# Sobolev spaces in one dimension and absolutely continuous functions

Jordan Bell
October 21, 2015

## 1 Locally integrable functions and distributions

Let $\lambda$ be Lebesgue measure on $\mathbb{R}$. We denote by $\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ the collection of Borel measurable functions $f:\mathbb{R}\to\mathbb{R}$ such that for each compact subset $K$ of $\mathbb{R}$,

 $N_{K}(f)=\int_{K}|f|d\lambda=\int_{\mathbb{R}}1_{K}|f|d\lambda<\infty.$

We denote by $L_{\mathrm{loc}}^{1}(\lambda)$ the collection of equivalence classes of elements of $\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ where $f\sim g$ when $f=g$ almost everywhere.

Write $B(x,r)=\{y\in\mathbb{R}:|y-x|. For $f\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ and $x\in\mathbb{R}$, we say that $x$ is a Lebesgue point of $f$ if

 $\lim_{r\to 0}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|d\lambda(y)=0.$

It is immediate that if $f$ is continuous at $x$ then $x$ is a Lebesgue point of $f$. The Lebesgue differentiation theorem11 1 Walter Rudin, Real and Complex Analysis, third ed., p. 138, Theorem 7.7. states that for $f\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$, almost every $x\in\mathbb{R}$ is a Lebesgue point of $f$. A sequence of Borel sets $E_{n}$ is said to shrink nicely to $x$ if there is some $\alpha>0$ and a sequence $r_{n}\to 0$ such that $E_{n}\subset B(x,r_{n})$ and $\lambda(E_{n})\geq\alpha\cdot\lambda(B(x,r_{n}))$. The sequence $B(x,n^{-1})=(x-n^{-1},x+n^{-1})$ shrinks nicely to $x$, the sequence $[x,x+n^{-1}]$ shrinks nicely to $x$, and the sequence $[x-n^{-1},x]$ shrinks nicely to $x$. It is proved that if $f\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ and for each $x\in\mathbb{R}$, $E_{n}(x)$ is a sequence that shrinks nicely to $x$, then

 $f(x)=\lim_{n\to\infty}\frac{1}{\lambda(E_{n})}\int_{E_{n}(x)}fd\lambda$

at each Lebesgue point of $f$.22 2 Walter Rudin, Real and Complex Analysis, third ed., p. 140, Theorem 7.10.

For a nonempty open set $\Omega$ in $\mathbb{R}$, we denote by $C_{c}^{k}(\Omega)$ the collection of $C^{k}$ functions $\phi:\mathbb{R}\to\mathbb{R}$ such that

 $\mathrm{supp}\,\phi=\overline{\{x\in\mathbb{R}:\phi(x)\neq 0\}}$

is compact and is contained in $\Omega$. We write $\mathscr{D}(\Omega)=C_{c}^{\infty}(\Omega)$, whose elements are called called test functions. The following statement is called the fundamental lemma of the calculus of variations or the Du Bois-Reymond Lemma.33 3 Lars Hörmander, The Analysis of Linear Partial Differential Operators I, second ed., p. 15, Theorem 1.2.5.

###### Theorem 1.

If $f\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ and $\int_{\mathbb{R}}f\phi d\lambda=0$ for all $\phi\in\mathscr{D}(\mathbb{R})$, then $f=0$ almost everywhere.

###### Proof.

There is some $\eta\in\mathscr{D}(-1,1)$ with $\int_{\mathbb{R}}\eta d\lambda=1$. We can explicitly write this out:

