Céa and Aubin–Nitsche error estimates

Fix $v_h\in V_h$. Since $V_h\subseteq V$, the continuous equation is valid at $v_h$, so $a(u,v_h)=F(v_h)$.

The discrete equation gives $a(u_h,v_h)=F(v_h)$. Subtracting the two equalities and using the stated linearity of $a$ in its first slot yields $a(u-u_h,v_h)=0$.

Define the adjoint form by $a^{†}(z,v):=\overline{a(v,z)}$. The sesquilinearity of $a$ shows that $a^{†}$ is linear in $z$ and conjugate-linear in $v$. Also $\mathrm{\mid }{a}^{†}(z,v)\mathrm{\mid }=\mathrm{\mid }a(v,z)\mathrm{\mid }\le M\mathrm{\parallel }z{\mathrm{\parallel }}_V\mathrm{\parallel }v{\mathrm{\parallel }}_V$, and $\mathrm{Re}{a}^{†}(z,z)=\mathrm{Re}a(z,z)\ge \alpha \mathrm{\parallel }z{\mathrm{\parallel }}_V^2$.

Set $G^{†}(v):=\overline{G(v)}$. Since $G$ is linear, $G^{†}$ is conjugate-linear, and $\mathrm{\mid }{G}^{†}(v)\mathrm{\mid }=\mathrm{\mid }G(v)\mathrm{\mid }\le \mathrm{\parallel }G{\mathrm{\parallel }}_{V^{\ast }}\mathrm{\parallel }v{\mathrm{\parallel }}_V$; hence $\mathrm{\parallel }{G}^{†}{\mathrm{\parallel }}_{V^{\ast }}=\mathrm{\parallel }G{\mathrm{\parallel }}_{V^{\ast }}$.

Apply [F2] to $a^{†}$ and $G^{†}$. It gives a unique $z\in V$ with $a^{†}(z,v)=G^{†}(v)$ and $\alpha \mathrm{\parallel }z{\mathrm{\parallel }}_V\le \mathrm{\parallel }{G}^{†}{\mathrm{\parallel }}_{V^{\ast }}$. Conjugating the equation gives $a(v,z)=G(v)$ for every $v$, and the norm equality in step 1.2 gives the claimed estimate and uniqueness.

Test the weak equation with $v=z$ and take real parts. By [F3] and [A1], $\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_{L^2}^2=\mathrm{Re}{\int }_{\mathrm{\Omega }}g\bar{z}dx\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }z{\mathrm{\parallel }}_{L^2}\le C_P\mathrm{\parallel }g{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_{L^2}\mathrm{.}$ If $\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_{L^2}=0$, [A1] gives $z=0$ in $L^2$; otherwise divide by $\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_{L^2}$. In either case $\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_{L^2}\le C_P\mathrm{\parallel }g{\mathrm{\parallel }}_{L^2}$ and $\mathrm{\parallel }z{\mathrm{\parallel }}_{L^2}\le C_P^2\mathrm{\parallel }g{\mathrm{\parallel }}_{L^2}$.

The operator $-\mathrm{\Delta }$ has principal matrix $a^{ij}={\delta }_{ij}$, ellipticity constant $1$, coefficients in $W^{1,\mathrm{\infty }}$, and zero lower order coefficients. Thus [F2] applies to the given weak solution and yields $\mathrm{\parallel }z{\mathrm{\parallel }}_{H^2}\le C_{\mathrm{r}\mathrm{e}\mathrm{g}}(\mathrm{\parallel }g{\mathrm{\parallel }}_{L^2}+\mathrm{\parallel }z{\mathrm{\parallel }}_{L^2})\mathrm{.}$ Insert the bound from step 1.1 to obtain the asserted estimate.

