1 Introduction and main result

In this paper we discuss the relation between the map \(u\mapsto |u|\) and the Dirichlet Laplacian. Recall that the Dirichlet Laplacian of order \(s>0\) of a function \(u\in L^2({\mathbb {R}}^n)\), \(n\ge 1\), is the distribution

where

is the Fourier transform in \({\mathbb {R}}^n\). The Sobolev–Slobodetskii space

$$\begin{aligned} H^s({\mathbb {R}}^n) =\{u\in L^2({\mathbb {R}}^n)~|~\left( -\Delta _{{\mathbb {R}}^n}\!\right) ^{\!\frac{s}{2}}\!u\in L^2({\mathbb {R}}^n)~\} \end{aligned}$$

naturally inherits an Hilbertian structure from the scalar product

The standard reference for the operator and functions in \(H^s({\mathbb {R}}^n)\) is the monograph [8] by Triebel.

For any positive order \(s\notin \mathbb N\) we introduce the constant

(1)

Notice that

$$\begin{aligned} C_{n,s}>0\quad \text {if }\lfloor s\rfloor \text { is even}; \qquad C_{n,s}<0\quad \text {if }\lfloor s\rfloor \text { is odd,} \end{aligned}$$
(2)

where \( \lfloor s\rfloor \) stands for the integer part of s. It is well known that for \(s\in (0,1)\) and \(u,v\in H^s({\mathbb {R}}^n)\) one has

(3)

Let us recall some known facts about the Nemytskii operator \(|\cdot |: u\mapsto |u|\).

  1. 1.

    \(|\cdot |\) is a Lipschitz transform of \(H^0({\mathbb {R}}^n)\equiv L^2({\mathbb {R}}^n)\) into itself.

  2. 2.

    Let \(0<s\le 1\). Then \(| \cdot |\) is a continuous transform of \(H^s({\mathbb {R}}^n)\) into itself, by general results about Nemytskii operators in Sobolev/Besov spaces, see [7, Theorem 5.5.2/3]. Also it is obvious that for \(u\in H^1({\mathbb {R}}^n)\)

    Here and elsewhere \(u^{\pm }=\max \{\pm u,0\}=\frac{1}{2}(|u|\pm u)\), so that \(u=u^+-u^-\), \(|u|=u^++u^-\). On the other hand, for \(s\in (0,1)\) and \(u\in H^s({\mathbb {R}}^n)\) formula (3) gives

    (4)

    From (4) we infer by the polarization identity

    that if u changes sign then

    (5)

    We mention also [4, Theorem 6] for a different proof and explanation of (5), that includes the case when is replaced by the Navier (or spectral Dirichlet) Laplacian on a bounded Lipschitz domain \(\Omega \subset {\mathbb {R}}^n\).

  3. 3.

    Let \(1<s<\frac{3}{2}\). The results in [2] and [6] (see also Section 4 of the exhaustive survey [3]) imply that \(| \cdot |\) is a bounded transform of \(H^s({\mathbb {R}}^n)\) into itself. That is, there exists a constant c(n, s) such that

In particular, \(|\cdot |\) is continuous at \(0\in H^s({\mathbb {R}}^n)\).

It is easy to show that the assumption \(s<\frac{3}{2}\) can not be improved, see Example 1 below and [2, Proposition p. 357], where a more general setting involving Besov spaces \(B^{s,q}_p({\mathbb {R}}^n)\), \(s\ge 1+\frac{1}{p}\), is considered.

At our knowledge, the continuity of \(|\cdot |: H^s({\mathbb {R}}^n)\rightarrow H^s({\mathbb {R}}^n)\), \(s\in (1,\frac{3}{2})\), is an open problem. We can only point out the next simple result.

Proposition 1

Let \(0<\tau<s<\frac{3}{2}\). Then \(|\cdot |:H^s({\mathbb {R}}^n)\rightarrow H^\tau ({\mathbb {R}}^n)\) is continuous.

Proof

Recall that \(H^s({\mathbb {R}}^n)\hookrightarrow H^\tau ({\mathbb {R}}^n)\) for \(0<\tau <s\). Actually, the Hölder inequality readily gives the well known interpolation inequality

Since \(|\cdot |\) is continuous \(L^2({\mathbb {R}}^n)\rightarrow L^2({\mathbb {R}}^n)\) and bounded \(H^s({\mathbb {R}}^n)\rightarrow H^s({\mathbb {R}}^n)\), the statement follows immediately. \(\square \)

Now we formulate our main result. It provides the complete proof of [5, Theorem 1] for s below the threshold \(\frac{3}{2}\) and gives a positive answer to a question raised in [1, Remark 4.2] by Nicola Abatangelo, Sven Jahros and Albero Saldaña.

