prove that $u$ is equal a.e. to an absolutely continuous function

Question

Prove that if $n=1$ and $u\in W^{1,p}(0,1) $ for some $1\leq p<\infty$, then $u$ is equal a.e. to an absolutely continuous function,and $u'$ (which exists a.e.) belongs to $L^{p}(0,1)$.

My idea:due to $(0,1)$ is a bounded and we can think that $u$ has a compact support,if $u\in W^{1,p}((0,1))$,then the $u^{\varepsilon}=\eta_{\varepsilon}*u$ is a smooth function with $ u^{\varepsilon}\rightrightarrows u $ when $\varepsilon\to 0^{+}$,and $\|Du^{\varepsilon}\|_{p}\leq \|Du\|_{p} $,then for $x,y\in (0,1),x\neq y$, we have \begin{align*} u^{\varepsilon}(x)-u^{\varepsilon}(y)&= \int_{0}^{1} (x-y)\cdot (Du^{\varepsilon}(tx+(1-t)y))dt \end{align*} then we have $$ |u^{\varepsilon}(x)-u^{\varepsilon}(y)|\leq \|Du^{\varepsilon}\|_{p}|x-y|\leq \|Du\|_{p}|x-y| $$ Let $\varepsilon\to 0$,we get $$ |u(x)-u(y)|\leq \|Du\|_{p}|x-y| $$

Answer 1

A graph will really help. I've sketched a graph of a function $F_{a,b,h,k}$ that cannot be used directly as a test function, but it can after mollification. The derivative $F'$ of $F$ is shown below the graph of $F$.

The exact assumptions are that $0 < a-h < a \le b < b+k < 1$. That way, after you mollify with a compactly supported $C^{\infty}$ function $\eta_{\epsilon}$, you end up with a function $\eta_{\epsilon}\star F_{a,b,h,k}$ that is compactly supported in $(0,1)$. And, $$ (\eta_{\epsilon}\star F_{a,b,h,k})'=\eta_{\epsilon}\star F_{a,b,h,k}'. $$ For fixed $a,b,h,k$ the functions $\eta_{\epsilon}\star F_{a,b,h,k}$ and $\eta_{\epsilon}\star F_{a,b,h,k}'$ remain uniformly bounded and converge pointwise a.e. to $F_{a,b,h,k}$ and $F_{a,b,h,k}'$, respectively as $\epsilon\rightarrow 0$. That's how you can extend the weak relation to include $F$. Indeed, suppose that $u \in L^{p}$ with weak derivative $Du \in L^{p}$. Then, $$ \int_{0}^{1}u(x)(\eta_{\epsilon}\star F_{a,b,h,k}')dx=-\int_{0}^{1}(Du) (\eta_{\epsilon}\star F_{a,b,h,k})dx $$ Letting $\epsilon\rightarrow 0$ and using the properties of the mollified $F,F'$ stated above gives \begin{align} \int_{0}^{1}(Du)F_{a,b,h,k}dx & = -\int_{0}^{1}uF_{a,b,h,k}'dx \\ & = -\frac{1}{h}\int_{a-h}^{a}u(x)dx+\frac{1}{k}\int_{b}^{b+k}u(x)dx \end{align} You can see that as $h\rightarrow 0$ the graph of $F_{a,b,h,k}$ remains bounded and converges pointwise everywhere; the same is true of $k$. So the integral on the left above converges as $h\rightarrow 0$ and/or $k\rightarrow 0$. Because $a \le b$ you can conclude that $\int_{0}^{t}u(x)dx$ has a derivative at every $t \in (0,1)$ because the left- and right-hand derivatives exist, and they're equal because, if $a=b$, the left side converges to $0$ as $h,k\rightarrow 0$. Hence, $$ \int_{a}^{b}Du dx = \left.\frac{d}{dt}\int_{0}^{t}udx\right|_{t=a}^{b} $$ The right side also equals $u(b)-u(a)$ for all $a,b \in [0,1]\setminus E$, where $E$ is a set of measure $0$, because of Lebesgue differentiation theorem. The left side is a continuous function of $a$, $b$. Therefore it is possible to change $u$ on a set of measure $0$ in order to assume that $u$ is continuous on $[0,1]$. After doing so, $$ u(b)-u(a)=\int_{a}^{b}Du dx. $$ Therefore, the modified $u$ is absolutely continuous on $[0,1]$ with $u'=Du$ a.e., where $Du$ is the weak derivative.

Answer 2

Yes, this is one way to do it. I would suggest you to read this book, chapter 8 for a complete introduction on 1 dimensional Sobolev space.

Also, this note collects more results, but without proof, of Sobolev function properties in one dimensions.

Answer 3

Another approach without convolutions:

Since $Du$ is an $L^p(0,1)$ function, hence $L^1(0,1)$ in particular, for each $x \in (0,1)$ the integral $w(x) = \int_0^x Du(t)\,dt$ makes sense. By the fundamental theorem of calculus for Lebesgue integrals, $w$ is absolutely continuous and $w' = Du$ almost everywhere. Intuitively, we should expect that $w$ differs from $u$ by a constant. To find the constant, set $c = \int_0^1 (u(t) - w(t))\,dt$. Then set $v = w+c$, which is again absolutely continuous and has $v' = Du$ almost everywhere. Now for any test function $\varphi \in C^\infty_c((0,1))$, use the definition of weak derivative and integration by parts to show that $\int_0^1 (u(t)-v(t)) \varphi(t)\,dt = 0$. Conclude that $u-v = 0$ almost everywhere.