Put $\zeta (t)=1-2\sigma (t)$. Formula [F4] gives $\sigma (1-t)=1-\sigma (t)$ on $[0,1]$, hence $\zeta (1-t)=-\zeta (t)$. Let $T(t)=1-t$ on $I$; it is a $C^1$ diffeomorphism with $\mathrm{\mid }{T}^{\mathrm{\prime }}(t)\mathrm{\mid }=1$. The continuous functions ${\zeta }_{+}=\max (\zeta ,0)$ and ${\zeta }_{-}=\max (-\zeta ,0)$ satisfy $0\le {\zeta }_{\pm }\le 1$, and [F10] gives ${\int }_I{\zeta }_{\pm }\le {\int }_I{1}_I={\lambda }_1(I)=1$, so both are integrable. Applying [F6] separately to these nonnegative functions gives ${\int }_I{\zeta }_{+}={\int }_I{\zeta }_{+}(1-t)={\int }_I{\zeta }_{-}$, where the last equality uses antisymmetry. By [F9], ${\int }_I\zeta ={\int }_I{\zeta }_{+}-{\int }_I{\zeta }_{-}=0$. For $0<\epsilon <1\mathrm{/}6$, define $g_{\epsilon }(x)=2\zeta ((x-1\mathrm{/}2+\epsilon )\mathrm{/}(2\epsilon ))$ and $q_{\epsilon }(x)={\int }_0^x{g}_{\epsilon }(t)dt$. By [F4]–[F5], $g_{\epsilon }$ is smooth, equals $2$ for $x\le 1\mathrm{/}2-\epsilon$ and $-2$ for $x\ge 1\mathrm{/}2+\epsilon$. Applying [F6] separately to ${\zeta }_{+}$ and ${\zeta }_{-}$ under $x=1\mathrm{/}2-\epsilon +2\epsilon t$, and subtracting by [F9], gives the transition integral $4\epsilon {\int }_0^1\zeta =0$; the two constant pieces cancel, so ${\int }_0^1{g}_{\epsilon }=0$. For $h\mathrm{\ne }0$ with $x+h\in [0,1]$, the difference quotient of $q_{\epsilon }$ is the average of $g_{\epsilon }$ between $x$ and $x+h$; its difference from $g_{\epsilon }(x)$ is bounded by ${\sup }_{\mathrm{\mid }t-x\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }}\mathrm{\mid }{g}_{\epsilon }(t)-g_{\epsilon }(x)\mathrm{\mid }$, which tends to zero by continuity. Thus $q_{\epsilon }^{\mathrm{\prime }}=g_{\epsilon }$ and $q_{\epsilon }$ is smooth. It vanishes at both endpoints and agrees with $\varphi$ outside $(1\mathrm{/}2-\epsilon ,1\mathrm{/}2+\epsilon )$. On that interval, $\mathrm{\mid }{q}_{\epsilon }^{\mathrm{\prime }}-{\varphi }^{\mathrm{\prime }}\mathrm{\mid }\le 4$ almost everywhere, so $\mathrm{\parallel }{q}_{\epsilon }^{\mathrm{\prime }}-{\varphi }^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\le 32\epsilon$. Also $\mathrm{\mid }{q}_{\epsilon }-\varphi \mathrm{\mid }\le 8\epsilon$ there and the difference vanishes outside, so $\mathrm{\parallel }{q}_{\epsilon }-\varphi {\mathrm{\parallel }}_2^2\le 128{\epsilon }^3$.

By [F3], each $q_{\epsilon }$ has weak derivative $g_{\epsilon }$. The convergence estimates in step 1.1 and [F8] allow passage to the limit in ${\int }_I{q}_{\epsilon }{\psi }^{\mathrm{\prime }}=-{\int }_I{g}_{\epsilon }\psi$ for every $\psi \in C_c^{\mathrm{\infty }}(I)$, giving ${\int }_I\varphi {\psi }^{\mathrm{\prime }}=-{\int }_I{\varphi }^{\mathrm{\prime }}\psi$, where ${\varphi }^{\mathrm{\prime }}=2$ on the left half and ${\varphi }^{\mathrm{\prime }}=-2$ on the right half. Thus $\varphi \in W^{1,2}(I)$. The polynomial $p$ is smooth with $p^{\mathrm{\prime }}=1\mathrm{/}2-x$ and both $p,p^{\mathrm{\prime }}\in L^2(I)$, so [F1] and [F3] give $p\in W^{1,2}(I)$.

