See this proof of existence of solution to nonhomogenous Dirichlet problem (from http://www.ann.jussieu.fr/~frey/cours/UdC/ma691/ma691_ch4.pdf page 7):
How is the inequality marked with the red dot derived from the previous inequality? How does manipulate the $\lVert u_g \rVert_{1,\Omega}$ term?
He does say on the line before "For every lift $u_g$", so I guess then $u_g$ is such that its trace is $g$, but otherwise it's an arbitrary member of $H^1$. Then the inequality follows if the infimum is achieved by some $u_g$, but how to show this??
Is there a way to show this inequality without recourse to the infimum norm?
Let $\mathcal{K}=\{v\in H^1(\Omega):\ \gamma_0 v=g\}$. Note that $\mathcal{K}$ is a closed convex subset of $H^1(\Omega)$.
Define $F:\mathcal{K}\to\mathbb{R}$ by $$F(v)=\|v\|_{1,\Omega}$$
I - $F$ is obviously coercive, i.e. if $\|v\|_{1,\Omega}\to\infty$, $v\in\mathcal{K}$ then, $F(v)\to\infty$.
II - $F$ is also weakly sequentially lower semi continuous,i.e. if $v_n\rightharpoonup v$ in $\mathcal{K}$ then, $F(v)\le\liminf F(v_n)$.
It follows from I, II and the properties of $\mathcal{K}$, that the infimum of $F$ in $\mathcal{K}$ is attained by some function $u\in\mathcal{K}$.