Let $u\in H^1(\mathbb R^2)$, where $H^1(\mathbb R^2)\equiv W^{1,2}(\mathbb R^2)$ is the usual Sobolev space with standard norm and inner product. I want to show that for any $1\le q <\infty$ (actually I believe it should be $2<q<\infty$), we have the inequality
$$\|u\|_{L^q(\mathbb R^2)}\le C\sqrt{q} \|u\|_{H^1(\mathbb R^2)} \tag1$$
For some constant $C>0$ which does not depend on $u$.
By Sobolev embedding theorem, we know that for $s$ satisfying $1/q = 1/2 – s/2 $, we have the continuous embedding $W^{s,2}(\mathbb R^2)=H^s(\mathbb R^2) \hookrightarrow L^q(\mathbb R^2)$, hence there exists $C_1>0$ such that
$$\|u\|_{L^q(\mathbb R^2)}\le C_1 \|u\|_{H^{s}(\mathbb R^2)} \tag{1a} $$
Next, we need to find a constant $C_2>0$ such that
$$\|u\|_{H^{s}(\mathbb R^2)}\le C_2 \|u\|_{H^1(\mathbb R^2)} \tag{1b} $$
This answer claims that $(1\text{b})$ is also a consequence of Sobolev embedding. By recursively applying the usual Sobolev inequality to $D^{s-1}u,D^{s-2}u,\ldots$ and so on, we can indeed show that the embedding $W^{s,2}(\mathbb R^2)\hookrightarrow W^{1,2_s^{*}}(\mathbb R^2)$ (where $2_s^{*}$ is the $(s-1)$th Sobolev conjugate of $2$) is continuous. However $W^{1,2_s^{*}}(\mathbb R^2)\ne W^{1,2}(\mathbb R^2)$, so I still fail to see why such a $C_2$ exists.
Regardless, assuming that $(1\text{b})$ holds, it can then be combined with $(1\text{a})$ and the fact that $1\le\sqrt q $ to conclude that
$$\|u\|_{L^q(\mathbb R^2)}\le C'\sqrt{q} \|u\|_{H^1(\mathbb R^2)} \tag{1'}$$
where $C'\equiv C_1 C_2$, which is "technically" identity $(1)$.
However this is not satisfactory at all since the dependence of $C'$ on $q$ is not made explicit, and I think the point is rather to show that
$$C' \equiv C'(q) = C\sqrt q \tag2$$
Now I am a bit stuck in trying to prove $(2)$, I have tried explictly writing down the $L^q$-norm of $u$, playing with spherical coordinates and Young's inequality, but I can't seem to get the desired conclusion. I suspect the proof follows from the way inequality $(1\text{b})$ is obtained, but I am clueless so far.
Hence I ask for your help : Why does the embedding $\mathbf{(1b)}$ hold ? And how to show $\mathbf{(2)} $ ?
Thanks in advance.
Best Answer
cs89 answers your first question well, but the paper attached there seems to indicate that $C(q)^2\sim8\pi\mathrm{e}q$, which fits your desired result (2) well. This answer tries to verify and generalize the inequality (1) in a relatively elementary way.
I saw this inequality first in a note written by Steve Shkoller (see Theorem 2.30 in Notes on $L^p$ and Sobolev Spaces), and later a generalization to the $n$-dimensional case in the well-known textbook, A First Course on Sobolev Spaces, written by Giovanni Leoni (see Theorem 12.33 in the second edition). Both proofs in the above two references seem sort of technical at first sight. Nevertheless, a slight modification of the argument used by Lawrence Evans in his famous textbook Partial Differential Equations to prove Morrey's inequality can show that there exists a constant $C$ depending only on $n$ such that \begin{equation} ||u||_{L^q(\mathbb{R}^n)}\le Cq^{1-1/n}||u||_{W^{1,n}(\mathbb{R}^n)} \tag{*} \end{equation} for any $u\in W^{1,n}(\mathbb{R}^n)$ and any $n\le q<\infty$.
First, we claim that there exists a constant $C$ depending on $n$ such that \begin{equation} |u(x)|\le C\int_{B(x,1)}\frac{|u(y)|+|Du(y)|}{|y-x|^{n-1}}dy \tag{**} \end{equation} for any $u\in C^1(\mathbb{R}^n)$ and any $x\in\mathbb{R}^n$. It is clear that $$ |u(x)| = \frac{1}{\alpha(n)}\int_{B(x,1)}|u(x)|dy \le \frac{1}{\alpha(n)}\int_{B(x,1)}|u(y)-u(x)|+|u(y)|dy , $$ and $$ \int_{B(x,1)}|u(y)|dy \le \int_{B(x,1)}\frac{|u(y)|}{|y-x|^{n-1}}dy. $$ Also recall that it is shown in Evans' book that $$ \int_{B(x,1)}|u(y)-u(x)|dy \le C\int_{B(x,1)}\frac{|Du(y)|}{|y-x|^{n-1}}dy . $$ Indeed, let $w$ be a point of the unit sphere $\partial B(0,1)$ and let $0<s<1$. Then $$ |u(x+sw)-u(x)| \le \int_0^s|Du(x+tw)|dt . $$ Hence, $$ \int_{\partial B(0,1)}|u(x+sw)-u(x)|dS(w) \le \int_{\partial B(0,1)}\int_0^s|Du(x+tw)|dtdS(w) . $$ Now $$\int_{\partial B(0,1)}\int_0^s|Du(x+tw)|dtdS(w) = \int_{B(x,s)}\frac{|Du(y)|}{|y-x|^{n-1}}dy $$ and $$ \int_{\partial B(0,1)}|u(x+sw)-u(x)|dS(w)=s^{-(n-1)}\int_{\partial B(x,s)}|u(y)-u(x)|dS(y) $$ We obtain $$ \int_{\partial B(x,s)}|u(y)-u(x)|dS(y) \le s^{n-1}\int_{B(x,1)}\frac{|Du(y)|}{|y-x|^{n-1}}dy , $$ and integrating with respect to $s$ from $0$ to $1$ yields $$ \int_{B(x,s)}|u(y)-u(x)|dy \le \frac{1}{n}\int_{B(x,1)}\frac{|Du(y)|}{|y-x|^{n-1}}dy . $$
From (**) and Young's inequality of convolutions, we get \begin{align*} ||u||_{L^q(\mathbb{R}^n)} &\le C\left(\int_{B(0,1)}|x|^{-\frac{n(n-1)q}{(n-1)q+n}}dx\right)^{1-1/n+1/q}||u||_{W^{1,n}(\mathbb{R}^n)} \\ &\le C\left(\frac{(n-1)q+n}{n^2}\right)^{1-1/n+1/q}||u||_{W^{1,n}(\mathbb{R}^n)} . \end{align*} Notice that $q^{1/q}\to1$ as $q\to\infty$, from which (*) follows.