Test functions, distributions, and Sobolev’s lemma

1 Introduction

If $X$ is a topological vector space, we denote by $X^{*}$ the set of continuous linear functionals on $X$. With the weak-* topology, $X^{*}$ is a locally convex space, whether or not $X$ is a locally convex space. (But in this note, we only talk about locally convex spaces.)

The purpose of this note is to collect the material given in Walter Rudin, Functional Analysis, second ed., chapters 6 and 7, involved in stating and proving Sobolev’s lemma.

2 Test functions

Suppose that $\mathrm{\Omega }$ is an open subset of ${ℝ}^n$. We denote by $𝒟(\mathrm{\Omega })$ the set of all $\varphi \in C^{\mathrm{\infty }}(\mathrm{\Omega })$ such that $\mathrm{supp}\varphi$ is a compact subset of $\mathrm{\Omega }$. Elements of $𝒟(\mathrm{\Omega })$ are called test functions. For $N=0,1,\mathrm{\dots }$ and $\varphi \in 𝒟(\mathrm{\Omega })$, write

$$\parallel \varphi {\parallel }_N=sup\{|(D^{\alpha }\varphi )(x)|:x\in \mathrm{\Omega },|\alpha |\le N\},$$

where

$$D^{\alpha }=D_1^{{\alpha }_1}\mathrm{\cdots }{D}_n^{{\alpha }_n},|\alpha |={\alpha }_1+\mathrm{\cdots }+{\alpha }_n.$$

For each compact subset $K$ of $\mathrm{\Omega }$, we define

$${𝒟}_K=\{\varphi \in 𝒟(\mathrm{\Omega }):\mathrm{supp}\varphi \subseteq K\},$$

and define ${\tau }_K$ to be the locally convex topology on ${𝒟}_K$ determined by the family of seminorms $\{\parallel \cdot {\parallel }_N:N\ge 0\}$. One proves that ${𝒟}_K$ with the topology ${\tau }_K$ is a Fréchet space. As sets,

$$𝒟(\mathrm{\Omega })=\bigcup _K{𝒟}_K.$$

Define $\beta$ to be the collection of all convex balanced subsets $W$ of $𝒟(\mathrm{\Omega })$ such that for every compact subset $K$ of $\mathrm{\Omega }$ we have $W\cap {𝒟}_K\in {\tau }_K$; to say that $W$ is balanced means that if $c$ is a complex number with $|c|\le 1$ then $cW\subseteq W$. One proves that $\{\varphi +W:\varphi \in 𝒟(\mathrm{\Omega }),W\in \beta \}$ is a basis for a topology $\tau$ on $𝒟(\mathrm{\Omega })$, that $\beta$ is a local basis at $0$ for this topology, and that with the topology $\tau$, $𝒟(\mathrm{\Omega })$ is a locally convex space. For each compact subset $K$ of $\mathrm{\Omega }$, one proves that the topology ${\tau }_K$ is equal to the subspace topology on ${𝒟}_K$ inherited from $𝒟(\mathrm{\Omega })$.

We write ${𝒟}^{\prime }(\mathrm{\Omega })={(𝒟(\mathrm{\Omega }))}^{*}$, and elements of ${𝒟}^{\prime }(\mathrm{\Omega })$ are called distributions. With the weak-* topology, ${𝒟}^{\prime }(\mathrm{\Omega })$ is a locally convex space.

It is a fact that a linear functional $\mathrm{\Lambda }$ on $𝒟(\mathrm{\Omega })$ is continuous if and only if for every compact subset $K$ of $\mathrm{\Omega }$ there is a nonnegative integer $N$ and a constant $C$ such that $|\mathrm{\Lambda }\varphi |\le C{\parallel \varphi \parallel }_N$ for all $\varphi \in {𝒟}_K$.

