TITLE: A property of sobolev spaces
QUESTION [2 upvotes]: Let $W^{k,p}(\Omega):=\{y\in L^p(\Omega) : D^{\alpha}y\in L^p(\Omega)$ for all $|\alpha|\leq k\}$
I want to prove now that:
(1) $u \in W^{1,2}(\mathbb R)$ 
is equivalent to
(2) $u \in L^2(\mathbb R)$ and there exists $v\in L^2(\mathbb R)$ for all $\phi(x)\in C^{\infty}_c(\mathbb R)$ such that
$\int_{\mathbb R}\frac{u(x+h)-u(x)}{h}\phi(x)dx\rightarrow\int_{\mathbb R}v(x)\phi(x)dx$ for $h \rightarrow 0$
I actually don't know how to start.. So any hints which can help me is much appreciated

REPLY [5 votes]: This just uses the definition of weak derivative. The trick is to notice that if $u,\phi$ are both in $L^2(\mathbb R)$ then
$$ \int_{\mathbb R} \frac{u(x+h) - u(x)}{h} \phi(x) \, dx =  \int_{\mathbb R} u(x) \frac{\phi(x-h) - \phi(x)}{h} \, dx \tag{$\ast$}.$$
If (2) holds, you can use the dominated convergence theorem and ($\ast$) to see that
$$ \int_{\mathbb R} u(x) \phi'(x) \, dx = - \lim_{h \to 0} \int_{\mathbb R} u(x) \frac{\phi(x-h) - \phi(x)}{h} \, dx = - \int_{\mathbb R} v(x) \phi(x) \, dx.$$
This implies that $u' = v$ in the weak sense and thus $u \in W^{1,2}(\mathbb R)$.
On the other hand, if $u \in W^{1,2}(\Omega)$ and $\phi \in C_c^\infty(\mathbb R)$ then $$\int_{\mathbb R} u'(x) \phi(x) \, dx = - \int_{\mathbb R} u(x) \phi'(x) \, dx = \lim_{h \to 0} \int_{\mathbb R} u(x) \frac{\phi(x-h) - \phi(x)}{h} \, dx.$$
In this case, ($\ast$) implies
$$ \int_{\mathbb R} u'(x) \phi(x) \, dx = \lim_{h \to 0} \int_{\mathbb R} \frac{u(x+h) - u(x)}{h} \phi(x) \, dx.$$
{"set_name": "stack_exchange", "score": 2, "question_id": 1060597}