# $W_0^{1,p}$ norm bounded by norm of Laplacian

Let $f\in W_0^{1,p}(U)$, for $U$ a bounded domain and $p < n/(n-1)$. I am trying to prove that there is an inequality of the form

$$\|f\|_{W^{1,p}} \leq C \int_{\Omega} |\Delta f|$$

where $\Delta$ is the Laplacian. I tried applying the Sobolev embedding theorem, followed by the inequality

$$\|D^2 f\|_p \leq C \|\Delta f\|_p$$

but this does not hold for $p = 1$. I also tried using elliptic estimates, also to no avail. Does anyone have any suggestions?

EDIT: Perhaps it is possible to deduce this from the $L^p$ inequality

$$\|D^2 f\|_{L^p} \leq C \|\Delta f\|_{L^p}$$

for $u\in W_0^{2,p}(\Omega)$, $1 < p < \infty$?

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Do you mean $\|f\|_{W^{1,n/(n-1)}}$? –  saz Nov 11 '12 at 9:23
Yes, sorry. Fixed. –  user15464 Nov 11 '12 at 12:48
I still don't see why there should exist an inequality of this form. Let $U:=(0,1)$ and $f(x) := x$, then $\Delta f = 0$, but $\|f\|_{W^{1,\frac{n}{n-1}}} >0$ ...? –  saz Nov 11 '12 at 14:52
Some condition is clearly needed so that the solution of $\Delta u = f$ becomes unique. –  Hans Engler Nov 11 '12 at 14:57
Oh, sorry, my mistake. I have corrected the question to state that $f\in W_0^{1,p}(\Omega)$ for $p < n/(n-1)$. –  user15464 Nov 11 '12 at 15:13
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## 2 Answers

I don't think this estimate is correct. Consider the case where $U$ is the unit disk $U = B_1(0)$ in $\mathbb{R}^2$ and set $f_\epsilon(x) = \log \frac{|x| + \epsilon}{ 1 + \epsilon}$ for $\epsilon > 0$. These functions all vanish on the boundary of $U$.

Setting $r = |x|$, one computes $\nabla f_\epsilon(x) = \frac{x}{r(r+\epsilon)}$ and $\Delta f_\epsilon(x) = \frac{\epsilon}{r(r+\epsilon)^2}$. Using polar coordinates, it is straight forward to show that $\|\Delta f_\epsilon\|_{L^1}$ remains bounded but $\|f_\epsilon\|_{W^{1,2}} \sim \sqrt{\log |\epsilon|}$ as $\epsilon \to 0$.

$L^p$ theory for elliptic equations does not extend to the case $p = 1$.

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I modified the statement to require $p < n/(n-1)$. Is it still false in this case? –  user15464 Nov 11 '12 at 15:34
In that case it's a subelliptic estimate and most likely correct. –  Hans Engler Nov 11 '12 at 15:48
Any ideas how to prove it? –  user15464 Nov 11 '12 at 15:59
See estimate (1.1) in www.math.unipr.it/~lunardi/Files/MetaLuna.pdf . –  Hans Engler Nov 11 '12 at 17:26
Can someone give a more detailed explanation of how to derive this estimate on an arbitrary domain (not all of $\mathbb R^n$)? –  user15464 Nov 11 '12 at 23:36
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The domain (variously called $U$ and $\Omega$ in the question) is not very important here. Since $f\in W^{1,p}_0(U)$, the extension of $f$ by zero belongs to $W^{1,p}_0(\mathbb R^n)$. Let $k(x)=c_n|x|^{2-n}$ be the fundamental solution of $\Delta$ on $\mathbb R^n$. The function $g=\Delta f$ belongs to $L^1(\mathbb R^n)$ by assumption. Therefore, the convolution $g*k$ vanishes at infinity. Since $f-g*k$ is harmonic, we have $f=g*k$.

Next, differentiation $\nabla f = g*(\nabla k)$ is legitimate because $\nabla k$ is also an integrable kernel (we do not get into singular integrals here). Most importantly, $|\nabla k|\approx |x|^{1-n}\in L^p$ in the specified range of $p$. Young's inequality for convolution yields $\|\nabla f\|_p\le \|g\|_1 \|\nabla k\|_p$. $\quad\Box$

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