Sobolev embedding when $p=n$: $W^{1,p}(\mathbb{R}^{n}) \hookrightarrow L^{q}(\mathbb{R}^{n})$ for $q: p \leqslant q < \infty$

Question

In class, aside from the standard Gagliardo-Nirenberg-Sobolev and Morrey inequalities, my professor also covered the case when $p=n$. In particular, if $p=n$, then $W^{1,p}(\mathbb{R}^{n}) \hookrightarrow L^{q}(\mathbb{R}^{n})$ for all $q$ such that $p \leqslant q < \infty$. We did not prove this case in class.

I can see that this would be true trivially whenever $q=p$: $||u||_{L^{p}} \leqslant ||u||_{W^{1,p}}$. If it was true for $q=\infty$, then I could interpolate to immediately get the result for $p < q < \infty$. But I know it's definitely not true when $q=\infty$ (consider $\log\log(1+|x|^{-1})$, which is in $W^{1,n}(B(0,1))$ but not $L^{\infty}(B(0,1))$).

Can anyone provide some guidance on this for me? At least if $u \in C_{c}^{\infty}(\mathbb{R}^{n})$ and $u\neq 0$, if I define $u_{\lambda}(x) = u(\lambda x)$ and assume that the inequality $||u_{\lambda}||_{L^{q}} \leqslant C||u_{\lambda}||_{W^{1,n}}$ holds, I cannot get a contradiction anymore as in the case when $1 \leqslant p < n$, where the contradiction is that $u=0$ a.e. by sending $\lambda$ to either $0$ or $\infty$.

Edit: I found that the proof is actually done by Brezis (Corollary 9.11). I may transcribe the answer later if I have time.

Answer 1

On a bounded domain $\Omega$, one can do the following: For $q \in [p,\infty)$ you can find $\tilde p < p = n$ with $$\frac1q = \frac1{\tilde p} - \frac1n.$$ Now, you can use the 'standard' Sobolev embedding for $\tilde p < n$ to obtain $$ W^{1,p}(\Omega) \hookrightarrow W^{1,\tilde p}(\Omega) \hookrightarrow L^q(\Omega). $$

Answer 2

Actually, in the limiting case for $p>1$ one can prove a stronger result, known as Trudinger inequality $$ W^{1,p}(\Omega) \hookrightarrow L^\varphi(\Omega) $$ where $\varphi$ is defined as follows $$ \varphi(t) = \exp\left(t^{\frac{p}{p-1}}\right) - 1. $$

Link to the original paper by N.Trudinger.