Finding weak derivative of a piecewise function

Question

Consider the following family of functions \begin{equation} f_n(x)=\left\{\begin{array} \{ 0 & if\;x<0\\ nx & if\;0\leq x\leq 1/n\\ 1\;& if \;1/n<x \end{array}\right. \end{equation} defined for any $n\in\mathbb{N}$. Show that the weak derivative $fn'$ of $f_n$ satisfies \begin{equation} f_n'(x)=\left\{\begin{array} \{ 0 & if\;x<0\\ n & if\;0\leq x\leq 1/n\\ 0\;& if \;1/n<x \end{array}\right. \end{equation} Now, I know that $f$ admits a weak derivative of order $\alpha$ iff there exists a function $g$ such that $\forall$ $\phi\in C_{0}^{|\alpha|}(\Omega)$, $$\int_{\Omega}f\frac{\partial^{|\alpha|}\phi}{\partial x_1^{\alpha_1}\cdot\cdot\cdot\partial x_{N}^{\alpha_N}}dx=(-1)^{|\alpha|}\int_{\Omega}g\phi dx.$$ So to find $f_n'$, we let $\phi\in C_{0}^1([-1,1])$. We can define the domain $\Omega$ into two subdomains, $\Omega_1=[-1,0]$, and $\Omega_2=[0,1]$. Now if support of $\phi\subset\Omega_1$ then \begin{equation} \int_{\Omega}f\frac{\partial \phi}{\partial x}dx=\int_{-1}^{0}f\frac{\partial \phi}{\partial x}dx=0 \end{equation} And if support of $\phi\subset\Omega_1$ then we have \begin{equation} \int_{\Omega}f\frac{\partial \phi}{\partial x}dx=\int_{0}^{1}f\frac{\partial \phi}{\partial x}dx=\int_{0}^{1}\frac{\partial \phi}{\partial x}dx=\phi(1)-\phi(0)=0 \end{equation} Here I am not sure if the last line is justified, since $\phi\in C_{0}^1([-1,1])$, it should be that $\phi(1)=0$, but I am not sure about $\phi(0)$. As well, for the in between case $0\leq x\leq 1/n$, do I consider this case as $supp(\phi)\cap\Omega_1\neq0$ and $supp(\phi)\cap\Omega_2\neq0$? Or am I overcomplicating it, and should just consider the endpoints of the integral to be how $x$ is defined piecewise.

Answer 1

No you don't want to split $\Omega$ into two separate region. If $\phi \in C^\infty_0((-1,1))$ then the definition of $f_n$ and integration by parts give \begin{align*} \int_{-1}^1 f_n(x) \phi' (x) \,dx &= n\int_0^{1/n} x \phi '(x) \,dx+\int_{1/n}^1 \phi '(x) \,dx \\ &= -n\int_0^1 \phi (x) \,dx + n\big ( (1/n)\cdot \phi(1/n)-0\cdot \phi(0)\big ) + \phi(1)-\phi(1/n)\\ &=-n\int_0^{1/n} \phi (x) \,dx + \phi(1/n)-\phi(1/n) \\ &=-n\int_0^{1/n} \phi (x) \,dx \\ &= -\int_{-1}^1 g (x) \phi(x)\, dx \end{align*} where $$g(x) = \begin{cases} 0, &\text{if } -1<x<0 \\ n, &\text{if } 0\leqslant x<1/n \\ 0, &\text{if } 1/n\leqslant x<1. \end{cases} $$ Notice that the middle $\phi$ terms 'cancel out' which is important and is the reason why the weak derivative of $f_n$ exists (and equals $g$). If you try to replicate this proof to find the second order weak derivative of $f_n$, you will find that $\phi$ terms don't cancel out so $f_n$ is not twice weakly differentiable.