# Question about step in the proof of standard discrete trace inequality

I'm studying from Guermond lecture notes available at https://www.math.tamu.edu/~guermond/M661_FALL_2019/chap12.pdf (see Lemma 12.8( Discrete trace inequality).)

Consider the simple case $$p=r$$, i.e. we wanna prove that $$||v||_{L^p(F)} \leq C h_K^{-1/p}||v||_{L^p(K)}$$

The proof is really short, as can be seen in the link. However, after standard scaling arguments, he arrives at $$||v||_{L^p(F)} \leq c'||A_K^{-1}|| ||A_K|| \Bigl(\frac{|F|}{\hat{|F|}} \frac{\hat{|K|}}{|K|} \Bigr)^{1/p} ||v||_{L^p(K)}$$

where

• $$A_k$$ is the matrix such that the local finite element triple $$(K,P_K,\sum)$$ is generated by $$\Psi_K(v)=A_k (v \circ T_K)$$. Here $$T_K: \hat{K} \rightarrow K$$ is the usual mapping for every mesh cell. I know that by shape regularity $$||A_K^{-1}|| ||A_K|| \leq c$$.

• $$|\hat{F}|$$ is the measure of the face in the reference element, and in the same way $$|\hat{K}|$$ is the measure of the reference element

so the bound is:

$$||v||_{L^p(F)} \leq C \Bigl(\frac{|F|}{\hat{|F|}} \frac{\hat{|K|}}{|K|} \Bigr)^{1/p} ||v||_{L^p(K)}$$

Question: how can we get to the result, i.e. how can we obtain that $$h_K^{-1/p}$$ power that we have in the statement of the lemma? Of course I'm missing some property regarding the $$\frac{|F|}{\hat{|F|}} \frac{\hat{|K|}}{|K|}$$ term. Any help is highly appreciated!

• So others don't have to look up Guermon's lectures, can you add what $F,\hat F$ are, and how $A_K$ is defined? Aug 31 at 4:04
• Just to be clear, for linear mappings you will have something like $|K| = h_K^d |\hat K|$. But that's presumably the easy part. Aug 31 at 4:06
• @WolfgangBangerth Thanks, I edited my question with those definitions. Uhm, then I'd be tempted to say $|F| = h_F^{d-1} |\hat{F}|$, and with this I could obtain the result since $h_F \leq h_K$. Why couldn't I do this? Aug 31 at 7:32
• Ah yes, if $F$ is the face, then you're right and you've got your proof :-) Aug 31 at 16:22

I assume we have a simplex $$K$$ with a face $$F \subset \partial K$$ and that $$K\subset \mathbb{R}^d, F\subset\mathbb{R}^{d-1}$$ are both full-dimensional.
I show in the following only the scaling $$|F|\leq ch^{d-1}|\hat{F}|$$ but note that $$|K|\leq ch^{d}|\hat{K}|$$ follows similarly.
First define the linear affine and inavertible map $$\psi:\hat{F}\rightarrow F$$ that maps the reference face $$\hat{F}$$ to the physical face $$F$$ by $$\psi(\hat{x})=A\hat{x}+b$$ with $$A\in\mathbb{R}^{d-1\times d-1}$$ and $$b\in \mathbb{R}^{d-1}$$. Then if we further define $$\rho:=sup\{diam(B):B\subset K \text{ is a Ball contained in }K \}$$ and $$h:=sup_{x,y\in K}||x-y||_2$$ we can find constants $$c_1,c_2>0$$ such that we have for the determinant of the jacobian $$\psi'$$ that $$c_1\rho^{d-1}\leq|det(\psi')|=|det(A)|\leq c_2 h^{d-1}$$.
\begin{align} |F|:=\int_{F=\psi(\hat{F})}dx=\int_{\hat{F}}|det(\psi')|d\hat{x}= |det(\psi')||\hat{F}|\leq c_2 h^{d-1}|\hat{F}| \end{align}
Therefore $$\frac{|F|}{|\hat{F}|}\leq c_2 h^{d-1}$$ with constant $$c_2>0$$ that doesnt depend on $$h$$.