Let $C_{\sigma }={\sup }_{[0,1]}\mathrm{\mid }{\sigma }^{\mathrm{\prime }}\mathrm{\mid }<\mathrm{\infty }$ by [F7], and define ${\eta }_{\epsilon }(x)=\sigma (x\mathrm{/}\epsilon -1)\sigma ((1-x)\mathrm{/}\epsilon -1)$. By [F4]–[F5], this smooth cutoff vanishes for $x\le \epsilon$ and $x\ge 1-\epsilon$, equals $1$ on $[2\epsilon ,1-2\epsilon ]$, and $\mathrm{\mid }{\eta }_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }\le 2C_{\sigma }\mathrm{/}\epsilon$. For the same $\epsilon$, $q_{\epsilon }=2x$ on $(0,2\epsilon )$ and $q_{\epsilon }=2(1-x)$ on $(1-2\epsilon ,1)$. Thus ${\eta }_{\epsilon }{q}_{\epsilon }\in C_c^{\mathrm{\infty }}(I)$, $\mathrm{\mid }{q}_{\epsilon }\mathrm{\mid }\le 4\epsilon$ and $\mathrm{\mid }{q}_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }=2$ on $E_{\epsilon }=(0,2\epsilon )\cup (1-2\epsilon ,1)$. The function error has squared norm at most $64{\epsilon }^3$. For the derivative difference $({\eta }_{\epsilon }-1)q_{\epsilon }^{\mathrm{\prime }}+{\eta }_{\epsilon }^{\mathrm{\prime }}{q}_{\epsilon }$, the first term has squared norm at most $16\epsilon$; the second has modulus at most $8C_{\sigma }$ on a set of length $2\epsilon$, so its squared norm is at most $128C_{\sigma }^2\epsilon$. Therefore ${\eta }_{\epsilon }{q}_{\epsilon }-q_{\epsilon }\to 0$ in $W^{1,2}(I)$. Together with step 1.1 this gives ${\eta }_{\epsilon }{q}_{\epsilon }\to \varphi$.

Using the cutoff ${\eta }_{\epsilon }$ from step 2.2, its product with $p$ lies in $C_c^{\mathrm{\infty }}(I)$. Since $\mathrm{\mid }p(x)\mathrm{\mid }\le \min (x,1-x)\mathrm{/}2$ and $\mathrm{\mid }{p}^{\mathrm{\prime }}(x)\mathrm{\mid }\le 1\mathrm{/}2$, $\mathrm{\parallel }{\eta }_{\epsilon }p-p{\mathrm{\parallel }}_2^2\le 4{\epsilon }^3$. The derivative difference is $({\eta }_{\epsilon }-1)p^{\mathrm{\prime }}+{\eta }_{\epsilon }^{\mathrm{\prime }}p$; the first term has squared norm at most $\epsilon$, and the second is supported on two intervals of total length $2\epsilon$ with modulus at most $2C_{\sigma }$, so its squared norm is at most $8C_{\sigma }^2\epsilon$. Hence ${\eta }_{\epsilon }p\to p$ in $W^{1,2}(I)$.

Take the explicit sequence ${\epsilon }_n=1\mathrm{/}(10(n+1))$, which lies in $(0,1\mathrm{/}6)$ and tends to zero. Steps 3.1 and 2.2 approximate $p$ and $\varphi$ by members of $C_c^{\mathrm{\infty }}(I)$ along this sequence. By the closure definition [F2], both classes belong to $W_0^{1,2}(I)$, as claimed.

Apply [F4] on $V$; its hypotheses are exactly the boundedness, coercivity, Hilbert, and datum assumptions in Given. This gives a unique $u\in V$ with $a(u,v)=F(v)$ for all $v\in V$.

Fix a finite ordered basis of $V_h$. By [F3], $V_h$ is closed in $V$. A Cauchy sequence in $V_h$ is Cauchy in $V$, so [F2] gives a limit in $V$; the closedness puts the limit in $V_h$. The restricted inner product is complete, so $V_h$ is Hilbert, also when $V_h=\{0\}$.

The restrictions of $a$ and $F$ to $V_h$ retain the same bounds $M$ and $\alpha$, and $F{\mathrm{\mid }}_{V_h}$ is bounded. Apply [F4] on the Hilbert space from step 1.2. This gives a unique $u_h\in V_h$ satisfying the discrete equations.

