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Let $W^{1,2}(\mathbb D)$ be the complex valued Sobolev space on $\mathbb D$ where $\mathbb D $ is the open unit disk of the complex plane. By definition, $W^{1,2} (\mathbb D)$ is the set of all complex valued locally integrable functions in the open unit disc such that $f \in L^2 (\mathbb D)$ and the weak derivatives $\frac{\partial f}{\partial z}$ , $\frac{\partial f}{\partial \bar z}$ are both in $L^2 (\mathbb D)$.

Since the boundary of $\mathbb D$ is the unit circle $\mathbb T = \{z \in \mathbb C : \lvert z \rvert=1\}$, there is a trace operator $T : W^{1,2} (\mathbb D) \to L^2 (\mathbb T)$ which is bounded operator and $Tf = f|_{\mathbb T}$ whenever $f\in C(\overline {\mathbb{D}}) \cap W^{1,2} (\mathbb D)$.

Is it true that the Poisson integral of $Tf$ is in $W^{1,2} (\mathbb D)$ whenever $f \in W^{1,2} (\mathbb D)$? I suspect that this is true, however, I have not been able to prove this. Hints would be appreciated.

Are there any references which involve the study of properties of traces and Poisson integrals? Any suggestions would be appreciated!

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  • $\begingroup$ This doesn't follow by uniqueness of the Dirichlet problem ? $\endgroup$
    – Ayman Moussa
    Commented Jul 20 at 10:18
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    $\begingroup$ @AymanMoussa I do not see it. Can you clariy? $\endgroup$
    – ash
    Commented Jul 20 at 10:50
  • $\begingroup$ Sorry, I think I missed part of the difficulty when answering the comment ! $\endgroup$
    – Ayman Moussa
    Commented Jul 20 at 21:31

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Yes, this is true. If $g=\sum a_n e^{in\varphi}$ is the Fourier expansion of a $g\in L^2(T)$, then $Pg=\sum a_n r^{|n|}e^{in\varphi}$. Observe first of all that $f\in W^{1,2}(D)$ if and only if $\partial f/\partial r$, $(1/r)\partial f/\partial\varphi\in L^2(D)$, as we see from the formulae $$ \frac{1}{r}\frac{\partial f}{\partial\varphi}=\frac{x}{r}\frac{\partial f}{\partial y}-\frac{y}{r}\frac{\partial f}{\partial x} $$ etc. relating these derivatives to those with respect to the Euclidean coordinates.

We can now do the area integration in polar coordinates, doing the $d\varphi$ integration first. The exponentials are orthogonal, so for example $$ \|(1/r)\partial_{\varphi} Pg\|^2 = \sum \frac{n^2}{|n|+1}|a_n|^2 , $$ and similarly $\|\partial_r Pg\|^2 = \sum_{n\not= 0} |n|\,|a_n|^2$.

Thus $Pg\in W^{1,2}$ if and only if $\sum |na_n^2|<\infty$, or, equivalently, $g\in W^{1/2,2}(T)$. Luckily for us, this space is exactly the image of the trace operator; see here.

So in fact $Pg\in W^{1,2}$ if and only if $g$ is the trace of an $f\in W^{1,2}$.

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  • $\begingroup$ What would be an explicit example of such an $W^{1,2} (\mathbb D)$? I have tried for quite a while now. $\endgroup$
    – ash
    Commented Jul 24 at 15:09
  • $\begingroup$ I'm not sure that can be done very explicitly. If you follow my answer, then you need to start out with a $g\in W^{1/2,2}(T)$, $g\notin W^{1,2}(T)$ and extend it to an $f\in W^{1,2}(D)$, or, put differently, apply the (right) inverse trace operator to $f$. If you search for these or similar keywords, you'll find more information online. $\endgroup$
    – Christian Remling
    Commented Jul 24 at 16:06
  • $\begingroup$ I did the computation that you suggested and it turns out that $\int_\mathbb{D} \lvert \partial _t Pg \rVert ^2 dA(r,t) = \sum_{n\in \mathbb Z} \frac{n^2}{\lvert n \rvert + 1} \abs{a_n}^2$. The aforementioned series converges iff $\sum \abs{n} \abs{a_n}^2$ converges. Can you please check? $\endgroup$
    – ash
    Commented Jul 24 at 18:49
  • $\begingroup$ @ash: Indeed, I miscalculated, fixed now. $\endgroup$
    – Christian Remling
    Commented Jul 24 at 21:41

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