T_0 u = T_0u\), we have \[\label{eq:3249518} \|Tu\|_{1,k}^2 = \Bigg\|T_0 u + \sum_{j=1}^{Q} E^{\ell}_j T^{\ell}_j u\Biggr\|_{1,k}^2 \le 2\|T_0 u\|_{1,k}^2 + 2\Bigg\| \sum_{j=1}^{Q} E^{\ell}_j T^{\ell}_j u\Bigg\|_{1,k}^2.\tag{80}\] For the first term on the right-hand side of 80 , Cauchy–Schwarz and ?? yield \[\begin{align} \|T_0 u\|_{1,k}^2 &= (T_0 u, T_0 u)_{1,k} = (T_0 u - u, T_0 u)_{1,k} + (u, T_0 u)_{1,k} \nonumber\\[0.5ex] &\le \|T_0 u - u\|_{1,k}\|T_0 u\|_{1,k} + \|u\|_{1,k}\|T_0 u\|_{1,k} \nonumber\\[0.5ex] &\le 2\|u\|_{1,k}\|T_0 u\|_{1,k} + \|u\|_{1,k}\|T_0 u\|_{1,k} = 3\|u\|_{1,k}\|T_0 u\|_{1,k}. \label{eq:3249519} \end{align}\tag{81}\] For the second term on the right-hand side of 80 we use 30 , Lemma 9 and 28 to obtain \[\label{eq:second} \Bigg\|\sum_{j=1}^{Q} E^{\ell}_j T^{\ell}_j u\Bigg\|_{1,k}^2 \le \Lambda_{\ell} \sum_{j=1}^{Q} \|T^{\ell}_j u\|_{1,k,\Omega^{\ell}_j}^2 \le \Lambda_{\ell} \sum_{j=1}^{Q} 4\big\|u|_{\Omega_i}\big\|_{1,k,\Omega^{\ell}_j}^2 \le 4 \Lambda_{\ell}^2\|u\|_{1,k}^2.\tag{82}\] Combining 81 and 82 with 80 we arrive at \[\|Tu\|_{1,k}^2 \le 2\times9\|u\|_{1,k}^2 + 2\times4 \Lambda_{\ell}^2\|u\|_{1,k}^2 = (18 + 8\Lambda_{\ell}^2)\|u\|^2_{1,k},\] which proves ?? . ◻

Proof of Theorem 10. Using ?? with ?? we obtain \[c_1 \|u\|_{1,k}^2 \le (Tu,u)_{1,k} = \bigl\langle \mathbf{M}_{AS,2}^{-1}\mathbf{B}\mathbf{u},\mathbf{u}\bigr\rangle_{\mathbf{D}_{k}},\] which can be written as \[c_1 \le \frac{\bigl\langle\mathbf{M}_{AS,2}^{-1}\mathbf{B}\mathbf{u},\mathbf{u}\bigr\rangle_{\mathbf{D}_{k}}}{\|\mathbf{u}\|_{\mathbf{D}_{k}}^2 }.\] Also, combining ?? with ?? , we get \[\|\mathbf{M}_{AS,2}^{-1}\mathbf{B}\mathbf{u}\|_{\mathbf{D}_{k}}^2 = \|Tu\|_{1,k}^2 \le c_2\|u\|_{1,k}^2 = c_2\|\mathbf{u}\|^2_{\mathbf{D}_{k}}, \qquad \text{i.e.} \quad \|\mathbf{M}_{AS,2}^{-1}\mathbf{B}\|_{\mathbf{D}_k}^2 \le c_2.\] The result on GMRES convergence now follows directly from the Elman theory [15] with \(\gamma=\frac{c_1}{c_2}\). ◻