Sobolev spaces in one dimension and absolutely continuous functions

1 Locally integrable functions and distributions

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

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

Write . For $f\in {ℒ}_{\mathrm{loc}}^1(\lambda )$ and $x\in ℝ$, we say that $x$ is a Lebesgue point of $f$ if

$$\underset{r\to 0}{lim}\frac{1}{\lambda (B(x,r))}{\int }_{B(x,r)}|f(y)-f(x)|𝑑\lambda (y)=0.$$

It is immediate that if $f$ is continuous at $x$ then $x$ is a Lebesgue point of $f$. The Lebesgue differentiation theorem states that for $f\in {ℒ}_{\mathrm{loc}}^1(\lambda )$, almost every $x\in ℝ$ 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 $\lambda (E_n)\ge \alpha \cdot \lambda (B(x,r_n))$. 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 $f\in {ℒ}_{\mathrm{loc}}^1(\lambda )$ and for each $x\in ℝ$, $E_n(x)$ is a sequence that shrinks nicely to $x$, then

$$f(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\lambda (E_n)}{\int }_{E_n(x)}f𝑑\lambda$$

at each Lebesgue point of $f$.

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

$$\mathrm{supp}\varphi =\overline{\{x\in ℝ:\varphi (x)\ne 0\}}$$

is compact and is contained in $\mathrm{\Omega }$. We write $𝒟(\mathrm{\Omega })=C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, 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.

Theorem 1.

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

Proof.

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

where

$$c={\int }_{-1}^1\exp \left(\frac{1}{y^2-1}\right)𝑑\lambda (y)=0.443994\mathrm{\dots }.$$

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

$f(x)$ $=f(x)\cdot {\displaystyle {\int }_{ℝ}}\eta (y)𝑑\lambda (y)$

$=f(x)\cdot {\displaystyle \frac{1}{r}}{\displaystyle {\int }_{ℝ}}\eta \left({\displaystyle \frac{y}{r}}\right)𝑑\lambda (y)$

$=f(x)\cdot {\displaystyle \frac{1}{r}}{\displaystyle {\int }_{ℝ}}\eta \left({\displaystyle \frac{x-y}{r}}\right)𝑑\lambda (y)$

$={\displaystyle \frac{1}{r}}{\displaystyle {\int }_{ℝ}}(f(x)-f(y))\eta \left({\displaystyle \frac{x-y}{r}}\right)𝑑\lambda (y)+{\displaystyle \frac{1}{r}}{\displaystyle {\int }_{ℝ}}f(y)\eta \left({\displaystyle \frac{x-y}{r}}\right)𝑑\lambda (y)$

$={\displaystyle \frac{1}{r}}{\displaystyle {\int }_{ℝ}}(f(x)-f(y))\eta \left({\displaystyle \frac{x-y}{r}}\right)𝑑\lambda (y)$

$={\displaystyle \frac{1}{r}}{\displaystyle {\int }_{(x-r,x+r)}}(f(x)-f(y))\eta \left({\displaystyle \frac{x-y}{r}}\right)𝑑\lambda (y).$

Then

$$|f(x)|\le {\parallel \eta \parallel }_{\mathrm{\infty }}\cdot \frac{1}{r}{\int }_{(x-r,x+r)}|f(y)-f(x)|𝑑\lambda (y)\to 0,r\to 0,$$

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

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

$${\mathrm{\Lambda }}_f(\varphi )={\int }_{ℝ}f\varphi 𝑑\lambda .$$

$𝒟(ℝ)$ is a locally convex space, and one proves that ${\mathrm{\Lambda }}_f$ is continuous and thus belongs to the dual space ${𝒟}^{\prime }(ℝ)$, whose elements are called distributions. We say that a distribution $\mathrm{\Lambda }$ is induced by $f\in {ℒ}_{\mathrm{loc}}^1(\lambda )$ if $\mathrm{\Lambda }={\mathrm{\Lambda }}_f$. For $\mathrm{\Lambda }\in {𝒟}^{\prime }(ℝ)$, we define $D\mathrm{\Lambda }:𝒟(ℝ)\to ℝ$ by

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

It is proved that $D\mathrm{\Lambda }\in {𝒟}^{\prime }(ℝ)$.

