On the maximum principle for the Riesz transform

TL;DR

This paper proves a maximum principle for the Riesz transform in certain cases and demonstrates the conjecture's failure for non-positive measures, advancing understanding of Riesz transforms in harmonic analysis.

Contribution

The paper establishes the maximum principle for the Riesz transform when 0<s<1 and for radial measures when 0<s<d, and shows the conjecture fails for non-positive measures.

Findings

Maximum principle proven for 0<s<1.

Maximum principle proven for radial measures when 0<s<d.

Conjecture invalid for non-positive measures.

Abstract

Let $\mu$ be a measure in $\mathbb R^d$ with compact support and continuous density, and let $$ R^s\mu(x)=\int\frac{y-x}{|y-x|^{s+1}}\,d\mu(y),\ \ x,y\in\mathbb R^d,\ \ 0<s<d. $$ We consider the following conjecture: $$ \sup_{x\in\mathbb R^d}|R^s\mu(x)|\le C\sup_{x\in\text{supp}\,\mu}|R^s\mu(x)|,\quad C=C(d,s). $$ This relation was known for $d-1\le s<d$, and is still an open problem in the general case. We prove the maximum principle for $0< s<1$, and also for $0<s<d$ in the case of radial measure. Moreover, we show that this conjecture is incorrect for non-positive measures.

Equations

$$R^s\mu (x)=\int \frac{y-x}{|y-x{|}^{s+1}}d\mu (y),x,y\in {\mathbb{R}}^d,0<s<d.$$

$$\underset{x\in {\mathbb{R}}^d}{\sup }|R^s\mu (x)|\le C\underset{x\in \text{supp}\mu }{\sup }|R^s\mu (x)|,C=C(d,s).$$

$$R_{\mu ,\epsilon }^{s}f(x)={\int }_{|y-x|>\epsilon }\frac{y-x}{|y-x{|}^{s+1}}f(y)d\mu (y),x,y\in {\mathbb{R}}^d,f\in L^2(\mu ),\epsilon >0.$$

$$R_{\mu }^{s}f(x)=\int \frac{y-x}{|y-x{|}^{s+1}}f(y)d\mu (y),$$

$$\Vert R_{\mu }^s\Vert :=\underset{\epsilon >0}{\sup }\Vert R_{\mu ,\epsilon }^s{\Vert }_{L^2(\mu )\to L^2(\mu )}<\infty .$$

$$\underset{x\in {\mathbb{R}}^d}{\sup }|R^s\mu (x)|\le C\underset{x\in \mathrm{supp}\mu }{\sup }|R^s\mu (x)|.$$

$$R^s\mu (x)=\nabla U_{\mu }^s(x),$$

$$U_{\mu }^s(x)=\frac{1}{s-1}\int \frac{d\mu (y)}{|y-x{|}^{s-1}},s\ne 1,U_{\mu }^1(x)=-\int \log |y-x|d\mu (y).$$

$$\underset{x\in {\mathbb{R}}^d}{\sup }|R_1^s\mu (x)|>M\underset{x\in \mathrm{supp}\mu }{\sup }|R_1^s\mu (x)|,$$

$$|R_1^s\mu (x^{\prime })|>A_d,\text{where }{x}^{\prime }=(1,0,\dots ,0),0<\delta <1/2.$$

$$|R_1^s\mu (x)|$$

$$=\frac{\mu (B(x,\delta ))}{{\delta }^s}+s{\int }_0^{\delta }\frac{\mu (B(x,r))}{r^{s+1}}dr+\delta (s+1){\int }_{\delta }^{\infty }\frac{\mu (B(x,r))}{r^{s+2}}dr$$

$$<C\frac{C_d}{\delta }\frac{{\delta }^d}{{\delta }^s}+\frac{Cs}{\delta }{\int }_0^{\delta }\frac{r^d}{r^{s+1}}dr+\frac{C\delta (s+1)}{\delta }{\int }_{\delta }^2\frac{r^{d-1}\delta }{r^{s+2}}dr+C\delta .$$

