# Test functions, distributions, and Sobolev’s lemma

Jordan Bell
May 22, 2014

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

 $\left\|\phi\right\|_{N}=\sup\{|(D^{\alpha}\phi)(x)|:x\in\Omega,|\alpha|\leq N\},$

where

 $D^{\alpha}=D_{1}^{\alpha_{1}}\cdots D_{n}^{\alpha_{n}},\qquad|\alpha|=\alpha_{% 1}+\cdots+\alpha_{n}.$

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

 $\mathscr{D}_{K}=\{\phi\in\mathscr{D}(\Omega):\mathrm{supp}\,\phi\subseteq K\},$

and define $\tau_{K}$ to be the locally convex topology on $\mathscr{D}_{K}$ determined by the family of seminorms $\{\left\|\cdot\right\|_{N}:N\geq 0\}$. One proves that $\mathscr{D}_{K}$ with the topology $\tau_{K}$ is a Fréchet space. As sets,

 $\mathscr{D}(\Omega)=\bigcup_{K}\mathscr{D}_{K}.$

Define $\beta$ to be the collection of all convex balanced subsets $W$ of $\mathscr{D}(\Omega)$ such that for every compact subset $K$ of $\Omega$ we have $W\cap\mathscr{D}_{K}\in\tau_{K}$; to say that $W$ is balanced means that if $c$ is a complex number with $|c|\leq 1$ then $cW\subseteq W$. One proves that $\{\phi+W:\phi\in\mathscr{D}(\Omega),W\in\beta\}$ is a basis for a topology $\tau$ on $\mathscr{D}(\Omega)$, that $\beta$ is a local basis at $0$ for this topology, and that with the topology $\tau$, $\mathscr{D}(\Omega)$ is a locally convex space.11 1 Walter Rudin, Functional Analysis, second ed., p. 152, Theorem 6.4; cf. Helmut H. Schaefer, Topological Vector Spaces, p. 57. For each compact subset $K$ of $\Omega$, one proves that the topology $\tau_{K}$ is equal to the subspace topology on $\mathscr{D}_{K}$ inherited from $\mathscr{D}(\Omega)$.22 2 Walter Rudin, Functional Analysis, second ed., p. 153, Theorem 6.5.

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

It is a fact that a linear functional $\Lambda$ on $\mathscr{D}(\Omega)$ is continuous if and only if for every compact subset $K$ of $\Omega$ there is a nonnegative integer $N$ and a constant $C$ such that $|\Lambda\phi|\leq C\left\|\phi\right\|_{N}$ for all $\phi\in\mathscr{D}_{K}$.33 3 Walter Rudin, Functional Analysis, second ed., p. 156, Theorem 6.8.

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

 $(D^{\alpha}\Lambda)(\phi)=(-1)^{|\alpha|}\Lambda(D^{\alpha}\phi),\qquad\phi\in% \mathscr{D}(\Omega).$

Let $K$ be a compact subset of $\Omega$. As $\Lambda$ is continuous, there is a nonnegative integer $N$ and a constant $C$ such that $|\Lambda\phi|\leq C\left\|\phi\right\|_{N}$ for all $\phi\in\mathscr{D}_{K}$. Then

 $|(D^{\alpha}\Lambda)(\phi)|=|\Lambda(D^{\alpha}\phi)|\leq C\left\|D^{\alpha}% \phi\right\|_{N}\leq C\left\|\phi\right\|_{N+|\alpha|},$

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

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

 $D^{\alpha}(fg)=\sum_{\beta\leq\alpha}\binom{\alpha}{\beta}(D^{\alpha-\beta}f)(% D^{\beta}g),$

where $\binom{\alpha}{\beta}$ are multinomial coefficients.

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

 $(f\Lambda)(\phi)=\Lambda(f\phi),\qquad\phi\in\mathscr{D}(\Omega);$

this makes sense because $f\phi\in\mathscr{D}(\Omega)$ when $\phi\in\mathscr{D}(\Omega)$. It is apparent that $f\Lambda$ is linear, and in the following lemma we prove that $f\Lambda$ is continuous.44 4 Walter Rudin, Functional Analysis, second ed., p. 159, §6.15.

