# Classical global estimate for $H^1$ error

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?

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.

• 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 Jul 10 at 22:45
• Well, you probably have some boundary condition to make the solution unique?
– knl
Jul 11 at 13:56
• 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. Jul 11 at 23:35
• @bobinthebox Does this answer the question?
– knl
Jul 14 at 10:10
• 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? Jul 14 at 17:58