For $\mathrm{\Lambda }\in {𝒟}^{\prime }(\mathrm{\Omega })$ and $\alpha$ a multi-index, we define

$$(D^{\alpha }\mathrm{\Lambda })(\varphi )={(-1)}^{|\alpha |}\mathrm{\Lambda }(D^{\alpha }\varphi ),\varphi \in 𝒟(\mathrm{\Omega }).$$

Let $K$ be a compact subset of $\mathrm{\Omega }$. As $\mathrm{\Lambda }$ is continuous, there is a nonnegative integer $N$ and a constant $C$ such that $|\mathrm{\Lambda }\varphi |\le C{\parallel \varphi \parallel }_N$ for all $\varphi \in {𝒟}_K$. Then

$$|(D^{\alpha }\mathrm{\Lambda })(\varphi )|=|\mathrm{\Lambda }(D^{\alpha }\varphi )|\le C{\parallel D^{\alpha }\varphi \parallel }_N\le C{\parallel \varphi \parallel }_{N+|\alpha |},$$

which shows that $D^{\alpha }\mathrm{\Lambda }\in {𝒟}^{\prime }(\mathrm{\Omega })$.

The Leibniz formula is the statement that for all $f,g\in C^{\mathrm{\infty }}({ℝ}^n)$,

$$D^{\alpha }(fg)=\sum _{\beta \le \alpha }\left(\genfrac{}{}{0pt}{}{\alpha }{\beta }\right)(D^{\alpha -\beta }f)(D^{\beta }g),$$

where $\left(\genfrac{}{}{0pt}{}{\alpha }{\beta }\right)$ are multinomial coefficients.

For $\mathrm{\Lambda }\in {𝒟}^{\prime }(\mathrm{\Omega })$ and $f\in C^{\mathrm{\infty }}(\mathrm{\Omega })$, we define

$$(f\mathrm{\Lambda })(\varphi )=\mathrm{\Lambda }(f\varphi ),\varphi \in 𝒟(\mathrm{\Omega });$$

this makes sense because $f\varphi \in 𝒟(\mathrm{\Omega })$ when $\varphi \in 𝒟(\mathrm{\Omega })$. It is apparent that $f\mathrm{\Lambda }$ is linear, and in the following lemma we prove that $f\mathrm{\Lambda }$ is continuous.

Lemma 1.

If $\mathrm{\Lambda }\mathrm{\in }{\mathrm{D}}^{\mathrm{\prime }}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$ and $f\mathrm{\in }{C}^{\mathrm{\infty }}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$, then $f\mathit{}\mathrm{\Lambda }\mathrm{\in }{\mathrm{D}}^{\mathrm{\prime }}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$.

Proof.

Suppose that $K$ is a compact subset of $\mathrm{\Omega }$. Because $\mathrm{\Lambda }$ is continuous, there is some nonnegative integer $N$ and some constant $C$ such that

$$|\mathrm{\Lambda }\varphi |\le C{\parallel \varphi \parallel }_N,\varphi \in {𝒟}_K.$$

For $|\alpha |\le N$, by the Leibniz formula, for all $\varphi \in {𝒟}_K$,

$$D^{\alpha }(f\varphi )=\sum _{\beta \le \alpha }\left(\genfrac{}{}{0pt}{}{\alpha }{\beta }\right)(D^{\alpha -\beta }f)(D^{\beta }\varphi ).$$

Because $f\in C^{\mathrm{\infty }}(\mathrm{\Omega })$, there is some $C_{\alpha }$ such that $|(D^{\alpha -\beta }f)(x)|\le C_{\alpha }$ for $\beta \le \alpha$ and for $x\in K$. Using $\varphi (x)=0$ for $x\notin K$, the above statement of the Leibniz formula, and the inequality just obtained, it follows that there is some $C_{\alpha }^{\prime }$ such that $|(D^{\alpha }(f\varphi ))(x)|\le C_{\alpha }^{\prime }{\parallel \varphi \parallel }_N$ for all $x\in \mathrm{\Omega }$. This gives

$${\parallel f\varphi \parallel }_N=\underset{|\alpha |\le N}{sup}\underset{x\in \mathrm{\Omega }}{sup}|(D^{\alpha }(f\varphi ))(x)|\le \underset{|\alpha |\le N}{sup}{C}_{\alpha }^{\prime }{\parallel \varphi \parallel }_N=C^{\prime }{\parallel \varphi \parallel }_N;$$

the last equality is how we define $C^{\prime }$, which is a maximum of finitely many $C_{\alpha }^{\prime }$ and so finite. Then,

$$|(f\mathrm{\Lambda })(\varphi )|=|\mathrm{\Lambda }(f\varphi )|\le C{\parallel f\varphi \parallel }_N\le C{C}^{\prime }{\parallel \varphi \parallel }_N,\varphi \in {𝒟}_K.$$

This bound shows that $f\mathrm{\Lambda }$ is continuous. ∎

The above lemma shows that $f\mathrm{\Lambda }\in {𝒟}^{\prime }(\mathrm{\Omega })$ when $f\in C^{\mathrm{\infty }}(\mathrm{\Omega })$ and $\mathrm{\Lambda }\in {𝒟}^{\prime }(\mathrm{\Omega })$. Therefore $D^{\alpha }(f\mathrm{\Lambda })\in 𝒟(\mathrm{\Omega })$, and the following lemma, proved in Rudin, states that the Leibniz formula can be used with $f\mathrm{\Lambda }$.