###### Lemma 1.

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

###### Proof.

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

 $|\Lambda\phi|\leq C\left\|\phi\right\|_{N},\qquad\phi\in\mathscr{D}_{K}.$

For $|\alpha|\leq N$, by the Leibniz formula, for all $\phi\in\mathscr{D}_{K}$,

 $D^{\alpha}(f\phi)=\sum_{\beta\leq\alpha}\binom{\alpha}{\beta}(D^{\alpha-\beta}% f)(D^{\beta}\phi).$

Because $f\in C^{\infty}(\Omega)$, there is some $C_{\alpha}$ such that $|(D^{\alpha-\beta}f)(x)|\leq C_{\alpha}$ for $\beta\leq\alpha$ and for $x\in K$. Using $\phi(x)=0$ for $x\not\in K$, the above statement of the Leibniz formula, and the inequality just obtained, it follows that there is some $C_{\alpha}^{\prime}$ such that $|(D^{\alpha}(f\phi))(x)|\leq C_{\alpha}^{\prime}\left\|\phi\right\|_{N}$ for all $x\in\Omega$. This gives

 $\left\|f\phi\right\|_{N}=\sup_{|\alpha|\leq N}\sup_{x\in\Omega}|(D^{\alpha}(f% \phi))(x)|\leq\sup_{|\alpha|\leq N}C_{\alpha}^{\prime}\left\|\phi\right\|_{N}=% C^{\prime}\left\|\phi\right\|_{N};$

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

 $|(f\Lambda)(\phi)|=|\Lambda(f\phi)|\leq C\left\|f\phi\right\|_{N}\leq CC^{% \prime}\left\|\phi\right\|_{N},\qquad\phi\in\mathscr{D}_{K}.$

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

The above lemma shows that $f\Lambda\in\mathscr{D}^{\prime}(\Omega)$ when $f\in C^{\infty}(\Omega)$ and $\Lambda\in\mathscr{D}^{\prime}(\Omega)$. Therefore $D^{\alpha}(f\Lambda)\in\mathscr{D}(\Omega)$, and the following lemma, proved in Rudin, states that the Leibniz formula can be used with $f\Lambda$.55 5 Walter Rudin, Functional Analysis, second ed., p. 160, §6.15.

###### Lemma 2.

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

 $D^{\alpha}(f\Lambda)=\sum_{\beta\leq\alpha}\binom{\alpha}{\beta}(D^{\alpha-% \beta}f)(D^{\beta}\Lambda).$

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

 $\Lambda\phi=\int_{\Omega}\phi(x)f(x)dx,\qquad\phi\in\mathscr{D}(\Omega).$

For $\phi\in\mathscr{D}_{K}$,

 $|\Lambda\phi|\leq\left\|\phi\right\|_{0}\int_{K}|f|dx,$

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

 $\Lambda\phi=\int_{\Omega}\phi d\mu,\qquad\phi\in\mathscr{D}(\Omega).$

For $\phi\in\mathscr{D}_{K}$,

 $|\Lambda\phi|\leq\left\|\phi\right\|_{0}|\mu|(K),$

from which it follows that $\Lambda$ 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 $\Lambda_{1},\Lambda_{2}\in\mathscr{D}^{\prime}(\Omega)$ and $\omega$ is an open subset of $\Omega$, we say that $\Lambda_{1}=\Lambda_{2}$ in $\omega$ if $\Lambda_{1}\phi=\Lambda_{2}\phi$ for all $\phi\in\mathscr{D}(\omega)$.

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

## 3 The Fourier transform

Let $C_{0}(\mathbb{R}^{n})$ be the set of those continuous functions $f:\mathbb{R}^{n}\to\mathbb{C}$ such that for every $\epsilon>0$, there is some compact set $K$ such that $|f(x)|<\epsilon$ for $x\not\in K$. With the supremum norm $\left\|\cdot\right\|_{\infty}$, $C_{0}(\mathbb{R}^{n})$ is a Banach space.

