The conforming Ritz matrix differs quadratically in the eigenspace gap

Put ${\varphi }_i:=e_i\mathrm{/}\sqrt{{\lambda }_i}$, $p_i:=P_h{\varphi }_i$, and $r_i:={\varphi }_i-p_i$. By [F1], the ${\varphi }_i$ are $a$-orthonormal. For any coefficient vector $c\in {\mathbb{K}}^m$, linearity of $P_h$ and [F2]–[F3] give ${\parallel {\sum }_{i=1}^m{c}_i{r}_i\parallel }_a={\parallel v-P_{h}v\parallel }_a\le {\delta }_h\mathrm{\parallel }v{\mathrm{\parallel }}_a={\delta }_h\mathrm{\parallel }c{\mathrm{\parallel }}_2,v:={\sum }_{i=1}^m{c}_i{\varphi }_i\mathrm{.}$ In particular, the error Gram matrix $G_{ij}:=a(r_j,r_i)$ is positive semidefinite and $\mathrm{\parallel }G{\mathrm{\parallel }}_2\le {\delta }_h^2$.

Since each $r_i$ is energy-orthogonal to $V_h$, it is orthogonal to every $p_j$. Therefore the stiffness Gram matrix $(A_h{)}_{ij}:=a(p_j,p_i)$ satisfies $(A_h{)}_{ij}=a({\varphi }_j-r_j,{\varphi }_i-r_i)={\delta }_{ij}-G_{ij}\mathrm{.}$ It follows that $\mathrm{\parallel }{A}_h-I{\mathrm{\parallel }}_2=\mathrm{\parallel }G{\mathrm{\parallel }}_2\le {\delta }_h^2$.

Let $b(u,v):=(Ju,Jv{)}_H$. The eigenrelation in [F1] gives $b({\varphi }_j,r_i)=a(r_j,r_i)\mathrm{/}{\lambda }_j$ and $b(r_j,{\varphi }_i)=a(r_j,r_i)\mathrm{/}{\lambda }_i$; in each identity, the energy orthogonality of $r_i$ or $r_j$ removes the projected component. By [F4], $\mathrm{\mid }a(r_j,r_i)\mathrm{\mid }\le \mathrm{\parallel }{r}_j{\mathrm{\parallel }}_a\mathrm{\parallel }{r}_i{\mathrm{\parallel }}_a\le {\delta }_h^2$ and $\mathrm{\mid }b(r_j,r_i)\mathrm{\mid }\le \mathrm{\parallel }J{r}_j{\mathrm{\parallel }}_H\mathrm{\parallel }J{r}_i{\mathrm{\parallel }}_H\le {\delta }_h^2\mathrm{/}{\lambda }_1$, where the last inequality uses [F1]. Expanding $b(p_j,p_i)-b({\varphi }_j,{\varphi }_i)$ into these two cross terms and $b(r_j,r_i)$ shows that every entry of $B_h-D_m$ has absolute value at most $3{\delta }_h^2\mathrm{/}{\lambda }_1$. If a matrix $M$ has entries bounded by $t$, then for $\mathrm{\parallel }c{\mathrm{\parallel }}_2=1$ each coordinate of $Mc$ is at most $t{\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\le t\sqrt{m}$, so $\mathrm{\parallel }M{\mathrm{\parallel }}_2\le mt$; hence $\mathrm{\parallel }{B}_h-D_m{\mathrm{\parallel }}_2\le 3m{\delta }_h^2\mathrm{/}{\lambda }_1$.

The least diagonal entry of $D_m$ is $1\mathrm{/}{\lambda }_m$. The assumed bound on ${\delta }_h$ and step 2.2 therefore give $c^{\ast }{B}_{h}c\ge (\frac{1}{{\lambda }_m}-\frac{3m{\delta }_h^2}{{\lambda }_1})\mathrm{\parallel }c{\mathrm{\parallel }}_2^2\ge \frac{1}{2{\lambda }_m}\mathrm{\parallel }c{\mathrm{\parallel }}_2^2\mathrm{.}$ In particular $B_h$ is positive definite and $P_h{\mathrm{\mid }}_{E_m}$ has $m$-dimensional image $U_h$.

Normalize $\mathrm{\parallel }c{\mathrm{\parallel }}_2=1$. The continuous quotient of $v={\sum }_i{c}_i{\varphi }_i$ is $R(v)=1\mathrm{/}(c^{\ast }{D}_{m}c)\le {\lambda }_m$ by [F1]. The quotient of $w_h=P_{h}v={\sum }_i{c}_i{p}_i$ is $R_h(w_h)=c^{\ast }{A}_{h}c\mathrm{/}(c^{\ast }{B}_{h}c)$. Steps 2.1, 2.2 and 3.1 give $\mathrm{\mid }{R}_h(w_h)-R(v)\mathrm{\mid }\le \frac{{\delta }_h^2}{1\mathrm{/}(2{\lambda }_m)}+\frac{3m{\delta }_h^2\mathrm{/}{\lambda }_1}{(1\mathrm{/}(2{\lambda }_m))(1\mathrm{/}{\lambda }_m)}=C_m{\delta }_h^2\mathrm{.}$ Thus $R_h(w_h)\le {\lambda }_m+C_m{\delta }_h^2$. Homogeneity removes the normalization of $c$, proving the bound for every nonzero $w_h\in U_h$.