Theorem 1

Let \(s\in (1,\frac{3}{2})\) and \(u \in H^s({\mathbb {R}}^n)\). Then formula (4) holds. In particular, if u changes sign then

Our proof is deeply based on the continuity result in Proposition 1. The knowledge of continuity of \(|\cdot |: H^s({\mathbb {R}}^n)\rightarrow H^s({\mathbb {R}}^n)\) could considerably simplify it.

We denote by c any positive constant whose value is not important for our purposes. Its value may change line to line. The dependance of c on certain parameters is shown in parentheses.

2 Preliminary results and proof of Theorem 1

We begin with a simple but crucial identity that has been independently pointed out in [5, Lemma 1] and [1, Lemma 3.11] (without exact value of the constant). Notice that it holds for general fractional orders \(s>0\).

Theorem 2

Let \(s>0\), \(s\notin \mathbb N\). Assume that \(v,w\in H^s({\mathbb {R}}^n)\) have compact and disjoint supports. Then

(6)

Proof

Let \(\rho _h\) be a sequence of mollifiers, and put \({w}_h:={w}*\rho _h\). Formula (3) gives

Since for large h the supports of v and \(w_h\) are separated, we have

Here we can integrate by parts. Using (1) one computes for \(a>0\)

$$\begin{aligned} \Delta \, \frac{C_{n,a}}{|x-y|^{n+2a}}=\frac{C_{n,a}(n+2a)(2a+2)}{|x-y|^{n+2a+2}}=-\,\frac{C_{n,a+1}}{|x-y|^{n+2(a+1)}} \end{aligned}$$

and obtains (6) with \(w_h\) instead of w.

Since the supports of v and w are separated, it is easy to pass to the limit as \(h\rightarrow \infty \) and to conclude the proof. \(\square \)

Remark 1

Motivated by (6) and (2), A.I. Nazarov conjectured in [5] that

for any not integer exponent \(s>0\) and for any changing sign function \(u\in H^s({\mathbb {R}}^n)\) such that \(u^\pm \in H^s({\mathbb {R}}^n)\).

Lemma 1

Let \(s\in (1,\frac{3}{2})\) and \(\varepsilon >0\). If a function \(u\in H^s({\mathbb {R}}^n)\) has compact support then \((u-\varepsilon )^+\in H^s({\mathbb {R}}^n)\), and

Proof

Take a nonnegative function \(\eta \in \mathcal{C}^\infty _0({\mathbb {R}}^n)\) such that \(\eta \equiv 1\) on \(\mathrm{supp}(u)\). Clearly \(u-\varepsilon \eta \in H^s({\mathbb {R}}^n)\). Hence, by Item 3 in the Introduction we have that \((u-\varepsilon \eta )^+=(u-\varepsilon )^+\in H^s({\mathbb {R}}^n)\) and

The proof is complete. \(\square \)

In order to simplify notation, for \(u:{\mathbb {R}}^n\rightarrow {\mathbb {R}}\) and \(s>0\) we put

Lemma 2

Let \(s\in (1,\frac{3}{2})\) and \(u\in H^s({\mathbb {R}}^n)\cap \mathcal{C}^0_0({\mathbb {R}}^n)\). Then (4) holds, and in particular \(\displaystyle {\Phi ^s_u\in L^1({\mathbb {R}}^n\times {\mathbb {R}}^n)}\).

Proof

Thanks to Lemma 1 we have that \((u^--\varepsilon )^+ \in H^s({\mathbb {R}}^n)\cap \mathcal{C}^0_0({\mathbb {R}}^n)\) for any \(\varepsilon >0\). Next, the supports of the functions \(u^+\) and \((u^--\varepsilon )^+\) are compact and disjoint. Thus we can apply Theorem 2 to get

(7)

Take a decreasing sequence \(\varepsilon \searrow 0\). From Lemma 1 we infer that \((u^--\varepsilon )^+\rightarrow u^-\) weakly in \(H^s({\mathbb {R}}^n)\), as \((u^--\varepsilon )^+\rightarrow u^-\) in \(L^2({\mathbb {R}}^n)\). Hence the duality product in (7) converges to the the duality product in (4). Next, the integrand in the right-hand side of (7) increases to \(\Phi ^s_u\) a.e. on \({\mathbb {R}}^n\times {\mathbb {R}}^n\). By the monotone convergence theorem we get the convergence of the integrals, and the conclusion follows immediately. \(\square \)

Lemma 3

Let \(s\in (1,\frac{3}{2})\) and \(u\in H^s({\mathbb {R}}^n)\). Then \(\displaystyle {\Phi ^s_u\in L^1({\mathbb {R}}^n\times {\mathbb {R}}^n)}\).