Let $m_{n}$ be normalized Lebesgue measure on $\mathbb{R}^{n}$:

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

Using $m_{n}$, we define

 $\left\|f\right\|_{L^{p}}=\left(\int_{\mathbb{R}^{n}}|f|^{p}dm_{n}\right)^{1/p}% ,\qquad 1\leq p<\infty$

and

 $(f*g)(x)=\int_{\mathbb{R}^{n}}f(x-y)g(y)dm_{n}(y).$

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

 $e_{t}(x)=\exp(it\cdot x),\qquad x\in\mathbb{R}^{n}.$

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

 $(\mathscr{F}f)(t)=\hat{f}(t)=\int_{\mathbb{R}^{n}}fe_{-t}dm_{n},\qquad t\in% \mathbb{R}^{n}.$

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

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

 $p_{N}(f)=\sup_{|\alpha|\leq N}\sup_{x\in\mathbb{R}^{n}}(1+|x|^{2})^{N}|(D^{% \alpha}f)(x)|,$

and let $\mathscr{S}_{n}$ be the set of those $f\in C^{\infty}(\mathbb{R}^{n})$ such that for every nonnegative integer $N$, $p_{N}(f)<\infty$. $\mathscr{S}_{n}$ is a vector space, and with the locally convex topology determined by the family of seminorms $\{p_{N}:N\geq 0\}$ it is a Fréchet space.66 6 Walter Rudin, Functional Analysis, second ed., p. 184, Theorem 7.4. Further, one proves that $\mathscr{F}:\mathscr{S}_{n}\to\mathscr{S}_{n}$ is a continuous linear map.77 7 Walter Rudin, Functional Analysis, second ed., p. 184, Theorem 7.4.

The Riemann-Lebesgue lemma is the statement that if $f\in L^{1}(\mathbb{R}^{n})$, then $\hat{f}\in C_{0}(\mathbb{R}^{n})$.88 8 Walter Rudin, Functional Analysis, second ed., p. 185, Theorem 7.5.

The inversion theorem99 9 Walter Rudin, Functional Analysis, second ed., p. 186, Theorem 7.7. is the statement that if $g\in\mathscr{S}_{n}$ then

 $g(x)=\int_{\mathbb{R}^{n}}\hat{g}e_{x}dm_{n},\qquad x\in\mathbb{R}^{n},$

and that if $f\in L^{1}(\mathbb{R}^{n})$ and $\hat{f}\in L^{1}(\mathbb{R}^{n})$, and we define $f_{0}\in C_{0}(\mathbb{R}^{n})$ by

 $f_{0}(x)=\int_{\mathbb{R}^{n}}\hat{f}e_{x}dm_{n},\qquad x\in\mathbb{R}^{n},$

then $f(x)=f_{0}(x)$ for almost all $x\in\mathbb{R}^{n}$. For $g\in\mathscr{S}_{n}$, as $\hat{g}\in\mathscr{S}_{n}$, the function $f(t)=\hat{g}(-t)$ belongs to $\mathscr{S}_{n}$. The inversion theorem tells us that for all $x\in\mathbb{R}^{n}$,

 $g(x)=\int_{\mathbb{R}^{n}}\hat{g}(t)e_{x}(t)dm_{n}(t)=\int_{\mathbb{R}^{n}}% \hat{g}(-t)e_{x}(-t)dm_{n}(t)=\int_{\mathbb{R}^{n}}f(t)e_{-x}(t)dm_{n}(t),$

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

 $\int_{\mathbb{R}^{n}}f\overline{g}dm_{n}=\int_{\mathbb{R}^{n}}\hat{f}\overline% {\hat{g}}dm_{n},\qquad f,g\in\mathscr{S}_{n},$

