I know that Trace operator $T : W^{1,p}(\Omega) \to L^p(\partial \Omega, H^{n-1})$ is well defined when $\Omega$ is a bounded Lipschitz domain. (here $H^n$ is the Hausdorff measure). What about only bounded domains?
In particular, consider this set $ A= \{ (x,y) \in \mathbb{R^2} : 0 \leq x\leq1 , 0 \leq y \leq x^2 \}$. We have "problems" near $(0,0)$ thus it is not Lipschitz. So, theorically, I can't define Trace operator. In order to prove this I'm trying to show that:
For every $ k \in \mathbb{N}$ there exists $\phi_k \in C^{\infty}(\mathbb{R^2})$ such that $\int_{\partial A} \vert \phi_k \vert dH^1 \geq k \Vert \phi_k \Vert_{W^{1,1}(A)}$
Clearly, if above statement holds than I can't define Trace operator on $A$. I chosen that set $A$ because it is quite simple and not Lipschitz, however I can't solve this exercise. Can you explain me how such $\phi_k $ should be? I think they could be bigger at the boundary. Maybe I should take advantage of problems near the origin but how?
Here $\int_{\partial A} \vert \phi_k \vert dH^1 = \int_0^1 \vert\phi_k(x,0) \vert dx + \int_0^1 \vert \phi_k(1,y) \vert dy +\int_0^1 \vert \phi_k(x,x^2) \vert \sqrt{1+4x^2} dx$
Consider $f(x,y)=\frac{1}{x} \in W^{1,1}(A)$.
This choice of $f$ is possible because $A$ is not Lipschitz and thus $\frac{\partial}{\partial{x}}f(x,y)\in L^{1}(A)$: $$\int_{0}^{1}\int_{0}^{x^{2}}|\frac{\partial}{\partial{x}}f(x,y)|\;dydx = \int_{0}^{1}\int_{0}^{x^{2}}\frac{1}{x^{2}}\;dydx = 1$$
But integrating over $\partial{A}$, we get $$\int_{0}^{1}\frac{1}{x}dx = \infty$$
EDIT: Since you are explicitely looking for a sequence of functions $f_{n} \in C^{\infty}(\mathbb{R}^{2})$:
Take the continuous functions $$g_{n}(x,y) = n \cdot 1_{(-\infty,-\frac{1}{n})\times \mathbb{R}} + \frac{1}{x+\frac{2}{n}} \cdot 1_{[-\frac{1}{n},\infty)\times \mathbb{R}}$$
Smooth them around their respective point of singularity at $-\frac{1}{n}$ to obtain $f_{n} \in C^{\infty}(\mathbb{R}^{2})$.
Then $\int_{0}^{1} f_{n}(x,0)dx \rightarrow \int_{0}^{1} f(x,0)dx = \infty$ but $||f_{n}||_{W^{1,1}(A)} \rightarrow ||f||_{W^{1,1}(A)} < \infty$.