Sobolev space is complete - confusion about one step of the proof

78 Views Asked by At

In the proof of why a Sobolev space $W^{k, p} (U)$ is complete (for example, in Evans), I understand why we have $||D^\alpha u|| \leq ||u||_{W^{k, p} (U)}$.

What I am confused about is how that implies that if $(u_i)_{i = 1}^\infty$ is Cauchy in $W^{k, p}(U)$, then $(D^\alpha u_i)_{i = 1}^\infty$ is Cauchy in $L^p(U)$ for $|\alpha| \leq k$.

1

There are 1 best solutions below

0
On BEST ANSWER

$||D^{\alpha}(u_m-u_n)||_{L^p}\leq ||u_m-u_n||_{W^{k,p}}<\varepsilon$ for m and n sufficiently large.