For positive integer k, how to show that there is no continuous trace operator mapping from $W^{k,2}(R^m)$ to $W^{k,2}(R^{m-1})$?
I want to prove this by contradiction, but have no idea what to do next. My first thought is to construct a sequence of functions in $C^{\infty}(R^m)\cap W^{k,2}(R^m)$ say $f_n$ which converges to f, then to show that the sequence $g_n = Tf_n$ doesn't converge to $Tf$ in $W^{k,2}(R^{m-1})$.
Update: Suppose that there exists such a continuous trace operator $T$. By the fact that $W_{0}^{k,2}(R^m) = W^{k,2}(R^m)$，for a function $f$ in $W^{k,2}(R^m)$, we can find a sequence of functions $f_n$ in $C^{\infty}_{c} (R^m)$ converges to $f$. Then $Tf_n \to Tf$. Since $Tf_n = 0$ for each $n$, $Tf = 0$.
My main problem is that I am not sure how to show $Tf = 0$ is not true for every $f \in W^{k,2}(R^m)$.
Any assistance will be appreciated