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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

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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
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

1 Answer 1

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.

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