Lemma 2.

If $f\mathrm{\in }{C}^{\mathrm{\infty }}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$ and $\mathrm{\Lambda }\mathrm{\in }{\mathrm{D}}^{\mathrm{\prime }}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$, then

$$D^{\alpha }(f\mathrm{\Lambda })=\sum _{\beta \le \alpha }\left(\genfrac{}{}{0pt}{}{\alpha }{\beta }\right)(D^{\alpha -\beta }f)(D^{\beta }\mathrm{\Lambda }).$$

If $f:\mathrm{\Omega }\to ℂ$ is locally integrable, define

$$\mathrm{\Lambda }\varphi ={\int }_{\mathrm{\Omega }}\varphi (x)f(x)𝑑x,\varphi \in 𝒟(\mathrm{\Omega }).$$

For $\varphi \in {𝒟}_K$,

$$|\mathrm{\Lambda }\varphi |\le {\parallel \varphi \parallel }_0{\int }_K|f|𝑑x,$$

from which it follows that $\mathrm{\Lambda }$ is continuous. If $\mu$ is a complex Borel measure on ${ℝ}^n$ or a positive Borel measure on ${ℝ}^n$ that assigns finite measure to compact sets, define

$$\mathrm{\Lambda }\varphi ={\int }_{\mathrm{\Omega }}\varphi 𝑑\mu ,\varphi \in 𝒟(\mathrm{\Omega }).$$

For $\varphi \in {𝒟}_K$,

$$|\mathrm{\Lambda }\varphi |\le {\parallel \varphi \parallel }_0|\mu |(K),$$

from which it follows that $\mathrm{\Lambda }$ is continuous. Thus, we can encode certain functions and measures as distributions. I will dare to say that we can encode most functions and measures that we care about as distributions.

If ${\mathrm{\Lambda }}_1,{\mathrm{\Lambda }}_2\in {𝒟}^{\prime }(\mathrm{\Omega })$ and $\omega$ is an open subset of $\mathrm{\Omega }$, we say that ${\mathrm{\Lambda }}_1={\mathrm{\Lambda }}_2$ in $\omega$ if ${\mathrm{\Lambda }}_1\varphi ={\mathrm{\Lambda }}_2\varphi$ for all $\varphi \in 𝒟(\omega )$.

Let $\mathrm{\Lambda }\in {𝒟}^{\prime }(\mathrm{\Omega })$ and let $\omega$ be an open subset of $\mathrm{\Omega }$. We say that $\mathrm{\Lambda }$ vanishes on $\omega$ if $\mathrm{\Lambda }\varphi =0$ for all $\varphi \in 𝒟(\omega )$. Taking $W$ to be the union of all open subsets $\omega$ of $\mathrm{\Omega }$ on which $\mathrm{\Lambda }$ vanishes, we define the support of $\mathrm{\Lambda }$ to be the set $\mathrm{\Omega }\setminus W$.

3 The Fourier transform

Let $C_0({ℝ}^n)$ be the set of those continuous functions $f:{ℝ}^n\to ℂ$ such that for every $ϵ>0$, there is some compact set $K$ such that for $x\notin K$. With the supremum norm $\parallel \cdot {\parallel }_{\mathrm{\infty }}$, $C_0({ℝ}^n)$ is a Banach space.