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

It is a fact that $\mathscr{D}(\mathbb{R}^{n})$ is a dense subset of $\mathscr{S}_{n}$ and that the identity map $i:\mathscr{D}(\mathbb{R}^{n})\to\mathscr{S}_{n}$ is continuous.1010 10 Walter Rudin, Functional Analysis, second ed., p. 189, Theorem 7.10. If $L_{1},L_{2}\in(\mathscr{S}_{n})^{*}$ are distinct, then there is some $f\in\mathscr{S}_{n}$ such that $L_{1}f\neq L_{2}f$, and as $\mathscr{D}(\mathbb{R}^{n})$ is dense in $\mathscr{S}_{n}$, there is a sequence $f_{j}\in\mathscr{D}(\mathbb{R}^{n})$ with $f_{j}\to f$ in $\mathscr{S}_{n}$. As

 $(L_{1}\circ i)(f_{j})-(L_{2}\circ i)(f_{j})=L_{1}f_{j}-L_{2}f_{j}\to L_{1}f_{j% }-L_{2}f_{j}\neq 0,$

there is some $f_{j}$ with $(L_{1}\circ i)(f_{j})\neq(L_{2}\circ i)(f_{j})$, and hence $L_{1}\circ i\neq L_{2}\circ i$. This shows that $L\mapsto L\circ i$ is a one-to-one linear map $(\mathscr{S}_{n})^{*}\to\mathscr{D}^{\prime}(\mathbb{R}^{n})$. Elements of $\mathscr{D}^{\prime}(\mathbb{R}^{n})$ of the form $L\circ i$ for $L\in(\mathscr{S}_{n})^{*}$ are called tempered distributions, and we denote the set of tempered distributions by $\mathscr{S}_{n}^{\prime}$. It is a fact that every distribution with compact support is tempered.1111 11 Walter Rudin, Functional Analysis, second ed., p. 190, Example 7.12 (a).

## 4 Sobolev’s lemma

Suppose that $\Omega$ is an open subset of $\mathbb{R}^{n}$. We say that a measurable function $f:\Omega\to\mathbb{C}$ is locally $L^{2}$ if $\int_{K}|f|^{2}dm_{n}<\infty$ for every compact subset $K$ of $\Omega$. We say that $\Lambda\in\mathscr{D}^{\prime}(\Omega)$ is locally $L^{2}$ if there is a function $g$ that is locally $L^{2}$ in $\Omega$ such that $\Lambda\phi=\int_{\Omega}\phi gdm_{n}$ for every $\phi\in\mathscr{D}(\Omega)$.

The following proof of Sobolev’s lemma follows Rudin.1212 12 Walter Rudin, Functional Analysis, second ed., p. 202, Theorem 7.25.

###### Theorem 3 (Sobolev’s lemma).

Suppose that $n,p,r$ are integers, $n>0$, $p\geq 0$, and

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

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

###### Proof.

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

 $D_{j}^{k}\Lambda_{f}=\Lambda_{g_{j,k}}.$

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

 $F(x)=\begin{cases}\psi(x)f(x)&x\in\Omega,\\ 0&x\not\in\Omega;\end{cases}$

in particular, for $x\in K$ we have $F(x)=f(x)$, and for $x\not\in\mathrm{supp}\,\psi$ we have $F(x)=0$. Because $\mathrm{supp}\,\psi\subset\Omega$ is compact and $f$ is locally $L^{2}$,

 $\left\|F\right\|_{L^{2}}=\left(\int_{\mathrm{supp}\,\psi}|\psi f|^{2}dm_{n}% \right)^{1/2}\leq\left\|\psi\right\|_{0}\left(\int_{\mathrm{supp}\,\psi}|f|^{2% }dm_{n}\right)^{1/2}<\infty,$

and using the Cauchy-Schwarz inequality, $\left\|F\right\|_{L^{1}}\leq\left\|F\right\|_{L^{2}}m_{n}(\mathrm{supp}\,\psi)% ^{1/2}<\infty$, so

 $F\in L^{2}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n}).$

Then,

 $\int_{\mathbb{R}^{n}}|\widehat{F}|^{2}dm_{n}<\infty.$ (1)

Because $\Lambda_{F}=\psi\Lambda_{f}$ in $\Omega$, the Leibniz formula tells us that in $\Omega$,