$$\delta {\int }_{\delta }^2\frac{r^{d-1}}{r^{s+2}}dr=\{\begin{array}{ll}\delta \ln \frac{2}{\delta },& s=d-2,\\ \frac{1}{d-s-2}(2^{d-s-2}\delta -{\delta }^{d-s-1}),& s\ne d-2.\end{array}$$

$$\bigg{|}\int_{\partial B}(R^{s}\mu\cdot\mathbb{n})\,d\sigma\bigg{|}\approx r^{d-s-1}\mu(B)+r^{d}\int_{r}^{\infty}\frac{d\mu(B(x_{0},t))}{t^{s+1}},$$

$$\int_{\partial B}(\widetilde{R}_{\varepsilon}^{s}\mu\cdot\mathbb{n})\,d\sigma=\int_{B}\nabla\cdot\widetilde{R}_{\varepsilon}^{s}\mu(x)\,dm_{d}(x)=\int_{B}\biggl{[}\int_{{\mathbb{R}}^{d}}\nabla\cdot\phi_{\varepsilon}(|y-x|)\frac{y-x}{|y-x|^{s+1}}\,d\mu(y)\biggr{]}dm_{d}(x).$$

$${\int }_{|y-x|\le 2\epsilon }\nabla \cdot {\varphi }_{\epsilon }(|y-x|)\frac{y-x}{|y-x{|}^{s+1}}d\mu (y)+{\int }_{|y-x|>2\epsilon }\nabla \cdot \frac{y-x}{|y-x{|}^{s+1}}d\mu (y)=:I_1(x)+I_2(x).$$

$$\bigg{|}\frac{\partial}{\partial x_{i}}\bigg{[}\phi_{\varepsilon}(|y-x|)\frac{y-x}{|y-x|^{s+1}}\bigg{]}\bigg{|}<C\bigg{[}\frac{1}{\varepsilon}\frac{1}{|y-x|^{s}}+\frac{1}{|y-x|^{s+1}}\bigg{]}<\frac{C}{|y-x|^{s+1}},\quad|y-x|\leq 2\varepsilon.$$

$$|I_1(x)|$$

$$\approx \frac{\mu (B(x,2\epsilon ))}{(2\epsilon {)}^{s+1}}+{\int }_0^{2\epsilon }\frac{\mu (B(x,t))}{t^{s+2}}dt.$$

$$\displaystyle\bigg{|}\int_{B}I_{2}(x)\,dm_{d}(x)\bigg{|}$$

$$\displaystyle=C\bigg{(}\int_{B(x_{0},r+2\varepsilon)}\bigg{[}\int_{B\cap\{|y-x|>2\varepsilon\}}\frac{dm_{d}(x)}{|y-x|^{s+1}}\bigg{]}d\mu(y)$$

$$\displaystyle+\int_{{\mathbb{R}}^{d}\setminus B(x_{0},r+2\varepsilon)}\bigg{[}\int_{B}\frac{dm_{d}(x)}{|y-x|^{s+1}}\bigg{]}d\mu(y)\bigg{)}=:C(J_{1}+J_{2}).$$

$${\int }_B\frac{d{m}_d(x)}{|y-x{|}^{s+1}}\approx \{\begin{array}{ll}{\int }_0^r\frac{t^{d-1}dt}{t^{s+1}}\approx r^{d-s-1},& |y-x_0|\le r,\\ \frac{r^d}{|y-x_0{|}^{s+1}},& |y-x_0|>r.\end{array}$$

$${\int }_{B\cap \{|y-x|>2\epsilon \}}\frac{d{m}_d(x)}{|y-x{|}^{s+1}}\approx {\int }_B\frac{d{m}_d(x)}{|y-x{|}^{s+1}}\approx r^{d-s-1},y\in B(x_0,r+2\epsilon ).$$