Let $m_n$ be normalized Lebesgue measure on ${\mathrm{R}}^n$:

$$d{m}_n(x)={(2\pi )}^{-n/2}dx.$$

Using $m_n$, we define

and

$$(f*g)(x)={\int }_{{ℝ}^n}f(x-y)g(y)𝑑m_n(y).$$

For $t\in {ℝ}^n$, define $e_t:{ℝ}^n\to ℂ$ by

$$e_t(x)=\exp (it\cdot x),x\in {ℝ}^n.$$

The Fourier transform of $f\in L^1({ℝ}^n)$ is the function $\widehat{f}:{ℝ}^n\to ℂ$ defined by

$$(ℱf)(t)=\widehat{f}(t)={\int }_{{ℝ}^n}f{e}_{-t}𝑑m_n,t\in {ℝ}^n.$$

Using the dominated convergence theorem, one shows that $\widehat{f}$ is continuous.

For $f\in C^{\mathrm{\infty }}({ℝ}^n)$ and $N$ a nonnegative integer, write

$$p_N(f)=\underset{|\alpha |\le N}{sup}\underset{x\in {ℝ}^n}{sup}{(1+{|x|}^2)}^N|(D^{\alpha }f)(x)|,$$

and let ${𝒮}_n$ be the set of those $f\in C^{\mathrm{\infty }}({ℝ}^n)$ such that for every nonnegative integer $N$, . ${𝒮}_n$ is a vector space, and with the locally convex topology determined by the family of seminorms $\{p_N:N\ge 0\}$ it is a Fréchet space. Further, one proves that $ℱ:{𝒮}_n\to {𝒮}_n$ is a continuous linear map.

The Riemann-Lebesgue lemma is the statement that if $f\in L^1({ℝ}^n)$, then $\widehat{f}\in C_0({ℝ}^n)$.

The inversion theorem is the statement that if $g\in {𝒮}_n$ then

$$g(x)={\int }_{{ℝ}^n}\widehat{g}{e}_x𝑑m_n,x\in {ℝ}^n,$$

and that if $f\in L^1({ℝ}^n)$ and $\widehat{f}\in L^1({ℝ}^n)$, and we define $f_0\in C_0({ℝ}^n)$ by

$$f_0(x)={\int }_{{ℝ}^n}\widehat{f}{e}_x𝑑m_n,x\in {ℝ}^n,$$

then $f(x)=f_0(x)$ for almost all $x\in {ℝ}^n$. For $g\in {𝒮}_n$, as $\widehat{g}\in {𝒮}_n$, the function $f(t)=\widehat{g}(-t)$ belongs to ${𝒮}_n$. The inversion theorem tells us that for all $x\in {ℝ}^n$,

$$g(x)={\int }_{{ℝ}^n}\widehat{g}(t)e_x(t)𝑑m_n(t)={\int }_{{ℝ}^n}\widehat{g}(-t)e_x(-t)𝑑m_n(t)={\int }_{{ℝ}^n}f(t)e_{-x}(t)𝑑m_n(t),$$

and hence that $g=\widehat{f}$. This shows that $ℱ:{𝒮}_n\to {𝒮}_n$ is onto. Using the inversion theorem, one checks that

$${\int }_{{ℝ}^n}f\overline{g}𝑑m_n={\int }_{{ℝ}^n}\widehat{f}\overline{\widehat{g}}𝑑m_n,f,g\in {𝒮}_n,$$

and so ${\parallel f\parallel }_{L^2}={\parallel ℱf\parallel }_{L^2}$ for $f\in {𝒮}_n$. It is a fact that ${𝒮}_n$ is a dense subset of the Hilbert space $L^2({ℝ}^n)$, and it follows that there is a unique bounded linear operator $L^2({ℝ}^n)\to L^2({ℝ}^n)$, that is equal to $ℱ$ on ${𝒮}_n$, and that is unitary. We denote this $ℱ:L^2({ℝ}^n)\to L^2({ℝ}^n)$.

It is a fact that $𝒟({ℝ}^n)$ is a dense subset of ${𝒮}_n$ and that the identity map $i:𝒟({ℝ}^n)\to {𝒮}_n$ is continuous. If $L_1,L_2\in {({𝒮}_n)}^{*}$ are distinct, then there is some $f\in {𝒮}_n$ such that $L_{1}f\ne L_{2}f$, and as $𝒟({ℝ}^n)$ is dense in ${𝒮}_n$, there is a sequence $f_j\in 𝒟({ℝ}^n)$ with $f_j\to f$ in ${𝒮}_n$. As

$$(L_1\circ i)(f_j)-(L_2\circ i)(f_j)=L_1{f}_j-L_2{f}_j\to L_1{f}_j-L_2{f}_j\ne 0,$$

there is some $f_j$ with $(L_1\circ i)(f_j)\ne (L_2\circ i)(f_j)$, and hence $L_1\circ i\ne L_2\circ i$. This shows that $L\mapsto L\circ i$ is a one-to-one linear map ${({𝒮}_n)}^{*}\to {𝒟}^{\prime }({ℝ}^n)$. Elements of ${𝒟}^{\prime }({ℝ}^n)$ of the form $L\circ i$ for $L\in {({𝒮}_n)}^{*}$ are called tempered distributions, and we denote the set of tempered distributions by ${𝒮}_n^{\prime }$. It is a fact that every distribution with compact support is tempered.