 $\eta(x)=\begin{cases}c^{-1}\exp\left(\frac{1}{x^{2}-1}\right)&|x|<1\\ 0&|x|\geq 1,\end{cases}$

where

 $c=\int_{-1}^{1}\exp\left(\frac{1}{y^{2}-1}\right)d\lambda(y)=0.443994\ldots.$

For $x$ a Lebesgue point of $f$ and for $0,

 $\displaystyle f(x)$ $\displaystyle=f(x)\cdot\int_{\mathbb{R}}\eta(y)d\lambda(y)$ $\displaystyle=f(x)\cdot\frac{1}{r}\int_{\mathbb{R}}\eta\left(\frac{y}{r}\right% )d\lambda(y)$ $\displaystyle=f(x)\cdot\frac{1}{r}\int_{\mathbb{R}}\eta\left(\frac{x-y}{r}% \right)d\lambda(y)$ $\displaystyle=\frac{1}{r}\int_{\mathbb{R}}(f(x)-f(y))\eta\left(\frac{x-y}{r}% \right)d\lambda(y)+\frac{1}{r}\int_{\mathbb{R}}f(y)\eta\left(\frac{x-y}{r}% \right)d\lambda(y)$ $\displaystyle=\frac{1}{r}\int_{\mathbb{R}}(f(x)-f(y))\eta\left(\frac{x-y}{r}% \right)d\lambda(y)$ $\displaystyle=\frac{1}{r}\int_{(x-r,x+r)}(f(x)-f(y))\eta\left(\frac{x-y}{r}% \right)d\lambda(y).$

Then

 $|f(x)|\leq\left\|\eta\right\|_{\infty}\cdot\frac{1}{r}\int_{(x-r,x+r)}|f(y)-f(% x)|d\lambda(y)\to 0,\qquad r\to 0,$

meaning that $f(x)=0$. This is true for almost all $x\in\mathbb{R}$, showing that $f=0$ almost everywhere. ∎

For $f\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$, define $\Lambda_{f}:\mathscr{D}(\mathbb{R})\to\mathbb{R}$ by

 $\Lambda_{f}(\phi)=\int_{\mathbb{R}}f\phi d\lambda.$

$\mathscr{D}(\mathbb{R})$ is a locally convex space, and one proves that $\Lambda_{f}$ is continuous and thus belongs to the dual space $\mathscr{D}^{\prime}(\mathbb{R})$, whose elements are called distributions.44 4 Walter Rudin, Functional Analysis, second ed., p. 157, §6.11. We say that a distribution $\Lambda$ is induced by $f\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ if $\Lambda=\Lambda_{f}$. For $\Lambda\in\mathscr{D}^{\prime}(\mathbb{R})$, we define $D\Lambda:\mathscr{D}(\mathbb{R})\to\mathbb{R}$ by

 $(D\Lambda)(\phi)=-\Lambda(\phi^{\prime}).$

It is proved that $D\Lambda\in\mathscr{D}^{\prime}(\mathbb{R})$.55 5 Walter Rudin, Functional Analysis, second ed., p. 158, §6.12.

Let $f,g\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$. If $D\Lambda_{f}=\Lambda_{g}$, we call $g$ a distributional derivative of $f$. In other words, for $f\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ to have a distributional derivative means that there is some $g\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ such that for all $\phi\in\mathscr{D}(\mathbb{R})$,

 $-\int_{\mathbb{R}}f\phi^{\prime}d\lambda=\int_{\mathbb{R}}g\phi d\lambda.$

If $g_{1},g_{2}\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ are distributional derivatives of $f$ then $\int_{\mathbb{R}}(g_{1}-g_{2})\phi d\lambda=0$ for all $\phi\in\mathscr{D}(\mathbb{R})$, which by Theorem 1 implies that $g_{1}=g_{2}$ almost everywhere. It follows that if $f$ has a distributional derivative then the distributional derivative is unique in $L_{\mathrm{loc}}^{1}(\lambda)$, and is denoted $Df\in L_{\mathrm{loc}}^{1}(\lambda)$:

 $-\int_{\mathbb{R}}f\phi^{\prime}d\lambda=\int_{\mathbb{R}}(Df)\cdot\phi d% \lambda,\qquad\phi\in\mathscr{D}(\mathbb{R}).$

## 2 The Sobolev space $H^{1}(\mathbb{R})$

We denote by $\mathscr{L}^{2}(\lambda)$ the collection of Borel measurable functions $f:\mathbb{R}\to\mathbb{R}$ such that $\int_{\mathbb{R}}|f|^{2}d\lambda<\infty$, and we denote by $L^{2}(\lambda)$ the collection of equivalence classes of elements of $\mathscr{L}^{2}(\lambda)$ where $f\sim g$ when $f=g$ almost everywhere, and write

 $\left\langle f,g\right\rangle_{L^{2}}=\int_{\mathbb{R}}fgd\lambda.$

It is a fact that $L^{2}(\lambda)$ is a Hilbert space.

