I want to find the weak derivative of $f(x)=1$ for $x\in(0,1)$ and $f(x)=0$ for $x\in(1,2)$. So basically it is constant ae.
I was expecting the weak derivative to be $0$. However, when calculating, I'm getting a different answer. Is the weak derivative of $f$ not $0$?
You can use the following
Lemma: Let $u'$ be the weak derivative of $u$ on $(a,b)$. Then for all intervals $(\alpha, \beta) \subset (a,b)$ it holds that $u'|_{(\alpha, \beta)}$ is also the weak derivative of $u|_{(\alpha, \beta)}$ on $(\alpha, \beta)$.
Proof. Let $(\alpha, \beta) \subset (a,b)$ and $\phi \in \mathcal{C}_{\text{c}}^{\infty}(\alpha, \beta)$ and define the trivial extension of $\phi$ by $\tilde{\phi} \in \mathcal{C}_{\text{c}}^{\infty}(a,b)$. Then, we conclude \begin{equation*} \int_{\alpha}^{\beta} u(x) \phi'(x) dx = \int_{a}^{b} u(x) \tilde{\phi}'(x) dx = - \int_{a}^{b} u'(x) \tilde{\phi}(x) dx = - \int_{\alpha}^{\beta} u'(x) \phi(x) dx, \end{equation*} which implies the proposition.$\ \square$
Back to your question: This implies, as you rightly point out, that the only candidate for the weak derivative of your function has to be \begin{equation*} v(x) = \begin{cases} 0, & \text{if } x \in (0,1), \\ 0, & \text{if } x \in (1, 2) \end{cases} \end{equation*} If $v$ is the weak derivative of $u$, for all test functions $\phi \in \mathcal{C}_{\text{c}}^{\infty}(0,2)$ the following has to hold: \begin{equation*} \int_{0}^{2} u(x) \phi'(x) dx = - \int_{0}^{2} v(x) \phi(x) dx. \end{equation*} Now, \begin{equation*} \int_{0}^{2} f(x) \phi'(x) = \int_{0}^{1} \phi'(x) = \phi(1) - \phi(0) = \phi(1) \neq 0 = - \int_{0}^{2} v(x) \phi(x). \end{equation*} for all test functions with $\phi(1) \neq 0$, which I'm sure you believe exist. Therefore, your function is not weakly differentiable.
