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
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 $$ \int_{\mathbb R} u'(x) \phi(x) \, dx = \lim_{h \to 0} \int_{\mathbb R} \frac{u(x+h) - u(x)}{h} \phi(x) \, dx.$$