# Bubble function property: how to prove it?

I'm studying error estimators and I need a check on an estimate. In our course, we've been given the following definition of bubble function (in 2D): It's a function defined on a triangle $$T$$ such that:

• $$b_T \in [0,1]$$
• $$b_T \in H_0^1(T)$$
• $$\exists D \subset T: |D|>0 , b_T \geq \frac{1}{2} \text{ in } D$$

Then, professor said that given a function $$\phi$$, say a polynomial of order $$k \geq 1$$ defined on the triangle $$T$$ , we have:

$$|b_T \phi|_{1,T} \leq h_T^{-1} ||\phi||_{0,T}$$

He said that it's almost trivial to prove this, and that one needs to come back and forth from the reference element to prove this. However, I'm not able to show this.

Coming back to reference element trought the affine map $$F(\hat{x})= B_T \hat{x} +b$$ such that $$F(\hat{T})=T$$, I have (by transformations of Sobolev seminorms) $$|b_T \phi|_{1,T} \leq C ||B_T|| J_T |\hat{b_T \phi}|_{1,\hat{T}}$$ Here we have $$b_T \in [0,1]$$ so $$\leq C ||B_T|| J_T |\hat{\phi}|_{1,\hat{T}} \leq C ||B_T|| \cdot ||B_T^{-1}|| \cdot |\phi|_{1,T} = (\star)$$ and now I'd like to apply a scaling argument, but I don't know how to get that $$h_T$$ negative power. Actually, I would apply an inverse estimate $$|v|_{l,T} \leq Ch^{r-l} |v|_{r} \text{ for r so I'd get $$\star \leq C ||B_T|| ||B_T^{-1}|| h_T^{-1} |\phi|_{0,T}$$

which would be the thesis, but I feel like I'm cheating, since the inverse estimate I'm using has been proven by using a scaling argument, and also I think that the quasi-uniformity of the mesh has been assumed, since I don't have any $$\rho$$ parameter coming from the triangles.

Could you please give me a check, or at least point out any flaw in my proof?

• I might be wrong because I just gave a quick glance at the question, but how is this related to computational science? I think the question might be more suitable for the math stackexchange. Jun 21 at 18:27
• I was thinking it's related, since it's a standard topic in finite element theory @Paddy Jun 21 at 18:57
• It is suitable for both SEs. I don't see a problem with it being here. Jun 22 at 6:07

I think there are two things here. First, if you have a discrete function, you can go between the different norms by scaling correspondingly with $$h_T$$, e.g., you have $$$$\| \psi \|_{1,T} \leq C h_T^{-1} \| \psi \|_{0,T}$$$$ for a discrete function $$\psi$$. You can get these exponents correctly by thinking in terms of physical units: when taking a spatial derivative of $$\psi$$ you divide with the length unit (the unit of $$h_T$$). Thus, we have inverse estimates like $$$$\| \psi \|_{k,T} \leq C h_T^{-k} \| \psi \|_{0,T}$$$$ for any $$k \geq 0$$. Indeed, the rigorous proof is through a scaling argument.

Secondly, these so-called bubble functions, if properly normalized, you can introduce without any dependency on $$h_T$$. So you'd have inequalities like $$$$c \| \psi \|_{0,T} \leq \| b_T \psi \|_{0,T} \leq C \| \psi \|_{0,T}.$$$$ Combining two of the above will lead to your result: first get rid of the derivative, and then get rid of the bubble function. I suggest that you approach the proofs separately.

If I recall correctly, when I took a course in finite elements 10 years ago, I've approached the second part by first proving that $$|||v||| = \|b_T v\|_{0,T}$$ is a norm in the polynomial space on $$T$$, scaling to a reference element, and then using the fact that all norms are equal in a finite dimensional space.

In your proof I don't understand how do you get rid of $$b_T$$. Remember that you have $$1$$-seminorm and so you are measuring the derivative. The values $$b_T \in [0, 1]$$ mean nothing. On the other hand, I've never seen such an abstract definition for the bubble function. I'm more familiar with bubble functions that are explicitly defined through some specific finite element basis. E.g., for a reference triangle, you'd use the canonical $$P_1$$ basis $$\psi_1 = \xi$$, $$\psi_2 = \eta$$, $$\psi_3 = 1 - \xi - \eta$$ and then define the global bubble as $$$$b_T(x,y) = \psi_1(F^{-1}(x,y)) \psi_2(F^{-1}(x,y)) \psi_3(F^{-1}(x,y)),$$$$ where $$F$$ is the local-to-global mapping.

• Thanks for your answer, especially for the "physical" meaning of the inverse estimate. Enlightening for a math student like me :-) May I ask if you could prove $$\| \psi \|_{1,T} \leq C h_T^{-1} \| \psi \|_{0,T}$$ That's the step I can't prove Jun 23 at 7:04
• I'd say that it can be showed as follows: $$|\phi|_{1,T} \leq C |det(BF)| ||BF^{-1}|| |\hat{\phi}|_{1,\hat{T}}$$ Now, $||BF^{-1}|| \leq \frac{\hat{h}}{\rho_T}$ and hence $$\leq |det(BF)| \frac{\hat{h}}{\rho_T} |\hat{\phi}|_{1,\hat{T}}$$ Now, by shape regularity, equivalence of norms and coming back to the reference element: $$\leq |det(BF)| C h_T^{-1} |\hat{\phi}|_{1,\hat{T}} \leq |det(BF)| C h_T^{-1} ||\hat{\phi}||_{1,\hat{T}} \leq |det(BF)| C h_T^{-1} ||\hat{\phi}||_{0,\hat{T}} \leq C h_T^{-1} ||\phi||_{0,T}$$ Jun 23 at 7:30
• Actually for the proof I just need the bound on $|\phi|_{1,T}$ and not on the full norm. Could you confirm my argument for the proof of that inverse inequality? @knl Jun 23 at 7:42
• Maybe the exponent of $|det(BF)|$ should be $\pm 1/2$? Otherwise, looks just fine.
– knl
Jun 24 at 6:03
• Thanks @knl. I have one last question, that just came to my mind: so far I have a bound for the $|v|_{1,T}$. How can I get the bound on $||v||_{1,T}$? I mean, how can I go from seminorm to full norm? Jun 24 at 12:58