We define the Sobolev space $H^{1}(\mathbb{R})$ to be the set of $f\in L^{2}(\lambda)$ that have a distributional derivative that satisfies $Df\in L^{2}(\lambda)$. We remark that the elements of $H^{1}(\mathbb{R})$ are equivalence classes of elements of $\mathscr{L}^{2}(\lambda)$. We define

 $\left\langle f,g\right\rangle_{H^{1}}=\left\langle f,g\right\rangle_{L^{2}}+% \left\langle Df,Dg\right\rangle_{L^{2}}.$

Let $f,g\in H^{1}(\mathbb{R})$ and let $\phi\in\mathscr{D}(\mathbb{R})$. Because $f,g$ have distributional derivatives $Df,Dg$,

 $\displaystyle-\int_{\mathbb{R}}(f+g)\phi^{\prime}d\lambda$ $\displaystyle=-\int_{\mathbb{R}}f\phi^{\prime}d\lambda-\int_{\mathbb{R}}g\phi^% {\prime}d\lambda$ $\displaystyle=\int_{\mathbb{R}}Df\cdot\phi d\lambda+\int_{\mathbb{R}}Dg\cdot% \phi d\lambda$ $\displaystyle=\int_{\mathbb{R}}(Df+Dg)\phi d\lambda.$

This means that $f+g$ has a distributional derivative, $D(f+g)=Df+Dg$. Thus $H^{1}(\mathbb{R})$ is a linear space. If $\left\langle f,f\right\rangle_{H^{1}}=0$ then $\int_{\mathbb{R}}|f|^{2}d\lambda=0$, which implies that $f=0$ as an element of $L^{2}(\lambda)$. Therefore $\left\langle\cdot,\cdot\right\rangle_{H^{1}}$ is an inner product on $H^{1}(\mathbb{R})$.

If $f_{n}$ is a Cauchy sequence in $H^{1}(\mathbb{R})$, then $f_{n}$ is a Cauchy sequence in $L^{2}(\lambda)$ and $Df_{n}$ is a Cauchy sequence in $L^{2}(\lambda)$, and hence these sequences have limits $f,g\in L^{2}(\lambda)$. For $\phi\in\mathscr{D}(\mathbb{R})$,

 $\displaystyle-\int_{\mathbb{R}}f\phi^{\prime}d\lambda$ $\displaystyle=-\lim_{n\to\infty}\int_{\mathbb{R}}f_{n}\phi^{\prime}d\lambda$ $\displaystyle=\lim_{n\to\infty}\int_{\mathbb{R}}(Df_{n})\cdot\phi d\lambda$ $\displaystyle=\int_{\mathbb{R}}g\phi d\lambda.$

This means that $f$ has distributional derivative, $Df=g$. Because $f,Df\in L^{2}(\lambda)$ it is the case that $f\in H^{1}(\mathbb{R})$. Furthermore,

 $\left\|f_{n}-f\right\|_{H^{1}}^{2}=\left\|f_{n}-f\right\|_{L^{2}}^{2}+\left\|% Df_{n}-Df\right\|_{L^{2}}^{2}=\left\|f_{n}-f\right\|_{L^{2}}^{2}+\left\|Df_{n}% -g\right\|_{L^{2}}^{2}\to 0,$

meaning that $f_{n}\to f$ in $H^{1}(\mathbb{R})$, which shows that $H^{1}(\mathbb{R})$ is a Hilbert space.

