In my finite element notes, after the proof of the global estimate for the interpolation error, assuming a regular triangulation with triangles $T_m$:

$$\sum_m|v - \Pi_h^r v|_{s,p,T_m} \leq \sigma^{-s} h^{k+1-s}|u|_{k+1,\Omega}$$ for $u \in W^{k+1,p}(\Omega)$

where $\Pi_h^k:C^0(\Omega) \rightarrow X_h^k$ and $X_h^k= \{v \in C^0(\Omega): v_{|T_m} \in P^k(T_m) \}$

After that, the professor told that in general $$\sum_m |v - \Pi_h^r v|_{k,p,T_m}^p \ne |v - \Pi_h^r v|_{k,p,\Omega}^p$$ I'm trying to find an example of this All I know is that, for instance, the space $X_h =\{ v \in C^o (\Omega): v_{|T_m} \in P^1(T_m) \}$ is not a subspace of $H^2(\Omega)$ since the second derivative has delta distributions.

I'd like to find even 1D example for which the sum over the intervals of the norms is not equal to the norm on the whole space, but I can't come up with anything.


Given a function $v \in W^{k,p}(T)$, $$|v|_{k,p,T}^p = \int_{T})D^{k} v)^p dx$$

where $k$ is a multiindex. It's the usual Sobolev seminorm

  • $\begingroup$ I think this site concentrates on the numerical aspect. This question is suitable for asking in Mathematics site. $\endgroup$
    – Truong
    Jul 9, 2021 at 22:41
  • 4
    $\begingroup$ Assuming $\Omega = \bigcup_m T_m$, isn't this simply triangle inequality? I guess, since the basis functions have compact support, the equality may hold more commonly. Have you tried to write $|\cdot|_{k,p,\Omega}$ and see it for yourself? Maybe, that will reveal something. $\endgroup$ Jul 10, 2021 at 1:08
  • $\begingroup$ @AbdullahAliSivas I tried In 1D: I consider the usual hat functions, and as $\Omega = [0,1]$ and write $|v- \Pi v|_{2,2,\Omega}$. This is $$\int_0^1 ((v - \sum_i v_i \phi(x))'')^2$$ but this integral cannot be splitted since the second derivative is not well defined. Is this what you were trying to say? $\endgroup$
    – FEGirl
    Jul 10, 2021 at 10:36
  • $\begingroup$ Not exactly. It is okay if the second derivative does not exist at some points, since you are integrating and as long as the discontinuous are limited to a set of measure zero, it will be fine. You are forgetting a sqrt in the definition of the seminorm, I am saying that using that and the triangle inequality, you should be able to see why they are not equal in general. $\endgroup$ Jul 10, 2021 at 20:16
  • $\begingroup$ @AbdullahAliSivas I'm really sorry but I can't see the way to show this. I can't understand what you mean when you say I have to apply triangle inequality: maybe $|v - \Pi v|_{2,2,\Omega}^2 \leq |v|_{2,2,\Omega}^2 + |\Pi|_{2,2,\Omega}^2$, but I am honestly stucked $\endgroup$
    – FEGirl
    Jul 10, 2021 at 23:35

2 Answers 2


Split $[0,1]$ into two elements $[0,1/2]$ and $[1/2,1]$. Consider the function $f$ satisfying $f(x)=0$ for $x < 1/2$ and $f(x)=1$ otherwise. For this function the $H^1$ norm is infinity but if you calculate the $H^1$ norm over the two elements separately you get 0 and $1/2$, respectively. In particular, the derivative of the Heaviside function is a delta distribution which is not square integrable.

Now you did not define what is the projection operator. Maybe if it projects to the piecewise constant finite element space, then you could apply the above example?

  • $\begingroup$ Why do you say that the $H^1$ norm is infinity? I mean, I think you mean that $$\int_0^1 \delta_{\frac{1}{2}} dx = \infty$$, but I cannot see why $\endgroup$
    – FEGirl
    Jul 15, 2021 at 14:14
  • $\begingroup$ No, square integral, i.e. $\int_0^1 \delta(x - 1/2)^2 \,d\mathrm{x}$. $\endgroup$
    – knl
    Jul 15, 2021 at 15:22
  • $\begingroup$ Oops, you're right. But I still don't understand why that integral is $\infty$ $\endgroup$
    – FEGirl
    Jul 15, 2021 at 16:26
  • $\begingroup$ Try using the standard rule $\int_0^1 f(x) \delta(x - a) dx = f(a)$ with $f(x) = \delta(x -1/2)$ and $a=1/2$. $\endgroup$
    – knl
    Jul 15, 2021 at 20:11
  • $\begingroup$ Can you accept the solution? $\endgroup$
    – knl
    Jul 18, 2021 at 8:28

The statement as given is indeed correct (i.e., left and right hand side are different) for almost any function $f(x)$. What it says, once you write out what these norms are, is that $$ \sum_m \sqrt{\int_{T_m} f(x)^2 } \neq \sqrt{ \sum_m \int_{T_m} f(x)^2 }. $$

  • $\begingroup$ Thanks for your answer. There was a big typo in my formula (i.e. a power $p$ was missing). I edited it. Basically what professor said is that we need to have continuity across interelement boundaries if we want that equality to hold true, but I cannot find an example that shows this $\endgroup$
    – FEGirl
    Jul 13, 2021 at 7:17
  • $\begingroup$ Then you ought to define how exactly you define the operator $\Pi$. $\endgroup$ Jul 13, 2021 at 15:39
  • $\begingroup$ $\Pi$ here is the classical Lagrange interpolation operator. My prof. said that that equality holds whenever our discrete space $V_h$ is a subspace of the continuous one. $\endgroup$
    – FEGirl
    Jul 15, 2021 at 14:17
  • $\begingroup$ That's why I ask. Define onto what space precisely the operator interpolates. $\endgroup$ Jul 15, 2021 at 15:20
  • $\begingroup$ It's defined so that $\Pi:C^0(\Omega) \rightarrow X_h^k$, where $X_h^k= \{v \in C^0: v_{T_m} \in P^k(T_m) \text{ for every }m\}$ $\endgroup$
    – FEGirl
    Jul 15, 2021 at 16:28

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