Estimate on $H^1$ seminorm in finite element context 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 $$|v|_{l,T} \leq  Ch^{r-l} |v|_{r} \text{ for  r<l}$$ 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?
 A: You're pretty much there, the key thing is that $|\hat b_T\cdot|_{H^1} $ and $\|\hat b_T\cdot\|_{L^2} $ are both norms on the reference element, then using the fact that $\hat b_T\le 1$, you get
$$|\hat b_T\hat\phi|_{H^1}\le C\|\hat b_T\hat\phi\|_{L^2}\le\|\hat\phi\|_{L^2}. $$
Now the result follows from scaling. If you keep the local mesh size $h_T$, the argument is purely local, and this doesn't require quasi uniformity of the mesh - only shape regularlity is needed.

As a simple example, lets take the case where the reference element is the unit triangle with vertices $(0,0),(0,1),(1,0)$, and the physical element is the triangle with vertices $(0,0),(0,h_T),(h_T,0)$, so that the affine map $F(\hat x) = h_T\hat x$, and so $F^{-1}(x) = x/h_T$. Then, given a function $w$ on $T$, we define $\hat w(\hat x) = w(F(\hat x))=w(x)$, which also implies that $w(x) = \hat w(F^{-1}(x))$, so that
$$Dw(x) = (DF^{-1})D\hat w(F^{-1}(x)) = (1/h_T)D\hat w(\hat x).$$ Using the multivariable change of variables fomula, the $H^1$-semi norm and $L^2$-norm scale as follows:
$$|w|_{H^1(T)}^2 = \int_T|Dw(x)|^2\,dx = \int_T|(DF^{-1})D\hat w(F^{-1}(x))|^2\,dx$$
$$=h_T^{-2}\int_\hat T|Dw(\hat x)|^2|Det(DF^{-1})|\,d\hat x$$
$$=h_T^{-d-2}|w|_{H^1(\hat T)}^2,$$
and
$$\|\hat w\|_{L^2(\hat T)}^2 = \int_\hat T\hat w^2(\hat x)\,d\hat x=\int_Tw^2(x)|Det(DF)|\,dx =h_T^{d}\|w\|_{L^2(T)}^2.$$
Now we return to the bubble function. We first use the $H^1$-scaling with $w=b_T\phi$, followed by the equivalence of norms to get (note that then $\hat w=\hat b_T\hat \phi$)
$$|b_T\phi|_{H^1(T)}^2 = h_T^{-d-2}|\hat b_T\hat\phi|_{H^1(\hat T)}^2\le C h_T^{-d-2}\|\hat b_T\hat\phi\|_{L^2(\hat T)}^2.$$
We then use the fact that $\hat b\le 1$ followed by the $L^2$ scaling (or first use the scaling, then use $b\le 1$) to get
$$h_T^{-d-2}\|\hat b_T\hat\phi\|_{L^2(\hat T)}^2\le h_T^{-d-2}\|\hat\phi\|_{L^2(\hat T)}^2=h_T^dh_T^{-d-2}\|\phi\|_{L^2(T)}^2=h_T^{-2}\|\phi\|_{L^2(T)}^2.$$
Taking square roots we get
$$|b_T\phi|_{H^1(T)}\le Ch_T^{-1}\|\phi\|_{L^2(T)}.$$
Notice that the proof is almost identical to the proof of the standard $H^1$ to $L^2$ inverse estimate, the only difference is that one needs to prove that $|\hat b_T\cdot|_{H^1}$ and $\|\hat b_T\cdot\|_{L^2}$ are norms on the reference finite element,   which is why the lecturer said the result is "trivial".
The proof for general triangles is done noting that for $F(\hat x) = B_Tx +b_T$, one has $DF = B_T$ and $Ch_T^d\le |\det B_T|\le Ch_T^d$, and $Ch_T\le |B_T|\le Ch_T$. In Brenner and Scott they prove this provided that the triangulation is shape regular.
