Smooth lowest-order Nedelec and RT0 interpolation estimate

On a fixed reference tetrahedron, each canonical basis function has finite $L^2$ norm. Under the covariant Piola map, [F4] and [F5] give $\mathrm{\parallel }{\varphi }_{K,e}{\mathrm{\parallel }}_{L^2(K)}\le C_{\sigma }{h}_K^{-1}{h}_K^{3\mathrm{/}2}=C_{\sigma }{h}_K^{1\mathrm{/}2}$ for each edge basis function. Under the contravariant Piola map they give $\mathrm{\parallel }{\psi }_{K,F}{\mathrm{\parallel }}_{L^2(K)}\le C_{\sigma }{h}_K^{-3}{h}_K{h}_K^{3\mathrm{/}2}=C_{\sigma }{h}_K^{-1\mathrm{/}2}$ for each unit-face-flux RT0 basis function.

By [F3], $\mathrm{div}{I}_K^{2}v=I_K^3\mathrm{div}v$, and $I_K^3$ is the cell average because its defining degree of freedom is the cell integral. Thus $\mathrm{\mid }\mathrm{div}v(x)-I_K^3\mathrm{div}v\mathrm{\mid }\le h_K\mathrm{\parallel }\mathrm{\nabla }\mathrm{div}v{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$ throughout $K$, and $\mathrm{\parallel }\mathrm{div}(v-I_K^{2}v){\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{5\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }\mathrm{div}v{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$.

Fix $K$ and $x_K\in K$, and put $u_K=u(x_K)$, a constant vector in the local Nedelec space by [F2]. For each of its six edges, the mean-value bound gives $\mathrm{\mid }{\int }_e(u-u_K)\cdot tds\mathrm{\mid }\le h_K^2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$. By edge-moment unisolvence [F2], $I_K^1(u-u_K)$ is the sum of these six coefficients times the corresponding basis functions. Step 1.1 and $\mathrm{\parallel }u-u_K{\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{5\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{\mathrm{\infty }}$ therefore yield $\mathrm{\parallel }u-I_K^{1}u{\mathrm{\parallel }}_{L^2(K)}\le C_{\sigma }{h}_K^{5\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$.

Set $q=\mathrm{curl}u$ and $q_K=q(x_K)$, a constant RT0 field. By the curl commuting identity [F3], $\mathrm{curl}{I}_K^{1}u=I_K^{2}q$. Each face-flux coefficient of $q-q_K$ is at most $C{h}_K^3\mathrm{\parallel }\mathrm{\nabla }q{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$; multiplying the four coefficients by the RT0 basis bounds in step 1.1 and adding $\mathrm{\parallel }q-q_K{\mathrm{\parallel }}_{L^2(K)}$ gives $\mathrm{\parallel }\mathrm{curl}(u-I_K^{1}u){\mathrm{\parallel }}_{L^2(K)}\le C_{\sigma }{h}_K^{5\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }\mathrm{curl}u{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$.

For $v$, set $v_K=v(x_K)$, again a constant in RT0. Each of the four face-flux coefficients of $v-v_K$ is bounded by $C{h}_K^3\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$. The RT0 basis scaling in step 1.1 and $\mathrm{\parallel }v-v_K{\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{5\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_{\mathrm{\infty }}$ give $\mathrm{\parallel }v-I_K^{2}v{\mathrm{\parallel }}_{L^2(K)}\le C_{\sigma }{h}_K^{5\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$.

Square and sum the four local bounds. By [F5], $h_K^5\le C_{\sigma }{h}^2\mathrm{\mid }K\mathrm{\mid }$; the cell interiors are disjoint and sum to $\mathrm{\Omega }$. Taking square roots gives the displayed global estimate with the maximum cell derivative bound and the factor $\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}2}$. The matching traces and the conformity part of [F3] assemble the interpolants globally.