Consider the 1d domain: $\Omega=(0,1)$ and the function $f(x) = x$. Then I know that $f \in W^{1,2}(\Omega) = H^{1}(\Omega)$. Because clearly $f \in L^{2}(\Omega)$ and also $f$ is weakly differentiable with $f^{\prime} \in L^{2}(\Omega)$. (I´m using the definition of $H^{1}(\Omega)$ as being the subset of functions $f$ in $L^2(\Omega)$ such that $f$ and its weak derivatives up to order 1 have a finite $L^{2}$ norm. )
Now I want to extend $f$ to $\mathbb{R}$ in the following way:
$$\tilde{f}(x) = \begin{cases} \displaystyle f(x)~~\text{if}~~ x\in \Omega \\ 0~~\text{elsewhere} \end{cases}$$ I don´t understand why $\tilde{f}$ is not in $H^{1}(\mathbb{R})$. Can someone explain this to me please?
The weak derivative of $\tilde f$ is $$\langle \tilde f' , \varphi \rangle= -\langle \tilde f , \varphi' \rangle = - \int_0^1 x\varphi'(x) dx$$
w<here the first equaltiy comes from the definition and the second one from the fact that $\tilde f$ is locally integrable. Then we have
$$\langle \tilde f' , \varphi \rangle= -[x\varphi(x)]_0^1 +\int_0^1 \varphi(x) dx=-\delta_1(\varphi)+\int_0^1 \varphi(x) dx$$
with $\delta_a$ the Dirac distribution at $a$, which does not belong to $L^2$.
Further explanation : when you were working in $(0,1)$, the test functions $\varphi$ had to vanish before reaching $1$, with the extension, it is no longer the case and the discontinuity of $\tilde f$ introduces the Dirac when taking the weak derivative.