Why is the following true? (Cauchy sequence, Sobolev norm, weak derivative, $L^p$) If $v_j$ is a Cauchy sequence w.r.t. the Sobolev norm , then the weak derivative of the sequence , i.e. $D^\alpha v_j$ is a Cauchy sequence wrt $L^p$ norm . 
Can anyone tell me why it's true?
Here norm is defined as $\displaystyle||u||=(\sum{||D^\alpha u||}^p _{L^p(\Omega)})^{1/p}$

The Sobolev norm is more or less the sum of the $L^p$ norms of the function and it's weak derivatives (up to an order depending on the Sobolev space). Thus $||Du||_p \leq ||u||_{W^{k,p}}$, and so a Sobolev-cauchy sequence of functions have cauchy weak derivatives (up to an order depending on the Sobolev space).
In particular, $||Du - Dv||_p \leq ||u - v||_{W^{k,p}}$ if $k \geq 1$, so if the larger is cauchy, then the smaller is cauchy too.
