Why no trace operator in $L^2(\Omega)$?

We have trace operator which allows us to define boundary values of an $H^1$ function. This is because of the fact that $C^\infty$ is dense in $H^1$ under the $H^1$ norm, I believe.

I'm sure either $C^0$ or $C^\infty$ is also dense in $L^2$ in the $L^2$ norm, so why no trace operator in this case? Or am I wrong?

The problem is that even though you can of course define a trace $T: C^\infty(\overline \Omega) \to L^2(\partial \Omega)$, to be able to extend $T$ to all of $L^p(\Omega)$ in a meaningful way, it is not sufficient to have any old operator $T$, but you really want $T$ to be continuous, i.e. there would need to be a constant $C >0$ such that

$$\Vert Tf \Vert_{L^2(\partial \Omega)}\le C \Vert f \Vert_{L^2(\Omega)}$$

In this case we would be able extend $T$ to an operator $T: L^2( \Omega) \to L^2(\partial \Omega)$ nicely. This is the case if you take the $H^1$ norm instead of the $L^2$ norm. However, it's a good exercise to show that such a $C$ does not exist for $L^2$ (or indeed for any $L^p$ space). Consider something simple like $\overline \Omega = [0,1]$.

Let us consider the following sequence of continuous functions:

$$f_j(x)=\frac{1}{1+\color{red}jdist(x,\partial\Omega)},~~~~~~x\in \Omega.$$

We have, $$0\le f_j(x)\le 1,~~~~~and ~~~~~f_j(x)=1,~~~x\in \partial\Omega.$$ furthermore, for every $x \in\Omega $, we have $f_j(x)\to 0$ thus, by convergences dominated theorem we get $$\Vert f_j \Vert^p_{L^p(\Omega)}\to0 $$

Assume that $\Omega$ is bounded of class $C^1$ and there is a trace operator from $L^p(\Omega)~~$ to $~~L^p(\partial \Omega)$ then there exists a constant $C>0$ Such that for all $f \in L^p(\Omega)$ we have: $$\Vert Tf \Vert_{L^p(\partial \Omega)}\le C \Vert f \Vert_{L^p(\Omega)}.$$

In particular, for every $j$ we have $$Area(∂\Omega)=\Vert Tf_j \Vert^p_{L^p(\partial \Omega)}\le C^p \Vert f_j \Vert^p_{L^p(\Omega)}\to0. $$ $$Area(∂\Omega)=0$$

which is absurd since $\Omega$ is a bounded of class $C^1$.