$$J_2\approx {\int }_{{\mathbb{R}}^d\setminus B(x_0,r+2\epsilon )}\frac{r^d}{|y-x_0{|}^{s+1}}d\mu (y)=r^d{\int }_{r+2\epsilon }^{\infty }\frac{d\mu (B(x_0,t))}{t^{s+1}}.$$

$$⟨R^s(\psi \mu ),\phi {⟩}_{\mu }=\frac{1}{2}{\iint }_{{\mathbb{R}}^d\times {\mathbb{R}}^d}\frac{y-x}{|y-x{|}^{s+1}}(\psi (y)\phi (x)-\psi (x)\phi (y))d\mu (y)d\mu (x);$$

$$\begin{split}\bigg{|}\int_{{\mathbb{R}}^{d}}\frac{y-x}{|y-x|^{s+1}}\psi(y)\,d\mu(y)\bigg{|}&=\bigg{|}\int_{{\mathbb{R}}^{d}}\bigg{[}\frac{y-x}{|y-x|^{s+1}}+\frac{x}{|x|^{s+1}}\bigg{]}\psi(y)\,d\mu(y)\bigg{|}\\ &\leq\frac{C}{|x|^{s+1}}\int_{{\mathbb{R}}^{d}}|y\psi(y)|\,d\mu(y)=\frac{C_{\psi}}{|x|^{s+1}}.\end{split}$$

$$\langle R^{s}(\psi\mu),\varphi\rangle_{\mu}=\langle R^{s}(\psi\mu),\xi\varphi\rangle_{\mu}+\int_{{\mathbb{R}}^{d}}\bigg{[}\int_{{\mathbb{R}}^{d}}\frac{y-x}{|y-x|^{s+1}}\psi(y)\,d\mu(y)\bigg{]}(1-\xi(x))\varphi(x)\,d\mu(x).$$

$${\int }_{|x|>2R}\frac{d\mu (x)}{|x{|}^{s+1}}\le C{\int }_{2R}^{\infty }\frac{\mu (B(0,t))}{t^{s+2}}dt\le C{\int }_{2R}^{\infty }\frac{1}{t^2}dt.$$

$$\underset{x\in {\mathbb{R}}^d}{\sup }|R^s{\mu }_n(x)|=1,\underset{x\in \mathrm{supp}{\mu }_n}{\sup }|R^s{\mu }_n(x)|\le \frac{1}{n}.$$

$${\theta }_{\mu }(x,r):=\frac{\mu (B(x,r))}{r^s},{\theta }_{\mu }:=\underset{x,r}{\sup }{\theta }_{\mu }(x,r).$$

$$0<c<{\theta }_{{\mu }_n}<C.$$

Full text

[TABLE]

We consider the following conjecture:

[TABLE]

This relation was known for $d-1\leq s<d$, and is still an open problem in the general case. We prove the maximum principle for $0<s<1$, and also for $0<s<d$ in the case of radial measure. Moreover, we show that this conjecture is incorrect for non-positive measures.

1. Introduction

Let $\mu$ be a non-negative finite Borel measure with compact support in $\mathbb{R}^{d}$, and let $0<s<d$. The truncated Riesz operator $R_{\mu,\varepsilon}^{s}$ is defined by the equality

[TABLE]

For every $\varepsilon>0$ the operator $R_{\mu,\varepsilon}^{s}$ is bounded on $L^{2}(\mu)$. By $R_{\mu}^{s}$ we denote a linear operator on $L^{2}(\mu)$ such that

[TABLE]

whenever the integral exists in the sense of the principal value. We say that $R_{\mu}^{s}$ is bounded on $L^{2}(\mu)$ if

[TABLE]

In the case $f\equiv 1$ the function $R_{\mu}^{s}1(x)$ is said to be the $s$-Riesz transform (potential) of $\mu$ and is denoted by $R^{s}\mu(x)$. If $\mu$ has continuous density with respect to the Lebesgue measure $m_{d}$ in $\mathbb{R}^{d}$, that is if $d\mu(x)=\rho(x)\,dm_{d}(x)$ with $\rho(x)\in C(\mathbb{R}^{d})$, then $R^{s}\mu(x)$ exists for every $x\in{\mathbb{R}}^{d}$.

