In finite element books, we have estimates for $$||u-u_h||$$ and also estimates for $$||u - I_h(u)||$$ where $I_h(u): V \mapsto V_h$ projects a function from the infinite dimensional space to the finite dimensional one. For instance, $V = H^1$ and $V_h \subset V$ is the set of piecewise poly of degree $r$.

What is the difference between $u_h$ and $I_h(u)$ ?

Here's my attempt:

Usually the interpolant is defined as (whenever point evaluation makes sense) $$I_h(u)(x) = \sum_{i=1}^N u(x_i)\phi_i(x) $$

Also, we know that $u_h \in V_h$, so there must exists a set of coefficients $\{U_i\}_i$ (called DoFs) such that $$u_h(x)= \sum_{i=1}^N U_i \phi_i(x)$$

So it seems to me that $u_h$ and $I_h(u)$ are really similar, but the only difference is in the coefficients: if $U_i$ coincides with $u(x_i)$ for every $i$, then the interpolant coincides with the discrete solution $u_h$. In practice, we obtain $\{ U_i\}_i$ by solving linear systems, so they will not be equal to $u(x_i)$

  • 1
    $\begingroup$ Correct. In general, $U_i$ differs from $u(x_i)$. But, for some (rare) special cases, they can match. $\endgroup$
    – user7440
    Commented Jul 3, 2021 at 2:42
  • $\begingroup$ In my interpretation of the symbols, it is not necessary that the solution $u_h$ of the $h$-discretized problem is equal to the projection $I_h(u)$ of the solution $u$ of the continuous problem. $\endgroup$ Commented Jul 3, 2021 at 6:21
  • $\begingroup$ @LutzLehmann could you please elaborate on this? It is precisely the source of my doubts. I mean, it is not necessarily equal because of the the fact that we got $U_i$ by approximation on a computer? $\endgroup$
    – FEGirl
    Commented Jul 3, 2021 at 8:28
  • $\begingroup$ The numerical solution will contain perturbations due to the truncation error of the method. // Or in a variational context, the construction of space and scalar product is only loosely connected to the functional that gets minimized, so projection on subspaces will in general give different results than the minimum of the functional in that subspace. $\endgroup$ Commented Jul 3, 2021 at 9:17

2 Answers 2


The other answer has everything you already need, but it's also worth pointing out that $u_h$ is computable whereas $I_hu$ is not: The latter requires you to know the exact solution, which in general we of course don't know (and if we did, we didn't need to compute a Galerkin approximation $u_h$).


Let's assume $u$ is the solution to the variational problem and $u_h$ is the Galerkin approximation of the solution on the subspace $V_h \subset V$.

By Cea's lemma you have: \begin{equation} ||u-u_h||_V \leq C\inf\limits_{v_h \in V_h}||u-v_h||_V \end{equation} for some positive constant $C$. Now you define the projection $I_h:V \rightarrow V_h$. As you already mentioned if $I_h$ involves point evaluations you need that $u$ is smooth enough such that $I_h(u)$ is well defined. If we have enough smoothness, we can state the inequality: \begin{equation} \inf\limits_{v_h \in V_h}||u-v_h||_V \leq ||u-I_h(u)||_V \end{equation} because $I(u)\in V_h$ by definition. Therefore overall: \begin{equation} ||u-u_h||_V \leq C||u-I_h(u)||_V \end{equation} So you see that it might be possible from the estimate that the two norms are equal, but this is generally not always the case and depends on how exactly you define $I_h$.


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.