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

$$-{\int }_{ℝ}f{\varphi }^{\prime }𝑑\lambda ={\int }_{ℝ}g\varphi 𝑑\lambda .$$

If $g_1,g_2\in {ℒ}_{\mathrm{loc}}^1(\lambda )$ are distributional derivatives of $f$ then ${\int }_{ℝ}(g_1-g_2)\varphi 𝑑\lambda =0$ for all $\varphi \in 𝒟(ℝ)$, 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 $Df\in L_{\mathrm{loc}}^1(\lambda )$:

$$-{\int }_{ℝ}f{\varphi }^{\prime }𝑑\lambda ={\int }_{ℝ}(Df)\cdot \varphi 𝑑\lambda ,\varphi \in 𝒟(ℝ).$$

2 The Sobolev space $H^1(ℝ)$

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

$${⟨f,g⟩}_{L^2}={\int }_{ℝ}fg𝑑\lambda .$$

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

We define the Sobolev space $H^1(ℝ)$ 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 $H^1(ℝ)$ are equivalence classes of elements of ${ℒ}^2(\lambda )$. We define

$${⟨f,g⟩}_{H^1}={⟨f,g⟩}_{L^2}+{⟨Df,Dg⟩}_{L^2}.$$

Let $f,g\in H^1(ℝ)$ and let $\varphi \in 𝒟(ℝ)$. Because $f,g$ have distributional derivatives $Df,Dg$,

$-{\displaystyle {\int }_{ℝ}}(f+g){\varphi }^{\prime }𝑑\lambda$ $=-{\displaystyle {\int }_{ℝ}}f{\varphi }^{\prime }𝑑\lambda -{\displaystyle {\int }_{ℝ}}g{\varphi }^{\prime }𝑑\lambda$

$={\displaystyle {\int }_{ℝ}}Df\cdot \varphi 𝑑\lambda +{\displaystyle {\int }_{ℝ}}Dg\cdot \varphi 𝑑\lambda$

$={\displaystyle {\int }_{ℝ}}(Df+Dg)\varphi 𝑑\lambda .$

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

If $f_n$ is a Cauchy sequence in $H^1(ℝ)$, then $f_n$ is a Cauchy sequence in $L^2(\lambda )$ and $D{f}_n$ is a Cauchy sequence in $L^2(\lambda )$, and hence these sequences have limits $f,g\in L^2(\lambda )$. For $\varphi \in 𝒟(ℝ)$,

$-{\displaystyle {\int }_{ℝ}}f{\varphi }^{\prime }𝑑\lambda$ $=-\underset{n\to \mathrm{\infty }}{lim}{\displaystyle {\int }_{ℝ}}{f}_n{\varphi }^{\prime }𝑑\lambda$

$=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle {\int }_{ℝ}}(D{f}_n)\cdot \varphi 𝑑\lambda$

$={\displaystyle {\int }_{ℝ}}g\varphi 𝑑\lambda .$

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

$${\parallel f_n-f\parallel }_{H^1}^2={\parallel f_n-f\parallel }_{L^2}^2+{\parallel D{f}_n-Df\parallel }_{L^2}^2={\parallel f_n-f\parallel }_{L^2}^2+{\parallel D{f}_n-g\parallel }_{L^2}^2\to 0,$$

meaning that $f_n\to f$ in $H^1(ℝ)$, which shows that $H^1(ℝ)$ 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.

Lemma 2.

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

$${\int }_{ℝ}f{\varphi }^{\prime }𝑑\lambda =0,\varphi \in 𝒟(ℝ),$$

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

Proof.

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

$$h=w-\eta \cdot {\int }_{ℝ}w𝑑\lambda ,$$

which belongs to $𝒟(ℝ)$ and satisfies ${\int }_{ℝ}h𝑑\lambda =0$. Define $\varphi :ℝ\to ℝ$ by

$$\varphi (x)={\int }_{-\mathrm{\infty }}^{x}h𝑑\lambda .$$

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

$0$ $={\displaystyle {\int }_{ℝ}}fh𝑑\lambda$

$={\displaystyle {\int }_{ℝ}}\left(fw-f\eta \cdot {\displaystyle {\int }_{ℝ}}w𝑑\lambda \right)𝑑\lambda$

$={\displaystyle {\int }_{ℝ}}\left(f-{\displaystyle {\int }_{ℝ}}f\eta 𝑑\lambda \right)\cdot w𝑑\lambda .$

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

Lemma 3.

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

$$f(x)={\int }_a^{x}g(y)𝑑\lambda (y).$$

Then

$${\int }_{ℝ}f{\varphi }^{\prime }𝑑\lambda =-{\int }_{ℝ}g\varphi 𝑑\lambda$$

for all $\varphi \mathrm{\in }\mathrm{D}\mathit{}\mathrm{(}\mathrm{R}\mathrm{)}$.