By $C,c$, possibly with indexes, we denote various constants which may depend only on $d$ and $s$.

We consider the following well-known conjecture.

Conjecture 1.1**.**

Let $\mu$ be a nonnegative finite Borel measure with compact support and continuous density with respect to the Lebesgue measure in $\mathbb{R}^{d}$. There is a constant $C$ such that

[TABLE]

For $s=d-1$ the proof is simple. Obviously,

[TABLE]

where

[TABLE]

Thus each component of the vector function $R^{s}\mu(x),\ s=d-1$, is harmonic in ${\mathbb{R}}^{d}\setminus\operatorname{supp}\mu$. Applying the maximum principle for harmonic functions we get (1.1).

For $d-1<s<d$, the relation (1.1) was established in [2] under stronger assumption that $\rho\in C^{\infty}({\mathbb{R}}^{d})$. In fact it was proved that (1.1) holds for each component of $R^{s}\mu$ with $C=1$ as in the case $s=d-1$. The proof is based on the formula which recovers a density $\rho$ from $U_{\mu}^{s}$. But this method does not work for $s<d-1$.

The problem under consideration has a very strong motivation and also is of independent interest. In [2] it is an important ingredient of the proof of the following theorem. By ${\mathcal{H}}^{s}$ we denote the $s$-dimensional Hausdorff measure.

Theorem 1.2** ([2]).**

Let $d-1<s<d$, and let $\mu$ be a positive finite Borel measure such that ${\mathcal{H}}^{s}(\operatorname{supp}\mu)<\infty$. Then $\|R^{s}\mu\|_{L^{\infty}(m_{d})}=\infty$ (equivalently, $\|R^{s}_{\mu}\|=\infty$).

If $s$ is integer, the conclusion of Theorem 1.2 is incorrect. For $0<s<1$ Theorem 1.2 was proved by Prat [10] using different approach. The obstacle for extension of this result to all noninteger $s$ between 1 and $d-1$ is the lack of the maximum principle. The same issue concerns the quantitative version of Theorem 1.2 obtained by Jaye, Nazarov, and Volberg [3].

The maximum principle is important for other problems on the connection between geometric properties of a measure and boundedness of the operator $R_{\mu}^{s}$ on $L^{2}(\mu)$ – see for example [3], [5], [6], [7]. All these results are established for $d-1<s<d$ or $s=d-1$.

The problem of the lower estimate for $\|R^{s}_{\mu}\|$ in terms of the Wolff energy (a far going development of Theorem 1.2) which is considered in [3], [5], was known for $0<s<1$. And the results in [6], [7] are $(d-1)$-dimensional analogs of classical facts known for $s=1$ (in particular, [7] contains the proof of the analog of the famous Vitushkin conjecture in higher dimensions). For $0<s\leq 1$, the proofs essentially use the Melnikov curvature techniques and do not require the maximum principle. But this tool is absent for $s>1$.

At the same time the validity of the maximum principle itself remained open even for $0<s<1$. It is especially interesting because the analog of (1.1) does not hold for each component of $R^{s}\mu$ when $0<s<d-1$ unlike the case $d-1\leq s<d$ – see Proposition 2.1 below.

