# Proof of R. Verfürth paper on adaptive mesh and bubble functions

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.
• 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. Jul 26 at 7:49