Consider the function $\phi$ defined by $\phi(x) = (1 − |x|)_+$, I have been tasked by an exercise in my textbook to verify both $\phi,\phi'\in L^p([-2,2])$ and deduce $\phi\in W^{1,p}((−2, 2))$ for all $p \in [1, \infty]$.
Here's what I've done so far:
I've shown $\phi \in C([-2,2])$ and $\phi \notin C^1((-2,2))$, from there I deduced that the first (weak) derivative $\phi'=0$ for $x \in (-2,-1)$ and $x\in (1,2)$, $\phi'=-1$ for $x\in (0,1)$ and $\phi'=1$ for $x\in (-1,0)$.
Here is where I am stuck, I think it's pretty obvious that $\phi'$ is $L^p$ on $[-2,2]$ as it is constant at either $0$, $1$ and $-1$ a.e. across $[-2,2]$, hence to my understanding $\int_{-2}^{2}\vert\phi'\vert^pdx=0<+\infty$ for all $p$. For $\phi$, I first found $\int_{-2}^{2}|\phi|^pdx$ for $p=1$ (which should be $1$) however I don't know how to do this for $p$ going up to infinity, maybe as $\vert\phi\vert<1$ for all $x$ then hence $\vert \phi\vert^p<1$ for all $x$, would that suffice? I'm also unsure, given these things, how I would deduce $\phi\in W^{1,p}((−2, 2))$ for all $p \in [1, \infty]$ however I ought to understand showing $\phi$ is $L^p$ first.
As Abolfazl Charman motlagh pointed out in the comments, you can use that $0\leqslant \phi \leqslant 1$, to easily obtain that $\phi \in L^p((-2,2)$. Alternatively, you could just directly calculate that \begin{align*} \int_{-2}^2 \vert \phi(x)\vert^p \, dx &= 2\int_0^1 (1-x)^p \, dx = \frac 2{p+1}<+\infty. \end{align*} To show that $\phi$ is in $W^{1,p}((-2,2))$ from first principles, we first must find its weak derivative. To this end, let $\eta \in C^\infty_0((-2,2))$ and observe that, via integration by parts, \begin{align*} \int_{-2}^2 \phi(x) \eta'(x) \, dx &= \int_{-1}^0 (1+ x ) \eta'(x) \, dx + \int_0^1 (1- x ) \eta'(x) \, dx\\ &=-\int_{-1}^0 \eta(x) \, dx + \eta(0) + \int_0^1 \eta(x) \, dx - \eta(0)\\ &= \int_{-1}^1 \operatorname{sgn}(x) \cdot \eta(x) \, dx \\ &= \int_{-2}^2 \operatorname{sgn}(x) \chi_{(-1,1)}(x) (1-\vert x \vert)_+^{p-1} \eta(x) \, dx \end{align*} where $\operatorname{sgn}(x) = 1$ if $x>0$ and $\operatorname{sgn}(x)=-1$ if $x<0$, and $\chi_{(-1,1)}$ denotes the characteristic function of $(-1,1)$. Note, we had to split into two integrals since $1-\vert x \vert$ is not differentiable across $x=0$ and integration by parts requires the functions to differentiable. Hence, the weak derivative $\phi'$ is given by $-\operatorname{sgn}(x) \cdot \chi_{(-1,1)}. $ Finally, we check that $\phi' \in L^p((-2,2))$: \begin{align*} \int_{-2}^2 \vert \phi'(x)\vert^p \, dx &= \int_{-1}^1 \, dx = 2<+\infty \end{align*} as required.
For $p=\infty$, we have that $ \| \phi\|_{L^\infty((-2,2))} = \| \phi'\|_{L^\infty((-2,2))} =1$, so $\phi \in W^{1,\infty}((-2,2))$.