Why $ W^{1,n}\subset\subset L^{q} $ for $ q\in [n,\infty) $?

Question

I am reading the book named "Functional Analysis, Sobolev Spaces, and Partial differential Equations" by Brezis. In the book, the Rellich-Kondrachov theorem is stated as follows

Suppose that $ \Omega $ is bounded and of class $ C^1 $ in $ \mathbb{R}^n $. Then we have the following compact injections:
\begin{align} W^{1,p}(\Omega)&\subset L^q(\Omega)& \forall q&\in [1,np/(n-p)) \quad p<n,\\ W^{1,n}(\Omega)&\subset L^q(\Omega)& \forall q&\in [n,+\infty), \\ W^{1,p}(\Omega)&\subset C(\overline{\Omega}) & \forall p&>n. \end{align}

In the book, the author states that the case $ p=n $ can be reduced to the case $ p<n $. I am confused of it, can you give me some hints?

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Brezis, Haim, Functional analysis, Sobolev spaces and partial differential equations, Universitext. New York, NY: Springer (ISBN 978-0-387-70913-0/pbk; 978-0-387-70914-7/ebook). xiii, 599 p. (2011). ZBL1220.46002.

Answer 1

This is simply because on a bounded domain, if $t>0$ then $L^{p+t}\subset L^p$.

So if $f\in W^{1,n}$, then $f\in W^{1,p}$ for all $p<n$. For each such $p$, applying the $p<n$ result shows that $W^{1,n}\subset L^q$ for all $q\in [1, np/(n-p))$. (Also recall bounded composed with compact is compact)

But $np/(n-p)$ is arbitrarily large by taking $p\to n$, giving the full range $[1,\infty)$. (this is the same as saying $[n,\infty)$ because of the same $L^{p+t}\subset L^p$ reason as above)