 $D_{j}^{r}\Lambda_{F}=D_{j}^{r}(\psi\Lambda_{f})=\sum_{s=0}^{r}\binom{r}{s}(D_{% j}^{r-s}\psi)(D_{j}^{s}\Lambda_{f})=\sum_{s=0}^{r}\binom{r}{s}(D_{j}^{r-s}\psi% )(\Lambda_{g_{j,s}}),$

hence, defining $H_{j}:\mathbb{R}^{n}\to\mathbb{C}$ by

 $H_{j}(x)=\begin{cases}\sum_{s=0}^{r}\binom{r}{s}(D_{j}^{r-s}\psi)(x)g_{j,s}(x)% &x\in\Omega,\\ 0&x\not\in\Omega,\end{cases}$

we have $D_{j}^{r}\Lambda_{F}=\Lambda_{H_{j}}$ in $\Omega$. It is apparent that $H_{j}\in L^{2}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n})$.

Let $\phi\in\mathscr{D}(\mathbb{R}^{n})$. There are $\phi_{1},\phi_{2}\in\mathscr{D}(\mathbb{R}^{n})$ with $\phi=\phi_{1}+\phi_{2}$ and $\mathrm{supp}\,\phi_{1}\subset\Omega$, $\mathrm{supp}\,\phi_{2}\subset\mathbb{R}^{n}\setminus\mathrm{supp}\,\psi$.1313 13 $\phi_{1}$ and $\phi_{2}$ are constructed using a partition of unity. See Walter Rudin, Functional Analysis, second ed., p. 162, Theorem 6.20. We have just established that $(D_{j}^{r}\Lambda_{F})\phi_{1}=\Lambda_{H_{j}}\phi_{1}$. For $\phi_{2}$, it is apparent that

 $(D_{j}^{r}\Lambda_{F})\phi_{2}=\Lambda_{F}(D_{j}^{r}\phi_{2})=\int_{\mathbb{R}% ^{n}}(D_{j}^{r}\phi_{2})(x)F(x)dm_{n}(x)=0$

and

 $\Lambda_{H_{j}}\phi_{2}=\int_{\mathbb{R}^{n}}\phi_{2}(x)H_{j}(x)dm_{n}(x)=0.$

Hence $(D_{j}^{r}\Lambda_{F})(\phi)=\Lambda_{H_{j}}\phi$. It is apparent that $\Lambda_{H_{j}}$ has compact support, so $D_{j}^{r}\Lambda_{F}=\Lambda_{H_{j}}$ are tempered distributions. Let $\xi\in\mathscr{S}_{n}$, and take $\phi\in\mathscr{S}_{n}$ with $\xi=\hat{\phi}$. Then,

 $\displaystyle(D_{j}^{r}\Lambda_{F})\phi$ $\displaystyle=$ $\displaystyle\Lambda_{F}D_{j}^{r}\phi$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}(D_{j}^{r}\phi)(x)F(x)dm_{n}(x)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\mathscr{F}(D_{j}^{r}\phi)(y)\widehat{F}(y)% dm_{n}(y)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}(iy_{j})^{r}\xi(y)\widehat{F}(y)dm_{n}(y),$

and

 $\Lambda_{H_{j}}\phi=\int_{\mathbb{R}^{n}}\phi(x)H_{j}(x)dm_{n}(x)=\int_{% \mathbb{R}^{n}}\xi(y)\widehat{H_{j}}(y)dm_{n}(y).$

It follows that $(iy_{j})^{r}\widehat{F}(y)=\widehat{H_{j}}(y)$ for all $y\in\mathbb{R}^{n}$. But $\widehat{H_{j}}\in L^{2}(\mathbb{R}^{n})$, so

 $\int_{\mathbb{R}^{n}}y_{i}^{2r}|\widehat{F}(y)|^{2}dm_{n}(y)<\infty,\qquad 1% \leq i\leq n.$ (2)