By [F5], with $e:=u-u_h$, we have $a(e,w_h)=0$ for every $w_h\in V_h$.

Fix any $v_h\in V_h$. Since $u-v_h=e+(u_h-v_h)$ and $u_h-v_h\in V_h$, [F5] gives $a(e,e)=a(e,u-v_h)-a(e,u_h-v_h)=a(e,u-v_h)\mathrm{.}$ Coercivity and boundedness therefore give $\alpha \mathrm{\parallel }e{\mathrm{\parallel }}_V^2\le \mathrm{Re}a(e,e)\le \mathrm{\mid }a(e,u-v_h)\mathrm{\mid }\le M\mathrm{\parallel }e{\mathrm{\parallel }}_V\mathrm{\parallel }u-v_h{\mathrm{\parallel }}_V\mathrm{.}$

If $e=0$, the claimed estimate is immediate. Otherwise divide step 4.1 by $\alpha \mathrm{\parallel }e{\mathrm{\parallel }}_V$ and then take the infimum over all $v_h\in V_h$. This proves Céa's estimate.

Hermitian symmetry makes $a(v,v)$ real, and coercivity makes it positive for every nonzero $v$. Thus $a$ is an inner product and $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_a$ is its induced norm.

Put $e=u-u_h$ and, for arbitrary $v_h\in V_h$, put ${\eta }_h=u_h-v_h$. By [F2], $a(e,{\eta }_h)=0$; Hermitian symmetry gives $a({\eta }_h,e)=\overline{a(e,{\eta }_h)}=0$.

Since $u-v_h=e+{\eta }_h$, sesquilinearity and step 1.2 give $\mathrm{\parallel }u-v_h{\mathrm{\parallel }}_a^2=a(e+{\eta }_h,e+{\eta }_h)=\mathrm{\parallel }e{\mathrm{\parallel }}_a^2+\mathrm{\parallel }{\eta }_h{\mathrm{\parallel }}_a^2\ge \mathrm{\parallel }e{\mathrm{\parallel }}_a^2\mathrm{.}$ Therefore $\mathrm{\parallel }u-u_h{\mathrm{\parallel }}_a\le \mathrm{\parallel }u-v_h{\mathrm{\parallel }}_a$ for every $v_h\in V_h$.

Taking the infimum in step 2.1 gives one inequality. The choice $v_h=u_h$ gives the reverse inequality, so equality holds.

Define $G_e(v):=(Jv,e{)}_H$. By [F4] and the assumed continuity of $J$, $\mathrm{\mid }{G}_e(v)\mathrm{\mid }\le C_E\mathrm{\parallel }v{\mathrm{\parallel }}_V\mathrm{\parallel }e{\mathrm{\parallel }}_H,$ so $G_e$ is a bounded linear functional. Fact [F2] gives the unique $z_e$ with $a(v,z_e)=G_e(v)$.

Set $v=u-u_h$ in the adjoint equation. Then $\mathrm{\parallel }e{\mathrm{\parallel }}_H^2=(J(u-u_h),e{)}_H=a(u-u_h,z_e)\mathrm{.}$ For any $w_h\in V_h$, [F3] gives $a(u-u_h,w_h)=0$, hence $a(u-u_h,z_e)=a(u-u_h,z_e-w_h)$.

Since the scalar in step 2.1 equals the nonnegative real number $\mathrm{\parallel }e{\mathrm{\parallel }}_H^2$, boundedness of $a$ gives $\mathrm{\parallel }e{\mathrm{\parallel }}_H^2=\mathrm{\mid }a(u-u_h,z_e-w_h)\mathrm{\mid }\le M\mathrm{\parallel }u-u_h{\mathrm{\parallel }}_V\mathrm{\parallel }{z}_e-w_h{\mathrm{\parallel }}_V$ for every $w_h\in V_h$. Taking the infimum and using the approximation and regularity assumptions yields $\mathrm{\parallel }e{\mathrm{\parallel }}_H^2\le M{C}_I{C}_R{h}^s\mathrm{\parallel }u-u_h{\mathrm{\parallel }}_V\mathrm{\parallel }e{\mathrm{\parallel }}_H\mathrm{.}$

If $\mathrm{\parallel }e{\mathrm{\parallel }}_H=0$, the conclusion is immediate. Otherwise divide step 3.1 by $\mathrm{\parallel }e{\mathrm{\parallel }}_H$ to obtain the asserted Aubin–Nitsche estimate.

