Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

If $v_j$ is a cauchy sequence w.r. to the Sobolev norm , then the weak derivative of the sequence , ie $D^\alpha v_j$ is a cauchy sequence wrt $L^p$ norm . Can anyone tell me why its true?
Here norm is defined as $\displaystyle||u||=(\sum{||D^\alpha u||}^p _{L^p(\Omega)})^{1/p}$ Thanks

share|improve this question
You should probably add a few details on your definitions. What Sobolev space are you considering? Are you completing $C_{c}^\infty$ with respect to some norm (which is usually written as $H^{k,p}$ -- then there's actually something to prove here) or are you defining the Sobolev space as those $L^p$ functions whose weak derivatives exist and are $p$-integrable (usually written as $W^{k,p}$ and in that case there's not much to say except noting that the Sobolev norm dominates the $p$-norm of the weak derivatives)? – t.b. May 22 '12 at 5:58
@t.b. exactly , i am dealing with the weak derivative one . What do you mean by saying that Sobolev norm dominates the p-norm ? can u explain a bit . – Theorem May 22 '12 at 6:01
Well, write down the norm! – t.b. May 22 '12 at 6:01
1  
All I'm saying is that for $|\beta| \leq k$ we have $\|D^\beta u\|_{L^p} \leq \left(\sum_{|\alpha| \leq k} \|D^\alpha u\|_{L^p}^p \right)^{1/p} = \|u\|_{W^{k,p}}$. From this you should see that if $(u_n)_{n=1}^\infty$ is a Cauchy sequence with respect to the Sobolev norm then $\|D^\beta u_n - D^\beta u_m\|_{L^p} \leq \|u_n - u_m\|_{W^{k,p}}$, so the sequence $(D^\beta u_n)_{n=1}^\infty$ is an $L^p$-Cauchy sequence. – t.b. May 22 '12 at 6:12
@t.b. thank you . I am not able to edit the norm in the Latex properly . i am trying. – Theorem May 22 '12 at 6:34
show 2 more comments

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.