I am trying to show $\chi_{B_R(0)}(x) \notin H^1 (\mathbb{R}^n)$ , ∀R>0.
since $H^1 (\mathbb{R}^n) := W^{1,2}(\mathbb{R}^n)$
That is, I have to show that $\chi_{B_R(0)} (x) \notin L^2(\mathbb{R}^n)$ and $\nexists g_1,g_2....g_n \in L^2(\mathbb{R}^n)$ such that
$-\int_{B_R(0)} \chi_{B_R(0)}(x) \frac{\partial \phi(x)}{\partial x_i} dx= \int_{B_R(0)} \phi(x) g_i(x) dx, \forall \phi \in C_c^\infty(\mathbb{R}^n)$. any ideas..
Hint: What is the derivative of a step function? Is it a function? Can a functional act on a functional?
Example: Look at $\chi_{[-R,R]}$ for $\mathbb{R}$.
Clearly $ \chi \in L^2(\mathbb{R})$, since
$$ \int_\mathbb{R} \chi_{[-R,R]}^2 dx = \int_{-R}^R dx = 2R < \infty$$
But what about $\partial_x \chi$ ?
$$\int_\mathbb{R} (\partial_x \chi )^2 dx = \chi \partial_x \chi \big |_\mathbb{R} - \int_{\mathbb{R}} \chi \partial_x^2 \chi dx = \chi \underbrace{\partial_x \chi }_{\delta \, like}\big |_\mathbb{R}- \int_{-R}^R \partial_x^2 \chi dx = \infty$$
Therefore $\chi \not \in H^1(\mathbb{R})$