## 3 Absolutely continuous functions

We prove a lemma that gives conditions under which a function, for which integration by parts needs not make sense, is equal to a particular constant almost everywhere.66 6 Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, p. 204, Lemma 8.1.

###### Lemma 2.

If $f\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$ and

 $\int_{\mathbb{R}}f\phi^{\prime}d\lambda=0,\qquad\phi\in\mathscr{D}(\mathbb{R}),$

then there is some $c\in\mathbb{R}$ such that $f=c$ almost everywhere.

###### Proof.

Fix $\eta\in\mathscr{D}(\mathbb{R})$ with $\int_{\mathbb{R}}\eta d\lambda=1$. Let $w\in\mathscr{D}(\mathbb{R})$ and define

 $h=w-\eta\cdot\int_{\mathbb{R}}wd\lambda,$

which belongs to $\mathscr{D}(\mathbb{R})$ and satisfies $\int_{\mathbb{R}}hd\lambda=0$. Define $\phi:\mathbb{R}\to\mathbb{R}$ by

 $\phi(x)=\int_{-\infty}^{x}hd\lambda.$

Using $\phi^{\prime}(x)=h(x)$ for all $x$ and $\phi(x)\to\int_{\mathbb{R}}hd\lambda=0$ as $x\to\infty$, check that $\phi\in\mathscr{D}(\mathbb{R})$. Then by hypothesis, $\int_{\mathbb{R}}f\phi^{\prime}d\lambda=0$, i.e.

 $\displaystyle 0$ $\displaystyle=\int_{\mathbb{R}}fhd\lambda$ $\displaystyle=\int_{\mathbb{R}}\left(fw-f\eta\cdot\int_{\mathbb{R}}wd\lambda% \right)d\lambda$ $\displaystyle=\int_{\mathbb{R}}\left(f-\int_{\mathbb{R}}f\eta d\lambda\right)% \cdot wd\lambda.$

Because this is true for all $w\in\mathscr{D}(\mathbb{R})$, by Theorem 1 we get that $f=\int_{\mathbb{R}}f\eta d\lambda$ almost everywhere. ∎

###### Lemma 3.

Let $g\in\mathscr{L}_{\mathrm{loc}}^{1}(\lambda)$, let $a\in\mathbb{R}$, and define $f:\mathbb{R}\to\mathbb{R}$ by

 $f(x)=\int_{a}^{x}g(y)d\lambda(y).$

Then

 $\int_{\mathbb{R}}f\phi^{\prime}d\lambda=-\int_{\mathbb{R}}g\phi d\lambda$

for all $\phi\in\mathscr{D}(\mathbb{R})$.

###### Proof.

Using Fubini’s theorem,

 $\displaystyle\int_{\mathbb{R}}f(x)\phi^{\prime}(x)d\lambda(x)$ $\displaystyle=-\int_{-\infty}^{a}\left(\int_{x}^{a}g(y)d\lambda(y)\right)\phi^% {\prime}(x)d\lambda(x)$ $\displaystyle+\int_{a}^{\infty}\left(\int_{a}^{x}g(y)d\lambda(y)\right)\phi^{% \prime}(x)d\lambda(x)$ $\displaystyle=-\int_{-\infty}^{a}\left(\int_{-\infty}^{y}\phi^{\prime}(x)d% \lambda(x)\right)g(y)d\lambda(y)$ $\displaystyle+\int_{a}^{\infty}\left(\int_{y}^{\infty}\phi^{\prime}(x)d\lambda% (x)\right)g(y)d\lambda(y)$ $\displaystyle=-\int_{-\infty}^{a}\phi(y)g(y)d\lambda(y)-\int_{a}^{\infty}\phi(% y)g(y)d\lambda(y)$ $\displaystyle=-\int_{\mathbb{R}}g(y)\phi(y)d\lambda(y).$

For real numbers $a,b$ with $a, we say that a function $f:[a,b]\to\mathbb{R}$ is absolutely continuous if for all $\epsilon>0$ there is some $\delta>0$ such that whenever $(a_{1},b_{1}),\ldots,(a_{n},b_{n})$ are disjoint intervals each contained in $[a,b]$ with $\sum(b_{k}-a_{k})<\delta$ it holds that $\sum|f(b_{k})-f(a_{k})|<\epsilon$. We say that a function $f:\mathbb{R}\to\mathbb{R}$ is locally absolutely continuous if for each nonempty compact interval $[a,b]$, the restriction of $f$ to $[a,b]$ is absolutely continuous. We denote the collection of locally absolutely continuous by $AC_{\mathrm{loc}}(\mathbb{R})$.

