Is the line measure operator bounded in $W^{1,p}(\mathbb{R}^2)$

Question

Consider the linear funtional $S_l: W^{1,p}(\mathbb{R}^2) \to \mathbb{R}$ for some segment $l \in \mathbb{R}^2$ such that $S_l (\mu)$ measures the length of $l$ with respect to $\mu dx$. In other words, it performs the line integral of the function $\mu$ on $l$. I'm curious if $S_l$ is a bounded functional for all finite $l$ for some $p > 1$? I know that for point evaluation operator it is true, since for $p > n$, there is continuous injection of $W^{1,p}(\mathbb{R}^n)$ to $L^\infty(\mathbb{R}^n)$. I wonder if it's also the case for evaluating the function on a line segment.

Answer 1

I just realized it suffices to use the trace operator $T:W^{1,p}(\Omega) \to L^p(\partial\Omega )$ for this, and as $\mu(l)$ is finite apply Holder to get $|Tu|_{L^1(\partial\Omega)} \leq C|u|W^{1,p}(\Omega)$ for all $p \geq 1$. So as long as the set where the integral is taken is some bounded boundary then the result holds.

Answer 2

Not only does $W^{1, p}(\mathbb{R}^{n})$ embed continuosly in $L^{\infty}(\mathbb{R}^{n})$ for $p > n$, but we also have (due to Morrey) that it embeds continuously into the Holder space $C^{0, 1 - \frac{n}{p}}$. This should give us that $S_{l}$ is bounded for all finite $l$ whenever $p > 2$ (since in this case we're working in $\mathbb{R}^{2}$).