Proof.

Using Fubini’s theorem,

${\displaystyle {\int }_{ℝ}}f(x){\varphi }^{\prime }(x)𝑑\lambda (x)$ $=-{\displaystyle {\int }_{-\mathrm{\infty }}^a}\left({\displaystyle {\int }_x^a}g(y)𝑑\lambda (y)\right){\varphi }^{\prime }(x)𝑑\lambda (x)$

$+{\displaystyle {\int }_a^{\mathrm{\infty }}}\left({\displaystyle {\int }_a^x}g(y)𝑑\lambda (y)\right){\varphi }^{\prime }(x)𝑑\lambda (x)$

$=-{\displaystyle {\int }_{-\mathrm{\infty }}^a}\left({\displaystyle {\int }_{-\mathrm{\infty }}^y}{\varphi }^{\prime }(x)𝑑\lambda (x)\right)g(y)𝑑\lambda (y)$

$+{\displaystyle {\int }_a^{\mathrm{\infty }}}\left({\displaystyle {\int }_y^{\mathrm{\infty }}}{\varphi }^{\prime }(x)𝑑\lambda (x)\right)g(y)𝑑\lambda (y)$

$=-{\displaystyle {\int }_{-\mathrm{\infty }}^a}\varphi (y)g(y)𝑑\lambda (y)-{\displaystyle {\int }_a^{\mathrm{\infty }}}\varphi (y)g(y)𝑑\lambda (y)$

$=-{\displaystyle {\int }_{ℝ}}g(y)\varphi (y)𝑑\lambda (y).$

∎

For real numbers $a,b$ with , we say that a function $f:[a,b]\to ℝ$ is absolutely continuous if for all $ϵ>0$ there is some $\delta >0$ such that whenever $(a_1,b_1),\mathrm{\dots },(a_n,b_n)$ are disjoint intervals each contained in $[a,b]$ with it holds that . We say that a function $f:ℝ\to ℝ$ 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 $A{C}_{\mathrm{loc}}(ℝ)$.

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

$$h(x)={\int }_a^{x}Df𝑑\lambda .$$

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

$${\int }_{ℝ}h{\varphi }^{\prime }𝑑\lambda =-{\int }_{ℝ}(Df)\cdot \varphi 𝑑\lambda ={\int }_{ℝ}f{\varphi }^{\prime }𝑑\lambda ,\varphi \in 𝒟(ℝ).$$

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

$${\int }_{ℝ}\tilde{f}{\varphi }^{\prime }𝑑\lambda =-{\int }_{ℝ}{\tilde{f}}^{\prime }\varphi 𝑑\lambda ,$$

and by definition of a distributional derivative,

$${\int }_{ℝ}\tilde{f}{\varphi }^{\prime }𝑑\lambda =-{\int }_{ℝ}(D\tilde{f})\varphi 𝑑\lambda .$$

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

Theorem 4.

For $f\mathrm{\in }{H}^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{R}\mathrm{)}$, there is a function $\stackrel{\mathrm{~}}{f}\mathrm{\in }A\mathit{}{C}_{\mathrm{loc}}\mathit{}\mathrm{(}\mathrm{R}\mathrm{)}$ such that $\stackrel{\mathrm{~}}{f}\mathrm{=}f$ almost everywhere and ${\stackrel{\mathrm{~}}{f}}^{\mathrm{\prime }}\mathrm{=}D\mathit{}f$ almost everywhere. The function $\stackrel{\mathrm{~}}{f}$ is $\frac{\mathrm{1}}{\mathrm{2}}$-Hölder continuous.

Proof.

For $x,y\in ℝ$,

$$\tilde{f}(x)-\tilde{f}(y)={\int }_y^x{\tilde{f}}^{\prime }𝑑\lambda ,$$

and using the Cauchy-Schwarz inequality,

$|\tilde{f}(x)-\tilde{f}(y)|$ $\le {\displaystyle {\int }_y^x}|{\tilde{f}}^{\prime }|𝑑\lambda$

$\le {|x-y|}^{1/2}{\left({\displaystyle {\int }_y^x}{|{\tilde{f}}^{\prime }|}^2𝑑\lambda \right)}^{1/2}$

$\le {\parallel Df\parallel }_{L^2}{|x-y|}^{1/2}.$

∎