4 Sobolev’s lemma

Suppose that $\mathrm{\Omega }$ is an open subset of ${ℝ}^n$. We say that a measurable function $f:\mathrm{\Omega }\to ℂ$ is locally $L^{\mathrm{2}}$ if for every compact subset $K$ of $\mathrm{\Omega }$. We say that $\mathrm{\Lambda }\in {𝒟}^{\prime }(\mathrm{\Omega })$ is locally $L^{\mathrm{2}}$ if there is a function $g$ that is locally $L^2$ in $\mathrm{\Omega }$ such that $\mathrm{\Lambda }\varphi ={\int }_{\mathrm{\Omega }}\varphi g𝑑m_n$ for every $\varphi \in 𝒟(\mathrm{\Omega })$.

The following proof of Sobolev’s lemma follows Rudin.

Theorem 3 (Sobolev’s lemma).

Suppose that $n\mathrm{,}p\mathrm{,}r$ are integers, $n\mathrm{>}\mathrm{0}$, $p\mathrm{\ge }\mathrm{0}$, and

$$r>p+\frac{n}{2}.$$

Suppose that $\mathrm{\Omega }$ is an open subset of ${\mathrm{R}}^n$, that $f\mathrm{:}\mathrm{\Omega }\mathrm{\to }\mathrm{C}$ is locally $L^{\mathrm{2}}$, and that the distribution derivatives $D_j^k\mathit{}f$ are locally $L^{\mathrm{2}}$ for $\mathrm{1}\mathrm{\le }j\mathrm{\le }n$, $\mathrm{1}\mathrm{\le }k\mathrm{\le }r$. Then there is some $f_{\mathrm{0}}\mathrm{\in }{C}^p\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$ such that $f_{\mathrm{0}}\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}f\mathit{}\mathrm{(}x\mathrm{)}$ for almost all $x\mathrm{\in }\mathrm{\Omega }$.

Proof.

To say that the distribution derivative $D_j^{k}f$ is locally $L^2$ means that there is some $g_{j,k}:\mathrm{\Omega }\to ℂ$ that is locally $L^2$ such that

$$D_j^k{\mathrm{\Lambda }}_f={\mathrm{\Lambda }}_{g_{j,k}}.$$

Suppose that $\omega$ is an open subset of $\mathrm{\Omega }$ whose closure $K$ is a compact subset of $\mathrm{\Omega }$. There is some $\psi \in 𝒟(\mathrm{\Omega })$ with $\psi (x)=1$ for $x\in K$, and we define $F:{ℝ}^n\to ℂ$ by

in particular, for $x\in K$ we have $F(x)=f(x)$, and for $x\notin \mathrm{supp}\psi$ we have $F(x)=0$. Because $\mathrm{supp}\psi \subset \mathrm{\Omega }$ is compact and $f$ is locally $L^2$,

and using the Cauchy-Schwarz inequality, , so

$$F\in L^2({ℝ}^n)\cap L^1({ℝ}^n).$$

Then,

(1)

Because ${\mathrm{\Lambda }}_F=\psi {\mathrm{\Lambda }}_f$ in $\mathrm{\Omega }$, the Leibniz formula tells us that in $\mathrm{\Omega }$,

$$D_j^r{\mathrm{\Lambda }}_F=D_j^r(\psi {\mathrm{\Lambda }}_f)=\sum _{s=0}^r\left(\genfrac{}{}{0pt}{}{r}{s}\right)(D_j^{r-s}\psi )(D_j^s{\mathrm{\Lambda }}_f)=\sum _{s=0}^r\left(\genfrac{}{}{0pt}{}{r}{s}\right)(D_j^{r-s}\psi )({\mathrm{\Lambda }}_{g_{j,s}}),$$

hence, defining $H_j:{ℝ}^n\to ℂ$ by

we have $D_j^r{\mathrm{\Lambda }}_F={\mathrm{\Lambda }}_{H_j}$ in $\mathrm{\Omega }$. It is apparent that $H_j\in L^2({ℝ}^n)\cap L^1({ℝ}^n)$.

