5
$\begingroup$

In the context of a second-order linear elliptic PDE with smooth coefficients for a function $u: \mathbb{R}^n \to \mathbb{R}$, the interior $W^{k,p}$ regularity theorems I have seen in the literature apply only if $1<p<\infty$. I am interested in the case $p=1$.

Here is a simple example (essentially Theorem B.3.2 in McDuff/Salamon: $J$-holomorphic curves and symplectic topology), the PDE $\Delta u = \text{div}X$ for an $L^p$ vector field $X=(f_1,\dots,f_n)$ on $\Omega\subseteq\mathbb{R}^n$:

Theorem. Let $1<p<\infty$, let $\Omega\subseteq\mathbb{R}^n$ be open. Let $u\in L^1_{\text{loc}}(\Omega,\mathbb{R})$ and $f_1,\dots,f_n\in L^p_{\text{loc}}(\Omega,\mathbb{R})$ satisfy $$ \int_{\Omega}u(x)\Delta\phi(x)\textrm{d}x = -\sum_{i=1}^n\int_{\Omega}f_i(x)\partial_i\phi(x)\textrm{d}x $$ for all $\phi\in C^\infty_0(\Omega,\mathbb{R})$. Then $u\in W^{1,p}_{\text{loc}}(\Omega,\mathbb{R})$.

Is this still true for $p=1$? If not, what would be a counterexample?

[I have edited the rest of the question.]

If you prefer, pick your favourite integers $k\geq0$ and $n\geq2$ and prove or disprove that every $u\in W^{k,1}_{\text{loc}}(\mathbb{R}^n,\mathbb{R})$ with $\Delta u\in W^{k,1}_{\text{loc}}(\mathbb{R}^n,\mathbb{R})$ lies in $W^{k+2,1}_{\text{loc}}(\mathbb{R}^n,\mathbb{R})$.

(As proved in Ornstein 1962, the elliptic estimate $||u||_{W^{2,p}} \leq C(||u||_{L^p} +||\Delta u||_{L^p})$ fails for $p=1$. This suggests that there exists a $u\in L^1_{\text{loc}}$ with $\Delta u\in L^1_{\text{loc}}$ and $u\notin W^{2,1}_{\text{loc}}$, but it is not obvious to me how to show that. Another thing that fails for $p=1$ is the surjectivity of $\Delta: W^{2,p}\to L^p$; see 2.1 in Bourgain/Brezis 2002 for an even stronger statement.)

If (as I expect) $u\in W^{k,1}_{\text{loc}}$ and $\Delta u\in W^{k,1}_{\text{loc}}$ do not imply $u\in W^{k+2,1}_{\text{loc}}$, what is the best regularity of $u$ we can deduce in general?

For instance, the Sobolev embedding $W^{k,1}_{\text{loc}} \subset W^{k-1,n/(n-1)}_{\text{loc}}$ and the $p>1$ regularity theory imply $u\in W^{k+1,n/(n-1)}_{\text{loc}}$. I expect that the same idea with fractional Sobolev spaces works as well and yields $u\in W^{k+2-\varepsilon,n/(n-\varepsilon)}_{\text{loc}}$ for every $\varepsilon>0$. Unfortunately I have not found references for the theorems needed for this conclusion. Is it true? Can one get even slightly more regularity?

Where are all these questions discussed in the literature?

$\endgroup$
8
  • 2
    $\begingroup$ @MichaelRenardy Why not write an answer giving or sketching a counterexample, or addressing the follow-up questions? $\endgroup$
    – Yemon Choi
    Dec 27, 2016 at 5:04
  • 2
    $\begingroup$ Or provide a reference. $\endgroup$
    – Deane Yang
    Dec 27, 2016 at 5:55
  • 1
    $\begingroup$ @M. Renardy: That the elliptic estimates do not hold for p=1 is good to know; a reference would be very helpful. But this information alone does not even answer my first question. It is conceivable (but admittedly seems unlikely) that every solution u lies in $W^{1,1}$ even when no uniform elliptic estimate forces it to do so. It all depends on why/how the estimates fail. $\endgroup$ Dec 28, 2016 at 1:05
  • 2
    $\begingroup$ The reference for the failure of the $p=1$ elliptic estimate is D. Ornstein: A non-inequality for differential operators in the $L_1$ norm, doi:10.1007/BF00253928, link. The analogue for $p=\infty$ is K. de Leeuw / H. Mirkil: A priori estimates for differential operators in $L_\infty$ norm, link. $\endgroup$ Dec 29, 2016 at 2:07
  • 1
    $\begingroup$ @MarcNardmann If $\Delta$ is surjective, then there is a uniform estimate. This follows from applying the open mapping theorem (en.wikipedia.org/wiki/…) to $\Delta$. $\endgroup$
    – Fan Zheng
    Jul 28, 2017 at 6:19

1 Answer 1

1
$\begingroup$

The problem is indeed coming form the fact that singular integrals, such as the Hilbert transform, although bounded on $L^p$ for $1<p<+\infty$ are failing to be bounded on $L^1$ or $L^\infty$.

However, a good substitute for $L^1(\mathbb R^n)$ could be the Hardy space $\mathcal H^1(\mathbb R^n)$, defined as the subspace of $u\in L^1(\mathbb R^n)$ such that $$ R_j u\in L^1(\mathbb R^n),\quad\text{where $ R_j $ is the Fourier multiplier $\xi_j/\vert \xi\vert$.} $$ Then the singular integral $\Delta^{-1}\nabla \text{div}$ should send $\mathcal H^1(\mathbb R^n)$ into itself. Note that it is a non-trivial task to localize this to a proper open subset $\Omega$ of $\mathbb R^n$, since the Hardy space is not a local space: $u\in \mathcal H^1(\mathbb R^n)$ does NOT imply $\chi u\in \mathcal H^1(\mathbb R^n)$ for a smooth compactly supported $\chi$ (in fact, a function $u\in \mathcal H^1(\mathbb R^n)$ must have a zero integral, typically a non-local condition).

Last but not least, the 1D Hilbert transform $\mathfrak h$ is the Fourier multiplier $\text{sign } \xi$ and with $u(x)=e^{-π x^2}$, you find $$ (\text{Fourier}(\mathfrak h u))(\xi)=(\text{sign } \xi) e^{-π \xi^2} $$ which is not in Fourier$(L^1(\mathbb R))$ since it is discontinuous (at 0), implying that $\mathfrak h u$ does not belong to $L^1(\mathbb R).$

$\endgroup$
1
  • $\begingroup$ In the problem that motivated my question, I am unfortunately not able to choose a space with nice properties. The better behaviour of a Hardy space does therefore not help me, I must understand as precisely as possible what happens for $L^1$. $\endgroup$ Dec 28, 2016 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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