# Why no trace operator in $L^2$?

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?

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It is worth to note that there exist a continuous linear operator $T:L^2(\Omega)\to W^{1-1/2,2}(\partial\Omega)^\ast$ if $\Omega$ has some regularity. – Tomás May 30 '13 at 19:36
Hi @Tomás, can you give a reference for that? – tomglabst Jun 30 '15 at 13:42
Yes @tomglabst. Take a look in Lions-Magenes books: Non-Homogeneous Boundary Value Problems and Applications. For now, I do not remember the volume. – Tomás Jul 5 '15 at 10:36

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 $\overline \Omega = [0,1]$.
@soup: Sure. Consider a sequence $f_n$ of smooth functions that satisfy $f_n(1) = 1$ and such that the sequence goes to zero elsewhere (notice that for any such sequence, the first derivative will explode; this is essentially why $H^1$ gives us enough control over our functions to define a trace). But maybe the one-dimensional example is not optimal, as integrating over the boundary is the same as evaluating pointwise there. You can try the same idea on $\Omega = \mathbb R \times [0, \infty)$, though. Now integration over $\partial \Omega = \mathbb R\times \{0\}$ makes a bit more sense. =) – Sam Nov 18 '12 at 22:18