I know Lax-Milgram theorem is fundamental to FEM. But it did not explain what will happen if coercivity is not met.

My understanding is if it is met, eigen value of the operator (or its corresponding matrix) is always positive (may be degenerated). If it is not, with the refinement of grid, more and more eigen value will be put into the possible range of eigen values (from positive to negative) and there will be some one very close to zero. Then the conditional number would be high enough to ruin the result.

Is that a correct explanation?


1 Answer 1


The intuition is correct, although I would amend your statement that the smallest eigenvalue of the operator and its matrix discretizations is always strictly positive and hence never degenerated.

Putting it another way, this is why in infinite-dimensional spaces, it is not sufficient for a (self-adjoint) operator $A:V\to V^*$ to satisfy $\langle v,Av\rangle >0$ for every $v\in V$ but you need to have $\langle v,Av\rangle \geq c\|v\|^2$ for some $c>0$ independent of $v$.

But note that if what you want to show that a matrix $M$ is invertible (which is what the point of the Lax-Milgram is), positive-definiteness is not necessary, only sufficient -- you can have invertible matrices that are not positive definite. All that is required is that $M$ is injective and surjective.

The corresponding theorem for infinite-dimensional operators is the Banach-Necas-Babuska Theorem (see, e.g., Theorem 8.1 in my lecture notes), which replaces the Lax-Milgram Theorem when studying Petrov-Galerkin methods such as DG. Again, you need to make the assumption "quantitative" in infinite dimensions: instead of surjectivity, you need the so-called inf-sup condition.

  • $\begingroup$ I know little about infinite dimension function space. But If the matrix is only positive definite, there is no guarantee that the eigenvalue is not degenerate. Think about a simple positive Jordan block. I think a lot of material is focusing too much on Symmetric Positive Definite operator/matrix. It has a very good property as you mentioned. $\endgroup$
    – CatDog
    Aug 3, 2018 at 15:12
  • $\begingroup$ Just like LBB condition's case. The original weak form meet LBB condition and is uni-solvent. But the discrete form may not hold LBB condition and the result is problematic. $\endgroup$
    – CatDog
    Aug 3, 2018 at 15:18
  • $\begingroup$ @CatDog Then I don't know what you mean by degenerate. Note that if the property (coercivity/LBB) holds for $V$ / $W$, it also holds for any finite-dimensional subspace $V_h$ / $W_h$ with the same constants. Of course, if you use a non-conforming method where, e.g., $V_h$ is not a subspace, you have to prove existence, uniqueness and stability from scratch. $\endgroup$ Aug 3, 2018 at 15:58
  • $\begingroup$ Degenerate: like this: ch.ntu.edu.tw/~byjin/course/ChemMath/handouts/degeneracy.pdf I am sorry if I misused this term. It is something related to difference between algebraic and geometric multiplicity. $\endgroup$
    – CatDog
    Aug 3, 2018 at 16:20
  • 2
    $\begingroup$ I just read a Hartmann's notes: ganymed.math.uni-heidelberg.de/~hartmann/publications/2008/… . In page 28, it explains why coercivity implies A is not singular. I think it is more clear to me. $\endgroup$
    – CatDog
    Aug 5, 2018 at 4:38

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