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


  • $A_k$ 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!

  • $\begingroup$ 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? $\endgroup$ Commented Aug 31, 2021 at 4:04
  • $\begingroup$ 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. $\endgroup$ Commented Aug 31, 2021 at 4:06
  • $\begingroup$ @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? $\endgroup$
    – FEGirl
    Commented Aug 31, 2021 at 7:32
  • $\begingroup$ Ah yes, if $F$ is the face, then you're right and you've got your proof :-) $\endgroup$ Commented Aug 31, 2021 at 16:22

1 Answer 1


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 $c_1\rho^{d-1}\leq|det(\psi')|=|det(A)|\leq c_2 h^{d-1}$.

Finally, we have by the transformation theorem for integrals:

\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 $h$.


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