We prove Conjecture 1.1 for $0<s<1$ in Section 2 (Theorem 2.3). The proof is completely different from the proof in the case $d-1\leq s<d$. In Section 3 we prove Conjecture 1.1 in the special case of radial density of $\mu$ (that is when $d\mu=h(|x|)\,dm_{d}(x)$), but for all $s\in(0,d)$. Section 4 contains an example showing that Conjecture 1.1 is incorrect for non-positive measures, even for radial measures with $C^{\infty}$-density (note that in [14, Conjecture 7.3] Conjecture 1.1 was formulated for all finite signed measures with compact support and $C^{\infty}$-density).

2. The case $0<s<1$

We start with a statement showing that the maximum principle fails for every component of $R^{s}\mu$ if $0<s<d-1$.

Proposition 2.1**.**

For any $d\geq 2$, $0<s<d-1$, and any $M>0$, there is a positive measure $\mu$ in ${\mathbb{R}}^{d}$ with $C^{\infty}$-density such that

[TABLE]

where $R_{1}^{s}\mu$ is the first component of $R^{s}\mu$.

Proof.

Let $E=\{(x_{1},\dots,x_{d})\in{\mathbb{R}}^{d}:x_{1}=0,\ x_{2}^{2}+\dots+x_{d}^{2}\leq 1\}$, and let $E_{\delta}$, $\delta>0$, be a $\delta$-neighborhood of $E$ in ${\mathbb{R}}^{d}$. Let $\mu=\mu_{\delta}$ be a positive measure supported on $\overline{E_{\delta}}$ with $\mu(\overline{E_{\delta}})=1$ and with $C^{\infty}$-density $\rho(x)$ such that $\rho(x)<2/\text{vol}(E_{\delta})\leq C_{d}/\delta$. Then

[TABLE]

On the other hand, for $x\in\operatorname{supp}\mu$ integration by parts yields

[TABLE]

Here by $C$ we denote different constants depending only on $d$, and $B(x,r):=\{y\in{\mathbb{R}}^{d}:|y-x|<r\}$. We have

[TABLE]

Thus, all terms in the right-hand side of the estimate for $|R_{1}^{s}\mu(x)|$ tend to 0 as $\delta\to 0$, and we may choose $\delta$ and a corresponding measure $\mu$ satisfying (2.1). ∎

We need the following lemma. The notation $A\approx B$ means that $cA<B<CB$ with constants $c,C$ which may depend only on $d$ and $s$.

Lemma 2.2**.**

Let $\mu$ be a non-negative measure in ${\mathbb{R}}^{d}$ with continuous density and compact support. Let $0<s<d-1$. Then for every ball $B=B(x_{0},r)$,

[TABLE]

where $\mathbb{n}$ is the outer normal vector to $B$ and $\sigma$ is the surface measure on $\partial B$.

Proof.

We will use the Ostrogradsky-Gauss Theorem and differentiation under the integral sign. To justify these operations and make an integrand sufficiently smooth, we approximate $K(x)=x/|x|^{s+1}$ by the smooth kernel $K_{\varepsilon}$ in the following standard way. Let $\phi(t)$, $t\geq 0$, be a $C^{\infty}$-function such that $\phi(t)=0$ as $0\leq t\leq 1$, $\phi(t)=1$ as $t\geq 2$, and $0\leq\phi^{\prime}(t)\leq 2$, $t>0$. Let $\phi_{\varepsilon}(t):=\phi(\frac{t}{\varepsilon})$, $K_{\varepsilon}(x):=\phi_{\varepsilon}(|x|)K(x)$, and $\widetilde{R}_{\varepsilon}^{s}\mu:=K_{\varepsilon}\ast\mu$. We have

[TABLE]

The inner integral is equal to

[TABLE]

One can easily see that

[TABLE]

Hence,

[TABLE]

Since $\mu$ has a continuous density with respect to $m_{d}$, we have $\mu(B(x,t))<A_{\mu,B}t^{d}$ as $t\leq 2\varepsilon<1$, $x\in B$. Taking into account that $s<d-1$, we obtain the relation $\int_{B}I_{1}(x)\,dm_{d}(x)\to 0$ as $\varepsilon\to 0$.

