T is a isosceles right - angled triangle in 2-dim and F is any edge of T.
v $\in P^{k}(T)$ (polynomial space with degree at most k).
In the book --mathematical aspects of discontinuous galerkin method, it says:(I'v changed the way)
Since the unit sphere in $P^{k}(T)$ for the $L^{2}$-norm is a compact set, there is constant C, only depending on k,F, such that for all $v \in P^{k}(T)$, $||v||_{L^2(F)} \leqslant C||v||_{L^{2}(T)}$
I think this equality is easy to derive by directly computing, but i am wondering how the compactness work and the meaning of the authors.
$P^k(T)$ is finite-dimensional, then the $L^2$-unit ball in $P^k(T)$ is bounded and closed, consequently compact. Then the continuous function $v\mapsto \|v\|_{L^2(F)}$ attains its maximum on the ball. And the inequality is satisfied with $C$ equal to that maximum.
You are right that one could have computed $C$ explicitly. In particular to see its dependence on relevant quantities of the discretization (polynomial degree, mesh-size)