I am working through the nice Finite Element notes by Douglas N. Arnold at the moment. They can be found here. A teeny tiny detail in the proof of Lemma 7.4 gives me a headache and I am looking for some advice.
We consider a triangle $T$ with an edge $e$. We define $P_e:L^2(e) \rightarrow \mathbb{R}$ to be the projection of a function to its average value. Now we are considering the operator \begin{align}\hat{P}:~&H^1(T)\rightarrow L^2(e)\\&\phi\mapsto\left.\phi\right|_e-P_e(\left.\phi\right|_e).\end{align} The notation $\left.\phi\right|_e$ means the restriction of $\phi$ to $e$.
On page 114 in the first line of the proof to Lemma 7.4 it is stated that this operator is bounded (and linear) but I can't see a way to bound this nicely in $H^1$ norm $||\cdot||_{H^1}=\text{max }(||\cdot||_{L^2},||\frac{\delta\cdot}{\delta x}||_{L^2})$. The derivative in the norm is meant in the weak sense.
Note that it should be a projection (i.e. boundable by 1) if one follows the proof further.
Any ideas or literature recommendations? Thanks a lot!
The restriction part is bounded by the trace theorem. Now it remains to prove that $P_e$ is bounded. This follows from the common "Cauchy-Schwarz with 1" trick:
$$|P_e(\phi)| = \left | \int_e \phi \cdot 1 \right | \leq \| \phi \|_{L^2(e)} \| 1 \|_{L^2(e)} \leq \| \phi \|_{H^1(e)} m(e)^{1/2}$$
where $m(e)$ is the measure of the edge $e$.