On the reference cell write $\widehat{I}w={\sum }_{i=1}^N{\widehat{\sigma }}_i(w){\widehat{\varphi }}_i$. Each degree of freedom is bounded on $W^{r+1,2}$ by Given, and each fixed polynomial shape function has finite $W^{1,2}$ norm. Thus $\mathrm{\parallel }\widehat{I}w{\mathrm{\parallel }}_{W^{1,2}}\le C\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{r+1,2}}$. By [F1], $\widehat{I}q=q$ for every $q\in {\mathbb{P}}_r$. For such $q$, $w-\widehat{I}w=(w-q)-\widehat{I}(w-q),$ so $\mathrm{\parallel }w-\widehat{I}w{\mathrm{\parallel }}_{W^{m,2}}\le \mathrm{\parallel }w-q{\mathrm{\parallel }}_{W^{m,2}}+C\mathrm{\parallel }w-q{\mathrm{\parallel }}_{W^{r+1,2}}(m=0,1)\mathrm{.}$

Choose the polynomial $q\in {\mathbb{P}}_r$ supplied by [F2]. Since $m\le 1\le r+1$, [F2] bounds both terms in step 1.1 by $C\mathrm{\mid }w{\mathrm{\mid }}_{W^{r+1,2}}$. Hence the reference-cell interpolation error obeys $\mathrm{\parallel }w-\widehat{I}w{\mathrm{\parallel }}_{W^{m,2}}\le C\mathrm{\mid }w{\mathrm{\mid }}_{W^{r+1,2}}$ for $m=0,1$.

For a cell $K$ with affine matrix $A_K$, apply [F3] to transform the reference error and the order-$(r+1)$ derivative seminorm back to $K$. The determinant factors cancel; the remaining factor is bounded by $C\mathrm{\parallel }{A}_K^{-1}{\mathrm{\parallel }}^m\mathrm{\parallel }{A}_K{\mathrm{\parallel }}^{r+1}\le C{h}_K^{r+1-m}$ by [F4]. Therefore $\mathrm{\parallel }u-I_{K}u{\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{r+1}\mathrm{\mid }u{\mathrm{\mid }}_{W^{r+1,2}(K)},\mathrm{\mid }u-I_{K}u{\mathrm{\mid }}_{W^{1,2}(K)}\le C{h}_K^r\mathrm{\mid }u{\mathrm{\mid }}_{W^{r+1,2}(K)}\mathrm{.}$

Let $E_h:={\bigcup }_{K\in {\mathcal{T}}_h}\mathrm{\partial }K$, a finite union of simplex facets. Each facet lies in a bounded affine hyperplane; choose a coordinate with nonzero normal component and write that hyperplane as the graph of an affine function over a bounded box in ${\mathbb{R}}^{d-1}$. Partition the base box into at most $C_0{N}^{d-1}$ boxes of side $1\mathrm{/}N$. The affine function varies by at most $C_1\mathrm{/}N$ on each, so each graph piece lies in a slightly enlarged half-open $d$-box of volume at most $C_2{N}^{-d}$. These elementary-set cover members have total volume at most $C_0{C}_2\mathrm{/}N$; the outer-measure definition and box-volume formula give outer measure zero for the facet, and completeness makes it Lebesgue-null. Thus $E_h$ is null, and $\mathrm{\Omega }\setminus E_h$ is the disjoint union of the cell interiors by the mesh definition. Square and sum the estimates in step 3.1 over these interiors; the sums of squared Sobolev seminorms equal the corresponding global seminorms. Since $h_K\le h\le 1$, the order-zero contribution $h_K^{r+1}$ is at most $h^r$. The global interpolation map in Given takes values in $V_h$, yielding $\mathrm{\parallel }u-I_{h}u{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}\le C{h}^r\mathrm{\mid }u{\mathrm{\mid }}_{W^{r+1,2}(\mathrm{\Omega })}$.