I'm studying adaptive meshes, and my professor wrote the following property for a bubble function ( see this scicomp post for the definition I'm using)$b_T$ defined on a triangle $T$.

$$||b_T \phi ||_{0,T} \leq ||\phi||_{0,T} \color{green} \leq c||b_T^{\frac{1}{2}} \phi||_{0,T} $$

The proof of the green ineuqlity is given in the seminal work by R. Verfürth, which can be found here (you can access the PDF, page. 7). It's a short proof, but there's a detail that I'd like to be sure about:

After he writes $$||b_T^{1/2} \phi||_{0,T} = |\det(B_f)|^{1/2} ||\hat{b_T^{1/2}} \phi ||_{0,\hat{T}}$$ Then he says

the result follows from the fact that all norms are equivalent on finite dimensional spaces.

That's what I want to be sure about: I think he's using the fact that $||b_T^{1/2} \cdot||_{0,T}$ defines a norm on the space of polynomials, and hence he can write

$$ |\det(B_f)|^{1/2} ||\hat{b_T^{1/2}} \phi ||_{0,\hat{T}} \geq |\det(B_f)|^{1/2} ||\hat{ \phi} ||_{0,\hat{T}}$$

and now coming back to the triangle $T$ we have the thesis. Is that correct?


He is only going to the reference triangle to have the constant independent of $h$, as required. But yes, he is using the fact that having the bubble function inside the norm double bars does not stop it from being a norm.

  • $\begingroup$ Since this is a 100 bounty question, could you please show that the bubble function inside the norm doesn't prevent that norm from being a norm? It suffices to me only the triangular inequality @GuillermoBCN $\endgroup$ Jul 24 at 8:56
  • $\begingroup$ The triangular inequality holds because the functions defined by the multiplication with a bubble function still belong to $L_2(T)$, and so the norm operator is also a norm for them. $\endgroup$ Jul 26 at 7:49

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