Is the inclusion map in the Sobolev embedding theorem a surjective map?

Question

Let $W^{k,p}(\mathbb{R}^n)$ be the Sobolev space of all real valued functions on $\mathbb{R}^n$ whose first $k$ weak derivatives are in $L^p(\mathbb{R}^n)$. Assume that $$ \frac{1}{q} = \frac{1}{p} - \frac{k-l}{n}$$ and $k >l$. The Sobolev embedding theorem states that the inclusion map $$ i : W^{k,p} \rightarrow W^{l,q} $$ is a bounded linear operator. My question is the following: is $i$ surjective? At the very least, is the image of $i$ closed?

Answer 1

It's not surjective and the image is not closed.

Example: Take $n = 2, k = 1, p = 1$ and hence $q = 2$. Let $f$ be the indicator function of the unit square $[-1,1]^2 \subset \mathbb{R}^2$. It is in $L^\infty$ and hence in $L^2$, but clearly its distributional derivative is not in $L^1$ (it's a signed measure and supported on the boundary of the square). Since $C^\infty_0$ functions are dense in $L^2$ and belong to $W^{1,1}$, the image of the embedding isn't closed either. It is, however, dense.