How to show this inequality between $L^2$- and $H^2$-norms? Let $\partial_n$ denote the normal derivative. Let $f\in H_0^1(U)\cap H^2(U)$ and $g \in H_0^1(V) \cap H^2(V)$ where $U,V \in \mathbb{R}^d$ are two domains with $U \cap V \neq \emptyset$. Outside there domains both functions are extended to zero.
Use the trace inequality and Cauchy-Schwarz inequality to show $$||\partial_n(u-u_r)||_{L^2(\partial(U\cap V))} \leq C(||f||_{H^2(U)}+||g||_{H^2(V)}).$$
I am not very familiar with this topic but here are my thoughts:
\begin{align*}
||\partial_n(f-g)||_{L^2(\partial(U\cap V))}
&=||n\cdot\nabla(f-g)||_{L^2(\partial(U\cap V))}\\
&\stackrel{CS}{\leq}||\nabla(f-g)||_{L^2(\partial(U\cap V))}\\
&\stackrel{TrIneq}{\leq} C (\lVert \nabla f \rVert_{H^1(U \cap V)} + \lVert \nabla g \rVert_{H^1(U \cap V)}) \\
&\leq C (\lVert \nabla f \rVert_{H^1(U)} + \lVert \nabla g \rVert_{H^1(V)}) \\
&\leq C (\lVert  f \rVert_{H^2(U)} + \lVert  g \rVert_{H^2(V)}).
\end{align*}
But is this legal? Formally this looks good to me but technically I am not quite sure because e.g. in the third step we should actually have euclidean norm bars around $\nabla(f-g)$ but how would this look within the Sobolev norm in the subsequent steps? Do I have to consider the derivatives of the euclidean norm?
Can someone please help me?
 A: I dont know what is the 'third step' because i think your first and second steps are exactly the same. But you are free to expand out the euclidean norm then proceed as usual-
$$\| \nabla (f-g)\|^2_{L^2(\partial (U\cap V)} =\left\|\sqrt{\sum_{i=0}^d|\partial_if-\partial_ig|^2}\right\|^2_{L^2(\partial (U\cap V)}=\sum_{i=0}^d\|\partial_if-\partial_ig\|^2_{L^2(\partial (U\cap V)}  $$
and then apply the trace inequality and so on, to each summand.
response to comment - no, for an arbitrary function $F$, $$\| F \|^2_{H^1(U)} =  \| F \|^2_{L^2(U)} + \sum_{i=1}^d \| \partial_i F \|^2_{L^2(U)} $$
For vector valued functions the euclidean norm $|\cdot|_{\ell^2_d}$ only comes in at the application of the $L^2(U)$ norm, i.e. $$\| F \|^2_{L^2(U)} = \big\| |F|_{\ell^2_d} \big\|^2_{L^2(U)}$$
which is (by the earlier calculation) of course equivalent to just defining $\|F\|^2_{L^2} = \sum_i \|F_i\|^2_{L^2}$.
This definition is directly applied to the derivatives of $\nabla f$,
\begin{align} \| \nabla f \|^2_{H^1(U)} 
&=  \| \nabla f \|^2_{L^2(U)} 
+ \sum_{i=1}^d \| \partial_i \nabla f \|^2_{L^2(U)}  
\\
&=   \big\| |\nabla f|_{\ell^2_d} \big\|^2_{L^2(U)} + \sum_{i=1}^d \big\| |\partial_i (\nabla f)|_{\ell^2_d} \big\|^2_{L^2(U)} 
\\
&= \sum_{i=1}^d \|\partial_i f\|_{L^2}^2 + \sum_{i=1}^d\sum_{j=1}^d \|\partial_i\partial_j f\|_{L^2}^2
\end{align}