To estimate the integral of $I_{2}(x)$ we use the equality $\nabla\cdot\frac{x}{|x|^{s+1}}=\frac{d-s-1}{|x|^{s+1}}$. Thus,

[TABLE]

Obviously,

[TABLE]

In order to estimate $J_{1}$ we note that for sufficiently small $\varepsilon$,

[TABLE]

Hence, $J_{1}\approx r^{d-s-1}\mu(B(x_{0},r+2\varepsilon))$. Moreover,

[TABLE]

Passing to the limit as $\varepsilon\to 0$, we get (2.2) ∎

Now we are ready to prove our main result.

Theorem 2.3**.**

Let $\mu$ be a non-negative measure in ${\mathbb{R}}^{d}$ with continuous density and compact support. Let $0<s<1$. Then (1.1) holds with a constant $C$ depending only on $d$ and $s$.

Proof.

Let us sketch the idea of proof. Let a measure $\mu$ be such that $\mu(B(y,t))\leq Ct^{s}$, $y\in{\mathbb{R}}^{d}$, $t>0$. For Lipschitz continuous compactly supported functions $\varphi$, $\psi$, define the form $\langle R^{s}(\psi\mu),\varphi\rangle_{\mu}$ by the equality

[TABLE]

the double integral exists since $|\psi(y)\varphi(x)-\psi(x)\varphi(y)|\leq C_{\psi,\varphi}|x-y|$. If we assume in addition that $\int\psi\,d\mu=0$, we may define $\langle R^{s}(\psi\mu),\varphi\rangle_{\mu}$ for any (not necessarily compactly supported) bounded Lipschitz continuous function $\varphi$ on ${\mathbb{R}}^{d}$; here we follow [4]. Let $\operatorname{supp}\psi\in B(0,R)$. For $|x|>2R$ we have

[TABLE]

Choose a Lipschitz continuous compactly supported function $\xi$ which is identically 1 on $B(0,2R)$. Then we may define the form $\langle R^{s}(\psi\mu),\varphi\rangle_{\mu}$ as

[TABLE]

The repeated integral is well defined because

[TABLE]

Assuming that Theorem 2.3 is incorrect and using the Cotlar inequality we establish the existence of a positive measure $\nu$ such that $\nu$ has no point masses, the operator $R_{\nu}^{s}$ is bounded on $L^{2}(\nu)$, and $\langle R^{s}(\psi\nu),1\rangle_{\nu}=0$ for every Lipschitz continuous function $\psi$ with $\int\psi\,d\mu=0$. It means that $\nu$ is a reflectionless measure, that is a measure without point masses with the following properties: $R_{\nu}^{s}$ is bounded on $L^{2}(\nu)$, and $\langle R^{s}(\psi\nu),1\rangle_{\nu}=0$ for every Lipschitz continuous compactly supported function $\psi$ such that $\int\psi\,d\mu=0$. But according to the recent result by Prat and Tolsa [11] such measures do not exist for $0<s<1$. We remark that the proof of this result contains estimates of an analog of the Melnikov’s curvature of a measure. This is the obstacle to extent the result to $s\geq 1$. We now turn to the details.

Suppose that $C$ satisfying (1.1) does not exists. Then for every $n\geq 1$ there is a positive measure $\mu_{n}$ such that

[TABLE]

Let

[TABLE]

We prove that

[TABLE]

The estimate from above is a direct consequence of Lemma 2.2. Indeed, for any ball $B(x,r)$ (2.2) implies the estimate

[TABLE]

which implies the desired inequality.

The estimate from below follows immediately from a Cotlar-type inequality

[TABLE]

(see [8, Theorem 7.1] for a more general result).