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

 $(1+|y|)^{2r}<(2n+2)^{r}(1+y_{1}^{2r}+\cdots+y_{n}^{2r}),\qquad y\in\mathbb{R}^% {n},$

we get

 $J=\int_{\mathbb{R}^{n}}(1+|y|)^{2r}|\widehat{F}(y)|^{2}dm_{n}(y)<\infty.$

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

 $\displaystyle\left(\int_{\mathbb{R}^{n}}(1+|y|)^{p}|\widehat{F}(y)|dm_{n}(y)% \right)^{2}$ $\displaystyle=$ $\displaystyle\left(\int_{\mathbb{R}^{n}}(1+|y|)^{r}|\widehat{F}(y)|(1+|y|)^{p-% r}dm_{n}(y)\right)^{2}$ $\displaystyle\leq$ $\displaystyle J\int_{\mathbb{R}^{n}}(1+|y|)^{2p-2r}dm_{n}(y)$ $\displaystyle=$ $\displaystyle J(2\pi)^{-n/2}\int_{0}^{\infty}\int_{S^{n-1}}(1+t)^{2p-2r}t^{n-1% }d\sigma_{n-1}(u)dt$ $\displaystyle=$ $\displaystyle\frac{2J}{\Gamma(n/2)}\int_{0}^{\infty}(1+t)^{2p-2r}t^{n-1}dt.$

This integral is finite if and only if $2p-2r+n-1<-1$, and we have assumed that $r>p+\frac{n}{2}$. Therefore,

 $\int_{\mathbb{R}^{n}}(1+|y|)^{p}|\widehat{F}(y)|dm_{n}(y)<\infty,$

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

Define

 $F_{\omega}(x)=\int_{\mathbb{R}^{n}}\widehat{F}e_{x}dm_{n},\qquad x\in\mathbb{R% }^{n}.$

(Note that $F$ depends on $\omega$.) $F,\widehat{F}\in L^{1}(\mathbb{R}^{n})$ so by the inversion theorem we have $F(x)=F_{\omega}(x)$ for almost all $x\in\mathbb{R}^{n}$. $F_{\omega}\in C_{0}(\mathbb{R}^{n})$. If $p\geq 1$, then we shall show that $F_{\omega}\in C^{p}(\Omega)$. Take $e_{k}$ to be the standard basis for $\mathbb{R}^{n}$. For $1\leq k_{1}\leq n$ and $\epsilon\neq 0$,

 $\displaystyle\frac{F_{\omega}(x+\epsilon e_{k_{1}})-F_{\omega}(x)}{\epsilon}$ $\displaystyle=$ $\displaystyle\frac{1}{\epsilon}\int_{\mathbb{R}^{n}}\widehat{F}(y)\left(\exp(i% \epsilon e_{k_{1}}\cdot y)-1\right)\exp(ix\cdot y)dm_{n}(y)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}iy_{k_{1}}\widehat{F}(y)\frac{e^{i\epsilon y% _{k_{1}}}-1}{i\epsilon y_{k}}e_{x}(y)dm_{n}(y).$

But $\left|iy_{k_{1}}\widehat{F}(y)\frac{e^{i\epsilon y_{k_{1}}}-1}{i\epsilon y_{k_% {1}}}e_{x}(y)\right|\leq|y_{k_{1}}\widehat{F}(y)|$ and $y_{k_{1}}\widehat{F}(y)$ belongs to $L^{1}(\mathbb{R}^{n})$ (supposing $p\geq 1$) so we can apply the dominated convergence theorem, which gives us

 $(D_{k_{1}}F_{\omega})(x)=\lim_{\epsilon\to 0}\frac{F_{\omega}(x+\epsilon e_{k_% {1}})-F_{\omega}(x)}{\epsilon}=\int_{\mathbb{R}^{n}}iy_{k_{1}}\widehat{F}(y)e_% {x}(y)dm_{n}(y).$

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

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 $\Omega$ whose closure is a compact subset of $\Omega$ and $\omega\cap\omega^{\prime}\neq\emptyset$, then $F_{\omega},F_{\omega^{\prime}}\in C^{p}(\mathbb{R}^{n})$ 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 $\Omega$ 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 $f_{0}\in C^{p}(\Omega)$. ∎