Let $\varphi \in 𝒟({ℝ}^n)$. There are ${\varphi }_1,{\varphi }_2\in 𝒟({ℝ}^n)$ with $\varphi ={\varphi }_1+{\varphi }_2$ and $\mathrm{supp}{\varphi }_1\subset \mathrm{\Omega }$, $\mathrm{supp}{\varphi }_2\subset {ℝ}^n\setminus \mathrm{supp}\psi$. We have just established that $(D_j^r{\mathrm{\Lambda }}_F){\varphi }_1={\mathrm{\Lambda }}_{H_j}{\varphi }_1$. For ${\varphi }_2$, it is apparent that

$$(D_j^r{\mathrm{\Lambda }}_F){\varphi }_2={\mathrm{\Lambda }}_F(D_j^r{\varphi }_2)={\int }_{{ℝ}^n}(D_j^r{\varphi }_2)(x)F(x)𝑑m_n(x)=0$$

and

$${\mathrm{\Lambda }}_{H_j}{\varphi }_2={\int }_{{ℝ}^n}{\varphi }_2(x)H_j(x)𝑑m_n(x)=0.$$

Hence $(D_j^r{\mathrm{\Lambda }}_F)(\varphi )={\mathrm{\Lambda }}_{H_j}\varphi$. It is apparent that ${\mathrm{\Lambda }}_{H_j}$ has compact support, so $D_j^r{\mathrm{\Lambda }}_F={\mathrm{\Lambda }}_{H_j}$ are tempered distributions. Let $\xi \in {𝒮}_n$, and take $\varphi \in {𝒮}_n$ with $\xi =\widehat{\varphi }$. Then,

$(D_j^r{\mathrm{\Lambda }}_F)\varphi$ $=$ ${\mathrm{\Lambda }}_F{D}_j^r\varphi$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}(D_j^r\varphi )(x)F(x)𝑑m_n(x)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}ℱ(D_j^r\varphi )(y)\widehat{F}(y)𝑑m_n(y)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}{(i{y}_j)}^r\xi (y)\widehat{F}(y)𝑑m_n(y),$

and

$${\mathrm{\Lambda }}_{H_j}\varphi ={\int }_{{ℝ}^n}\varphi (x)H_j(x)𝑑m_n(x)={\int }_{{ℝ}^n}\xi (y)\widehat{H_j}(y)𝑑m_n(y).$$

It follows that ${(i{y}_j)}^r\widehat{F}(y)=\widehat{H_j}(y)$ for all $y\in {ℝ}^n$. But $\widehat{H_j}\in L^2({ℝ}^n)$, so

(2)

Using (1), (2), and the inequality

we get

Let ${\sigma }_{n-1}$ be surface measure on $S^{n-1}$, with ${\sigma }_{n-1}(S^{n-1})=\frac{2{\pi }^{n/2}}{\mathrm{\Gamma }(n/2)}$. Using the Cauchy-Schwarz inequality and the change of variable $y=tu$, $u\in S^{n-1}$, $t\ge 0$,

${\left({\displaystyle {\int }_{{ℝ}^n}}{(1+|y|)}^p|\widehat{F}(y)|𝑑m_n(y)\right)}^2$ $=$ ${\left({\displaystyle {\int }_{{ℝ}^n}}{(1+|y|)}^r|\widehat{F}(y)|{(1+|y|)}^{p-r}𝑑m_n(y)\right)}^2$

$\le$ $J{\displaystyle {\int }_{{ℝ}^n}}{(1+|y|)}^{2p-2r}𝑑m_n(y)$

$=$ $J{(2\pi )}^{-n/2}{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_{S^{n-1}}}{(1+t)}^{2p-2r}{t}^{n-1}𝑑{\sigma }_{n-1}(u)𝑑t$

$=$ ${\displaystyle \frac{2J}{\mathrm{\Gamma }(n/2)}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{(1+t)}^{2p-2r}{t}^{n-1}𝑑t.$

This integral is finite if and only if , and we have assumed that $r>p+\frac{n}{2}$. Therefore,

from which we get that $y^{\alpha }\widehat{F}(y)$ is in $L^1({ℝ}^n)$ for $|\alpha |\le p$.

