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I'm having lots of troubles in understanding the proof the estimation of the classical $H^1$ error using finite elements of degree $r$.

$$||u-u_h||_{H^1(\Omega)} \leq \frac{M}{\alpha} C h^r |u|_{H^{r+1}(\Omega)}$$ where $M$ and $\alpha$ are the operator norm and the coercivity constant, respectively and $C$ is a constand independent on $h$.

My book (Quarteroni - Numerical methods for differential problems) shows this by using the Cèa's lemma and the estimate on the seminorm of the interpolation error. More precisely, it states:

$$||u-u_h||_{H^1(\Omega)} \leq \frac{M}{\alpha} \inf_{v_h \in V_h} ||u - v_h||_{H^1(\Omega)} \leq \frac{M}{\alpha} ||u-\Pi^ru||_{H^1(\Omega)}$$ Now use the estimate $$|u-\Pi^r u|_{H^m(\Omega)} \leq Ch^{r+1-m} |u|_{H^{m}(\Omega)}$$ (valid for $m=0,1$ and $r\geq 1$. ) with $m=1$ to conclude

The very last step is what I cannot do: I can't see how to pass from the $H^1$ seminorm to the full $H^1$ norm. I mean, if I need to bound $||u-\Pi^ru||_{H^1(\Omega)}$, how can I use the bound on the seminorm?

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I think $u - \Pi u$ is zero on a part of the boundary. Then you can use Poincare inequality to bound $\| u - \Pi u \|_0 \leq C| u - \Pi u|_1$.

It could also be possible to look at the case $m = 0$ and argue the $L^2$ part is of higher order and, hence, smaller in the asymptotic limit.

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  • $\begingroup$ Poincarè inequality was my first guess indeed, but I wasn't able to say whether or not it's $0$ on the boundary. All I know is that $\Pi u$ is equal to $u$ on the dofs, by definition. Why do you think it can be $0$ on some non trivial part of the boundary? @knl $\endgroup$
    – FEGirl
    Commented Jul 10, 2021 at 22:45
  • $\begingroup$ Well, you probably have some boundary condition to make the solution unique? $\endgroup$
    – knl
    Commented Jul 11, 2021 at 13:56
  • $\begingroup$ This is usually done by assuming that the boundary values $u|_{\partial\Omega}$ are piecewise polynomial so that $(u-\Pi u)|_{\partial\Omega}=0$. The proof can be extended to arbitrary boundary conditions, but that adds a substantial level of complexity. $\endgroup$
    – Wolfgang Bangerth
    Commented Jul 11, 2021 at 23:35
  • $\begingroup$ @bobinthebox Does this answer the question? $\endgroup$
    – knl
    Commented Jul 14, 2021 at 10:10
  • $\begingroup$ Actually before accepting I was trying to verify whether or not that was an assumption on the book. I think that what the author is using is the Bramble-Hilbert lemma $$||\tau(u)||_{s,p,\Omega} \leq ||\tau|| |u|_{k+1,p,\Omega}$$ for $\tau:W^{k+1,p}(\Omega \rightarrow W^{s,p}(\Omega)$ and $s \leq k$. @knl Do you agree? $\endgroup$
    – FEGirl
    Commented Jul 14, 2021 at 17:58

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