Proof

Take a sequence of functions \(u_h\in \mathcal{C}^\infty _0({\mathbb {R}}^n)\) such that \(u_h\rightarrow u\) in \(H^s({\mathbb {R}}^n)\) and almost everywhere. Since \(\displaystyle {\Phi ^s_{u_h}\rightarrow \Phi ^s_u}\) a.e. on \({\mathbb {R}}^n\times {\mathbb {R}}^n\), Fatou’s Lemma, Lemma 2 for \(u_h\) and the boundeness of \(v\mapsto v^\pm \) in \(H^s({\mathbb {R}}^n)\) give

that concludes the proof. \(\square \)

Proof of Theorem 1

Take a sequence \(u_h\in \mathcal{C}^\infty _0({\mathbb {R}}^n)\) such that \(u_h\rightarrow u\) in \(H^s({\mathbb {R}}^n)\) and almost everywhere. Consider the nonnegative functions

$$\begin{aligned} v_h:=u_h^+\wedge u^+=u^+-(u^+-u^+_h)^+~,\quad w_h:=u_h^-\wedge u^-=u^--(u^--u^-_h)^+. \end{aligned}$$

Then \(v_h, w_h\in H^s({\mathbb {R}}^n)\). Next, take any exponent \(\tau \in (1,s)\). By Proposition 1 we have that \(u^\pm -u^\pm _h\rightarrow 0\) in \(H^\tau ({\mathbb {R}}^n)\); hence \((u^\pm -u^\pm _h)^+\rightarrow 0\) in \(H^\tau ({\mathbb {R}}^n)\) by Item 3 in the Introduction. Thus,

$$\begin{aligned} v_h\rightarrow u^+~,\quad w_h\rightarrow u^-\quad \text {in }H^\tau ({\mathbb {R}}^n)\text { and almost everywhere, } \ \text { as } \ h\rightarrow \infty . \end{aligned}$$
(8)

Now we take a small \(\varepsilon >0\). Recall that \((v_h-\varepsilon )^+ \in H^\tau ({\mathbb {R}}^n)\) by Lemma 1. Moreover, from \(0\le v_h \le u_h^+\), \(0\le w_h\le u_h^-\) it follows that

$$\begin{aligned} \text {supp}((v_h-\varepsilon )^+)\subseteq \{u_h\ge \varepsilon \};\qquad \text {supp}(w_h)\subseteq \text {supp}(u^-_h). \end{aligned}$$

In particular, the functions \((v_h-\varepsilon )^+, w_h\) have compact and disjoint supports. Thus we can apply Theorem 2 to infer

We first take the limit as \(\varepsilon \searrow 0\). The argument in the proof of Lemma 2 gives

(9)

Next we push \(h\rightarrow \infty \). By (8) we get

Further, since the integrand in the right-hand side of (9) does not exceed \(\Phi ^\tau _u(x,y)\), Lemma 3, (8) and Lebesgue’s theorem give

Thus, we proved (4) with s replaced by \(\tau \). It remains to pass to the limit as \(\tau \nearrow s\). By Lebesgue’s theorem, we have

Now we fix \(\tau _0\in (1,s)\) and notice that \(0\le \Phi ^\tau _u \le \max \{\Phi ^{\tau _0}_u,\Phi ^s_u\}\) for any \(\tau \in (\tau _0,s)\). Therefore, Lemma 3 and Lebesgue’s theorem give

The proof of (4) is complete. The last statement follows immediately from (4), polarization identity and (2). \(\square \)

Example 1

It is easy to construct a function \(u\in \mathcal{C}^\infty _0({\mathbb {R}}^n)\) such that \(u^+\in H^s({\mathbb {R}}^n)\) if and only if \(s<\frac{3}{2}\).

Take \(\varphi \in \mathcal{C}^\infty _0({\mathbb {R}})\) satisfying \(\varphi (0)=0, \varphi '(0)> 0\) and \(x\varphi (x)\ge 0\) on \({\mathbb {R}}\). By direct computation one checks that \(\varphi ^+=\chi _{(0,\infty )}\varphi \in H^s({\mathbb {R}})\) if and only if \(s<\frac{3}{2}\). If \(n=1\) we are done. If \(n\ge 2\) we take \(u(x_1,x_2,\dots , x_n)=\varphi (x_1)\varphi (x_2)\dots \varphi (x_n)\).