Define

$$F_{\omega }(x)={\int }_{{ℝ}^n}\widehat{F}{e}_x𝑑m_n,x\in {ℝ}^n.$$

(Note that $F$ depends on $\omega$.) $F,\widehat{F}\in L^1({ℝ}^n)$ so by the inversion theorem we have $F(x)=F_{\omega }(x)$ for almost all $x\in {ℝ}^n$. $F_{\omega }\in C_0({ℝ}^n)$. If $p\ge 1$, then we shall show that $F_{\omega }\in C^p(\mathrm{\Omega })$. Take $e_k$ to be the standard basis for ${ℝ}^n$. For $1\le k_1\le n$ and $ϵ\ne 0$,

${\displaystyle \frac{F_{\omega }(x+ϵe_{k_1})-F_{\omega }(x)}{ϵ}}$ $=$ ${\displaystyle \frac{1}{ϵ}}{\displaystyle {\int }_{{ℝ}^n}}\widehat{F}(y)\left(\exp (iϵe_{k_1}\cdot y)-1\right)\exp (ix\cdot y)𝑑m_n(y)$

$=$ ${\displaystyle {\int }_{{ℝ}^n}}i{y}_{k_1}\widehat{F}(y){\displaystyle \frac{e^{iϵy_{k_1}}-1}{iϵy_k}}{e}_x(y)𝑑m_n(y).$

But $\left|i{y}_{k_1}\widehat{F}(y)\frac{e^{iϵy_{k_1}}-1}{iϵy_{k_1}}{e}_x(y)\right|\le |y_{k_1}\widehat{F}(y)|$ and $y_{k_1}\widehat{F}(y)$ belongs to $L^1({ℝ}^n)$ (supposing $p\ge 1$) so we can apply the dominated convergence theorem, which gives us

$$(D_{k_1}{F}_{\omega })(x)=\underset{ϵ\to 0}{lim}\frac{F_{\omega }(x+ϵe_{k_1})-F_{\omega }(x)}{ϵ}={\int }_{{ℝ}^n}i{y}_{k_1}\widehat{F}(y)e_x(y)𝑑m_n(y).$$

From the above expression, it is apparent that $D_{k_1}{F}_{\omega }$ is continuous. This is true for all $1\le k_1\le n$, so $F_{\omega }\in C^1({ℝ}^n)$. If $p\ge 2$, then $y_{k_1}{y}_{k_2}\widehat{F}(y)$ is in $L^1({ℝ}^n)$ for any $1\le k_2\le n$, and repeating the above argument we get $F_{\omega }\in C^2({ℝ}^n)$. In this way, $F_{\omega }\in C^p({ℝ}^n)$.

For all $x\in \omega$, $f(x)=F(x)$, so $f(x)=F_{\omega }(x)$ for almost all $x\in \omega$. If ${\omega }^{\prime }$ is an open subset of $\mathrm{\Omega }$ whose closure is a compact subset of $\mathrm{\Omega }$ and $\omega \cap {\omega }^{\prime }\ne \mathrm{\varnothing }$, then $F_{\omega },F_{{\omega }^{\prime }}\in C^p({ℝ}^n)$ satisfy $f(x)=F_{\omega }(x)$ for almost all $x\in \omega$ and $f(x)=F_{{\omega }^{\prime }}(x)$ for almost all $x\in {\omega }^{\prime }$, so $F_{\omega }(x)=F_{{\omega }^{\prime }}(x)$ for almost all $x\in \omega \cap {\omega }^{\prime }$. Since $F_{\omega },F_{{\omega }^{\prime }}$ are continuous, this implies that $F_{\omega }(x)=F_{{\omega }^{\prime }}(x)$ for all $x\in \omega \cap {\omega }^{\prime }$. Thus, it makes sense to define $f_0(x)=F_{\omega }(x)$ for $x\in \omega$. Because every point in $\mathrm{\Omega }$ has an open neighborhood of the kind $\omega$ and the restriction of $f_0$ to each $\omega$ belongs to $C^p(\omega )$, it follows that $f_0\in C^p(\mathrm{\Omega })$. ∎