Let $f\in H^{1}(\mathbb{R})$, let $a\in\mathbb{R}$, and define $h:\mathbb{R}\to\mathbb{R}$ by

 $h(x)=\int_{a}^{x}Dfd\lambda.$

By Lemma 3 and by the definition of a distributional derivative,

 $\int_{\mathbb{R}}h\phi^{\prime}d\lambda=-\int_{\mathbb{R}}(Df)\cdot\phi d% \lambda=\int_{\mathbb{R}}f\phi^{\prime}d\lambda,\qquad\phi\in\mathscr{D}(% \mathbb{R}).$

Hence $\int_{\mathbb{R}}(f-h)\phi^{\prime}d\lambda=0$ for all $\phi\in\mathscr{D}(\mathbb{R})$, which by Lemma 2 implies that there is some $c\in\mathbb{R}$ such that $f-h=c$ almost everywhere. Let $\widetilde{f}=c+h$. On the one hand, the fact that $Df\in L_{\mathrm{loc}}^{1}(\lambda)$ implies that $h\in AC_{\mathrm{loc}}(\mathbb{R})$ and so $\widetilde{f}\in AC_{\mathrm{loc}}(\mathbb{R})$. On the other hand, $\widetilde{f}=f$ almost everywhere. Furthermore, because $\widetilde{f}$ is locally absolutely continuous, integration by parts yields

 $\int_{\mathbb{R}}\widetilde{f}\phi^{\prime}d\lambda=-\int_{\mathbb{R}}% \widetilde{f}^{\prime}\phi d\lambda,$

and by definition of a distributional derivative,

 $\int_{\mathbb{R}}\widetilde{f}\phi^{\prime}d\lambda=-\int_{\mathbb{R}}(D% \widetilde{f})\phi d\lambda.$

Therefore by Theorem 1, $\widetilde{f}^{\prime}=D\widetilde{f}$ almost everywhere. But the fact that $\widetilde{f}=f$ almost everywhere implies that $D\widetilde{f}=Df$ almost everywhere, so $\widetilde{f}^{\prime}=Df$ almost everywhere. In particular, $\widetilde{f}^{\prime}\in L^{2}(\lambda)$.

###### Theorem 4.

For $f\in H^{1}(\mathbb{R})$, there is a function $\widetilde{f}\in AC_{\mathrm{loc}}(\mathbb{R})$ such that $\widetilde{f}=f$ almost everywhere and $\widetilde{f}^{\prime}=Df$ almost everywhere. The function $\widetilde{f}$ is $\frac{1}{2}$-Hölder continuous.

###### Proof.

For $x,y\in\mathbb{R}$,77 7 cf. Giovanni Leoni, A First Course in Sobolev Spaces, p. 222, Theorem 7.13.

 $\widetilde{f}(x)-\widetilde{f}(y)=\int_{y}^{x}\widetilde{f}^{\prime}d\lambda,$

and using the Cauchy-Schwarz inequality,

 $\displaystyle|\widetilde{f}(x)-\widetilde{f}(y)|$ $\displaystyle\leq\int_{y}^{x}|\widetilde{f}^{\prime}|d\lambda$ $\displaystyle\leq|x-y|^{1/2}\left(\int_{y}^{x}|\widetilde{f}^{\prime}|^{2}d% \lambda\right)^{1/2}$ $\displaystyle\leq\left\|Df\right\|_{L^{2}}|x-y|^{1/2}.$