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?

  • 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
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    $\begingroup$ Or provide a reference. $\endgroup$
    – Deane Yang
    Dec 27, 2016 at 5:55
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ @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


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 $L^1(\mathbb R).$

  • $\begingroup$ 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

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