Exercise 8.7 on Haim Brezis' book

Question

I am doing the exercises about Sobolev spaces on the book of Haim Brezis, and I am geting stuck with this,

Let $I=(0,1)$, given a function $u$ defined on $I$,

$$\bar{u}(x)=\begin{cases} u(x), & x \in I \\ 0, & x \in \mathbb{R} \setminus I \end{cases}$$ Assume that $u\in L^p(I)$ with $1\leq p <\infty$, such that $\bar{u}\in W^{1,p}(\mathbb{R})$, prove that $u \in W^{1,p}_0(I)$

Since $\bar{u}$ is a continous function, so I am trying to use a Cutt-Off technic to get a sequence with a compac support converging to $u$ but I couldn't apply this idea, Any hint or intermediate question please?

Answer 1

Proof: $u \in W_0^{1,p}(0,1)$

• For all $\varphi \in C_c^\infty(0,1)$ (hence its $0$-extension $\tilde \varphi$ is in $C_c^\infty(\mathbb{R})$) we have $$ -\int_0^1 u(x) \partial_x \varphi(x) \text{ d}x = - \int_\mathbb{R} \tilde u(x) \partial_x \tilde \varphi(x) \text{ d}x = \int_\mathbb{R} \partial_x \tilde u(x) \tilde \varphi(x) \text{ d}x=\int_0^1 v(x) \varphi(x) \text{ d}x$$ where $v(x):=\begin{cases} \partial_x \tilde u(x), &\text{for } x \in (0,1) \\ 0, &\text{else} \end{cases}$ is the weak derivative of $u$. Showing that $v \in L^p(0,1)$ is straight forward.
• Since $u \in W^{1,p}(0,1)\subset C([0,1])$, it remains to show $u(0)=u(1)=0$. Also $\tilde u \in C(\mathbb{R})$, hence $u(x) \to 0$ for $x \to 0$ and $u(y) \to 0$ for $y \to 1$. Now use the continuity of $u$.