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

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

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

 $u(x)=\int_{-\infty}^{0}\frac{\partial u}{\partial x_{k}}(x+se_{k})\,ds.$

Therefore,

 $\lvert u(x)\rvert\leq S_{k}(x):=\int_{\mathbf{R}}\bigl{\lvert}\frac{\partial u% }{\partial x_{k}}(x_{1},\dots,x_{k-1},s,x_{k+1},\dots,x_{n})\bigr{\rvert}\,ds.$

Note that $S_{k}$ does not depend on $x_{k}$. One also has

 $\lvert u(x)\rvert^{n/(n-1)}\leq\prod_{k=1}^{n}\lvert S_{k}(x)\rvert^{1/(n-1)}.$
 $\int_{\mathbf{R}^{n}}|u(x)|^{n/(n-1)}\,dx\leq\int_{\mathbf{R}^{n}}\prod_{k=1}^% {n}\lvert S_{k}(x)\rvert^{1/(n-1)}\,dx.$

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:

 $\int_{\mathbf{R}^{n}}\lvert u(x)\rvert^{n/(n-1)}\,dx\leq\int_{\mathbf{R}^{n-1}% }S_{1}(x)\prod_{k=2}^{n}\Bigl{(}\int_{\mathbf{R}}S_{k}(x)\,dx_{1}\Bigr{)}^{1/(% n-1)}\,dx_{1}\dots dx_{n}.$

The repetition of this process for the variables $x_{2},\dots,x_{n}$ gives

 $\int_{\mathbf{R}^{n}}\lvert u(x)\rvert^{n/(n-1)}\,dx\leq\prod_{k=1}^{n}\Bigl{(% }\int_{\mathbf{R}^{n}}\bigl{\lvert}\frac{\partial u}{\partial x_{k}}\bigr{% \rvert}\,dx\Bigr{)}^{1/(n-1)}.$

By the arithmetic-geometric means inequality, one obtains

 $\int_{\mathbf{R}^{n}}\lvert u(x)\rvert^{n/(n-1)}\,dx\leq n^{-n/(n-1)}\Bigl{(}% \sum_{k=1}^{n}\Bigl{(}\int_{\mathbf{R}^{n}}\bigl{\lvert}\frac{\partial u}{% \partial x_{k}}\bigr{\rvert}\,dx\Bigr{)}\Bigr{)}^{n/(n-1)}.$

One finally concludes

 $\lVert u\rVert_{L^{n/(n-1)}}\leq n^{1/2-n/(n-1)}\lVert\nabla u\rVert_{L^{n/(n-% 1)}}.$

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

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

 $\lVert u_{m}-u_{\ell}\rVert_{L^{n/(n-1)}}\leq n^{1/2-n/(n-1)}\lVert\nabla u_{m% }-\nabla u_{\ell}\rVert_{L^{1}}.$

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

 $\lVert u\rVert_{L^{n/(n-1)}}\leq n^{1/2-n/(n-1)}\lVert\nabla u\rVert_{L^{n/(n-% 1)}}.$

## Step 3: $1 and $u$ is smooth

Suppose $1 and $u$ is a smooth compactly supported function. Let

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

and

 $v=\lvert u\rvert^{r}.$

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

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

One has, by the Hölder inequality,

 $\rVert\nabla v\lVert_{L^{1}(\mathbf{R}^{N})}\leq r\rVert\lvert u\rvert^{r}% \lVert_{L^{p/p-1}(\mathbf{R}^{N})}\rVert\nabla u\lVert_{L^{p}(\mathbf{R}^{N})}% =r\rVert u\lVert_{L^{np/(n-p)}(\mathbf{R}^{N})}^{r-1}\rVert\nabla u\lVert_{L^{% p}(\mathbf{R}^{N})}$

Therefore, the Sobolev inequality yields

 $\rVert u\lVert_{L^{np/(n-p)}(\mathbf{R}^{N})}^{r}=\rVert v\lVert_{L^{n/(n-1)}(% \mathbf{R}^{N})}\leq rn^{1/2-n/(n-1)}\rVert u\lVert_{L^{np/(n-p)}(\mathbf{R}^{% N})}^{r-1}\rVert\nabla u\lVert_{L^{p}(\mathbf{R}^{N})}.$

This yields

 $\rVert u\lVert_{L^{np/(n-p)}(\mathbf{R}^{N})}\leq n^{1/2-n/(n-1)}\frac{p(n-1)}% {n-p}\rVert\nabla u\lVert_{L^{p}(\mathbf{R}^{N})}.$

## Step 4: $1 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].

## References

Title proof of Sobolev inequality for $\Omega=\mathbf{R}^{n}$ ProofOfSobolevInequalityForOmegamathbfRn 2013-03-22 15:05:22 2013-03-22 15:05:22 vanschaf (8061) vanschaf (8061) 14 vanschaf (8061) Proof msc 46E35