I know that the piecewise linear finite element approximation $u_h$ of $$ \Delta u(x)=f(x)\quad\text{in }U\\ u(x)=0\quad\text{on }\partial U $$ satisfies $$ \|u-u_h\|_{H^1_0(U)}\leq Ch\|f\|_{L^2(U)} $$ provided that $U$ is smooth enough and $f\in L^2(U)$.

Question: If $f\in H^{-1}(U)\setminus L^2(U)$, do we have the following analogous estimate, in which one derivative is taken away on both sides: $$ \|u-u_{h}\|_{L^2(U)}\leq Ch\|f\|_{H^{-1}(U)}\quad? $$

Can you provide references?

Thoughts: Since we still have $u\in H^1_0(U)$, it should be possible to obtain convergence in $L^2(U)$. Intuitively, this should even be possible with piecewise constant functions.

  • $\begingroup$ I think that you get $\|u-u_h\|_0 \leq C h \|u-u_h\|_1$ from the standard Nitsche trick even for $u \in H^1$. You can find this e.g. in Braess - Finite elements. $\endgroup$
    – knl
    Commented Feb 3, 2017 at 14:53

1 Answer 1


Yes, this is the standard Aubin-Nitsche (or duality) trick. The idea is to use the fact that $L^2$ is its own dual space to write the $L^2$-norm as an operator norm $$\|u\|_{L^2} = \sup_{\phi\in L^2\setminus\{0\}} \frac{(u,\phi)}{\|\phi\|_{L^2}}.$$ We thus have to estimate $(u-u_h,\phi)$ for arbitrary $\phi\in L^2$. To do that, we "lift" $u-u_h$ to $H^1_0$ by considering first for arbitrary $\phi\in L^2$ the solution $w_\phi\in H^1_0$ of the dual problem $$\label{eq1}\tag{1} (\nabla w_\phi,\nabla v) =(\phi,v) \qquad\text{for all }v\in H^1_0. $$ Using the standard regularity of the Poisson equation, we know that $$ \|w_\phi\|_{H^2}\leq C \|\phi\|_{L^2}.$$

Inserting $v=u-u_h\in H^1_0$ in \eqref{eq1} and using Galerkin orthogonality for any finite element (in your case, piecewise linear) function $w_h$ yields the estimate $$ \begin{aligned} (\phi,u-u_h) &= (\nabla w_\phi,\nabla (u-u_h))\\ &= (\nabla w_\phi-\nabla w_h,\nabla (u-u_h))\\ &\leq C \|u-u_h\|_{H^1}\|w_\phi-w_h\|_{H^1}. \end{aligned} $$ Since this holds for all $w_h$, the inequality is still true if we take the infimum over all piecewise linear $w_h$. We therefore obtain $$\label{eq2}\tag{2} \|u-u_h\|_{L^2} = \sup_{\phi\in L^2\setminus\{0\}} \frac{(u-u_h,\phi)}{\|\phi\|_{L^2}} \leq C \|u-u_h\|_{H^1} \sup_{\phi\in L^2\setminus\{0\}} \frac{\inf_{w_h}\|w_\phi-w_h\|_{H^1}}{\|\phi\|_{L^2}}. $$ This is the Aubin-Nitsche-Lemma.

The next step is now to use standard error estimates for the best finite element approximation of solutions to the Poisson equation. Since $u$ is only in $H^1$, we don't get a better estimate than $$\label{eq3}\tag{3} \|u-u_h\|_{H^1} \leq \inf_{v_h}\|u-v_h\|_{H^1} \leq c \|u\|_{H^1} \leq C \|f\|_{H^{-1}}. $$ But fortunately, we can use the fact that $w_\phi$ has higher regularity since the right-hand side $\phi\in L^2$ instead of $H^{-1}$. In this case, we have $$\label{eq4}\tag{4} \inf_{w_h}\|w_\phi-w_h\|_{H^1} \leq c h \|w_\phi\|_{H^2} \leq C h \|\phi\|_{L^2} $$ Inserting \eqref{eq3} and \eqref{eq4} into \eqref{eq2} now yields the desired estimate.

(Note that the standard estimates require that the polynomial degree $k$ of the finite element approximation and the Sobolev exponent $m$ of the true solution satisfy $m<k+1$, so this argument doesn't work for piecewise constant ($k=0$) approximation. We also have used that $u-u_h\in H^1_0$ -- i.e., that we have a conforming approximation -- which is not true for piecewise constants.)

Since you asked for a reference: You can find a statement (even for negative Sobolev spaces $H^{-s}$ instead of $L^2$) in Theorem 5.8.3 (together with Theorem 5.4.8) in

Susanne C. Brenner and L. Ridgway Scott, MR 2373954 The mathematical theory of finite element methods, Texts in Applied Mathematics ISBN: 978-0-387-75933-3.

  • 1
    $\begingroup$ And I get to make use of our shiny new citation feature :) $\endgroup$ Commented Feb 3, 2017 at 12:55
  • $\begingroup$ Thanks for your answer, but continuous functions aren't embedded into $ H^1_0$ are they? $\endgroup$
    – Bananach
    Commented Feb 3, 2017 at 14:12
  • $\begingroup$ Yes, sorry, I stroked off there -- they're dense, but not embedded. The duality argument works the same, though (just work with $H^1_0$ and $H^{-1}$ directly). I'll edit my answer accordingly. $\endgroup$ Commented Feb 3, 2017 at 14:18
  • $\begingroup$ Thanks for the extensive update. And for finding another shiny citation $\endgroup$
    – Bananach
    Commented Feb 3, 2017 at 16:12
  • 1
    $\begingroup$ @Praveen I don't think you need any theory here. Simple choose $v_h$ to be constant zero. $\endgroup$
    – Bananach
    Commented Feb 4, 2017 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.