1
$\begingroup$

I apologise if this is indeed classical but my functional analysis is quite rusty...

My work recently led me to the norm: $(\|u\|_p)^p=\int_D (|u|^p+|\Delta u|^p)d\lambda$ where $D$ is the unit disk in the plane. Is the completion of the space of smooth functions on $D$ with respect to this norm a known Sobolev space ? If not is there a reference where this function space is studied ? I'm particularly interested into embeddings into other function spaces.

(Remark: I'm asking this because it works for $p=2$, using integration by parts and Cauchy-Schwarz inequality $\|\ \|_2$ is equivalent to the usual norm on $W^{2,2}(D)$.)

EDIT:

As pointed in the comments, my remark is not quite right. It however remains almost true: if one takes the completion of compactly supported functions on $D$, then one gets the Sobolev space $W^{2,2}_0(D)$. I should add that I am mostly interested in the case $p=1$.

Improve this question
$\endgroup$
5
  • $\begingroup$ These are just equivalent to $W^{2, p}$ by usual elliptic estimates in $L^p$, if I am not mistaken... $\endgroup$
    – Matthias Ludewig
    Commented Jun 21, 2014 at 7:49
  • 2
    $\begingroup$ I think your claim for $p=2$ is false. For instance let $ x_m ->x_0$ with $ x_0 \in \partial D$ with $|x_m|>1$. Set $ u_m(x)=|x|^{2-N}$ (assume $N >2$). Then for certain $N$ $ u_m$ is bounded in your Sobolev space but its not bounded in $W^{2,2}(D)$ or even in $H^1(D)$ (i think) depending on the dimension. Your claim is correct (via elliptic regularity) if you assume some appropriate boundary conditions $\endgroup$
    – Craig
    Commented Jun 21, 2014 at 8:52
  • 1
    $\begingroup$ edit. $u_m$ should be given by $u_m(x)=|x-x_m|^{2-N}$. $\endgroup$
    – Craig
    Commented Jun 21, 2014 at 8:54
  • $\begingroup$ edit (again..sorry) Just seeing the ``plane'' $N=2$ comment now. I think taking $ u_m(x)=\log(|x-x_m|)$ may still show the result is false but I am not sure now...sorry for the rapid fire comments $\endgroup$
    – Craig
    Commented Jun 21, 2014 at 12:25
  • $\begingroup$ @Craig : I believe your counterexample $u(x)=\ln |x-1|$ is correct: Clearly, $\|u\|_p<\infty$ for all $p<\infty$, but it seems $u_x, u_y\in L^p$ only if $p< 2$. I would suggest you make this an answer. (By the way, you can delete comments that are no longer relevant.) $\endgroup$
    – Christian Remling
    Commented Jun 21, 2014 at 17:04

0

Reset to default

You must log in to answer this question.