proof of Sobolev inequality for $\mathrm{\Omega }={𝐑}^n$

Step 1: $u$ is smooth and $p=1$

First suppose $u$ is a compactly supported smooth function, and let ${(e_k)}_{1\le k\le n}$ denote a basis of ${𝐑}^n$. For every $1\le k\le n$,

$$u(x)={\int }_{-\mathrm{\infty }}^0\frac{\partial u}{\partial x_k}(x+s{e}_k)𝑑s.$$

Therefore,

$$|u(x)|\le S_k(x):={\int }_{𝐑}\left|\frac{\partial u}{\partial x_k}(x_1,\mathrm{\dots },x_{k-1},s,x_{k+1},\mathrm{\dots },x_n)\right|𝑑s.$$

Note that $S_k$ does not depend on $x_k$.One also has

$${|u(x)|}^{n/(n-1)}\le \prod _{k=1}^n{|S_k(x)|}^{1/(n-1)}.$$

The integration of this inequality yields,

$${\int }_{{𝐑}^n}{|u(x)|}^{n/(n-1)}𝑑x\le {\int }_{{𝐑}^n}\prod _{k=1}^n{|S_k(x)|}^{1/(n-1)}dx.$$

Since $S_1$ does not depend on $x_k$, we can apply the generalized Hölder inequality with $n-1$ for the integration with respect to $x_1$ in order to obtain:

$${\int }_{{𝐑}^n}{|u(x)|}^{n/(n-1)}𝑑x\le {\int }_{{𝐑}^{n-1}}{S}_1(x)\prod _{k=2}^n{\left({\int }_{𝐑}{S}_k(x)𝑑x_1\right)}^{1/(n-1)}d{x}_1\mathrm{\dots }d{x}_n.$$

The repetition of this process for the variables $x_2,\mathrm{\dots },x_n$ gives

$${\int }_{{𝐑}^n}{|u(x)|}^{n/(n-1)}𝑑x\le \prod _{k=1}^n{\left({\int }_{{𝐑}^n}\left|\frac{\partial u}{\partial x_k}\right|𝑑x\right)}^{1/(n-1)}.$$

By the arithmetic-geometric means inequality, one obtains

$${\int }_{{𝐑}^n}{|u(x)|}^{n/(n-1)}𝑑x\le n^{-n/(n-1)}{\left(\sum _{k=1}^n\left({\int }_{{𝐑}^n}\left|\frac{\partial u}{\partial x_k}\right|𝑑x\right)\right)}^{n/(n-1)}.$$

One finally concludes

$${\parallel u\parallel }_{L^{n/(n-1)}}\le n^{1/2-n/(n-1)}{\parallel \nabla u\parallel }_{L^{n/(n-1)}}.$$

Step 2: general $u$ and $p=1$

In general if $u\in W^{1,1}({𝐑}^n)$. It can be approximated by a sequence of compactly supported smooth functions $(u_m)$. By step 1, one has

$${\parallel u_m-u_{\mathrm{\ell }}\parallel }_{L^{n/(n-1)}}\le n^{1/2-n/(n-1)}{\parallel \nabla u_m-\nabla u_{\mathrm{\ell }}\parallel }_{L^1}.$$

therefore $(u_m)$ is a Cauchy sequence in $L^{n/(n-1)}({𝐑}^n)$. Since it converges to $u$ in $L^1({𝐑}^n)$, the limit of $(u_m)$ is $u$ in $L^{n/(n-1)}({𝐑}^n)$ and one has

$${\parallel u\parallel }_{L^{n/(n-1)}}\le n^{1/2-n/(n-1)}{\parallel \nabla u\parallel }_{L^{n/(n-1)}}.$$

Step 3: and $u$ is smooth

Suppose and $u$ is a smooth compactly supported function.Let

$$r=\frac{p(n-1)}{n-p}$$

and

$$v={|u|}^r.$$

Since $u$ is smooth, $v\in W^{1,1}$ (It is however not necessarily smooth), and its weak derivative is

$$\nabla v=ru{|u|}^{r-2}\nabla u.$$

One has, by the Hölder inequality,

$$\parallel \nabla v{\parallel }_{L^1({𝐑}^N)}\le r\parallel |u|{}^r{\parallel }_{L^{p/p-1}({𝐑}^N)}\parallel \nabla u{\parallel }_{L^p({𝐑}^N)}=r\parallel u{\parallel }_{L^{np/(n-p)}({𝐑}^N)}^{r-1}\parallel \nabla u{\parallel }_{L^p({𝐑}^N)}$$

Therefore, the Sobolev inequality yields

$$\parallel u{\parallel }_{L^{np/(n-p)}({𝐑}^N)}^r=\parallel v{\parallel }_{L^{n/(n-1)}({𝐑}^N)}\le r{n}^{1/2-n/(n-1)}\parallel u{\parallel }_{L^{np/(n-p)}({𝐑}^N)}^{r-1}\parallel \nabla u{\parallel }_{L^p({𝐑}^N)}.$$

This yields

$$\parallel u{\parallel }_{L^{np/(n-p)}({𝐑}^N)}\le n^{1/2-n/(n-1)}\frac{p(n-1)}{n-p}\parallel \nabla u{\parallel }_{L^p({𝐑}^N)}.$$

Step 4: and $u\in W^{1,p}$

This is done as step 2.

This proof is due to Gagliardo and Nirenberg, who were the first to prove the inequality for $p=1$. This proof can be also found in [1, 2, 3].