Let $B(x_{n},r_{n})$ be a ball such that $\theta_{\mu_{n}}(x_{n},r_{n})>c=c(s,d)$, and let $\nu_{n}(\cdot)=r_{n}^{-s}\mu_{n}(x_{n}+r_{n}\cdot)$. Then

[TABLE]

In particular, $\nu_{n}(B(0,1))=\theta_{\mu_{n}}(x_{n},r_{n})>c$. Choosing a weakly converging subsequence of $\{\nu_{n}\}$, we obtain a positive measure $\nu$. If we prove that

(a) $\nu(B(y,t))\leq Ct^{s}$,

(b) $\langle R^{s}\nu,\psi\rangle_{\nu}=0$ for every Lipschitz continuous compactly supported function $\psi$ with $\int\psi\,d\nu=0$,

(c) the operator $R_{\nu}^{s}$ is bounded on $L^{2}(\nu)$,

then $\nu$ is reflectionless, and we come to contradiction with Theorem 1.1 in [11] mentioned above. Thus, the proof would be completed.

The property (a) follows directly from (2.3). For weakly converging measures $\nu_{n}$ with $\theta_{\mu_{n}}<C$ we may apply Lemma 8.4 in [4] which yields (b). To establish (c) we use the inequality

[TABLE]

for any positive Borel measure $\mu$ such that $\mu(B(x,r))\leq r^{s},\ x\in{\mathbb{R}}^{d},\ r>0$, – see [12, Lemma 2] or [13, p. 47], [1, Lemma 5.1] for a more general setting. Thus, $R_{\varepsilon}^{s}\nu_{n}(x):=R_{\nu_{n},\varepsilon}^{s}1(x)\leq C$ for every $\varepsilon>0$. Hence, $R_{\varepsilon}^{s}\nu(x)\leq C$ for $\varepsilon>0$, $x\in{\mathbb{R}}^{d}$, and the non-homogeneous $T1$-theorem [9] implies the boundedness of $R_{\nu}^{s}$ on $L^{2}(\nu)$. ∎

3. The case of radial density

Lemma 2.2 allows us to prove the maximum principle for all $s\in(0,d)$ in the special case of radial density.

Proposition 3.1**.**

Let $d\mu(x)=h(|x|)\,dm_{d}(x)$, where $h(t)$ is a continuous function on $[0,\infty)$, and let $s\in(0,d-1)$. Then (1.1) holds with a constant $C$ depending only on $d$ and $s$.

We remind that for $s\in[d-1,d)$ Conjecture 1.1 is proved in [2] for any compactly supported measure with $C^{\infty}$ density. Thus, for compactly supported radial measures with $C^{\infty}$ density (1.1) holds for all $s\in(0,d)$.

Proof.

Because $\mu$ is radial, by (2.2) we have

[TABLE]

Thus,

[TABLE]

Fix $w\not\in\operatorname{supp}\mu$, and let $r=|w|$. If

[TABLE]

then there is $r_{1}\in(0,r)$ such that $\{y:|y|=r_{1}\}\subset\operatorname{supp}\mu$ and $\mu(B(0,r))=\mu(B(0,r_{1}))$. Hence,

[TABLE]

If

[TABLE]

then there is $r_{2}>r$ such that $\{y:|y|=r_{2}\}\subset\operatorname{supp}\mu$ and $\mu(B(0,r))=\mu(B(0,r_{2}))$. Hence,

[TABLE]

and we have

[TABLE]

∎

4. Counterexample

Given $\varepsilon>0$, we construct a signed measure $\nu=\nu(\varepsilon)$ in ${\mathbb{R}}^{5}$ with the following properties:

(a) $\nu$ is a radial signed measure with $C^{\infty}$-density;

(b) $\operatorname{supp}\nu\in D_{\varepsilon}:=\{1-\varepsilon\leq|x|\leq 1+\varepsilon\}$;

(c) $|R^{2}\nu(x)|<\varepsilon$ for $x\in\operatorname{supp}\nu$; $|R^{2}\nu(x)|>a>0$ for $|x|=2$, where $a$ is an absolute constant. Here $R^{2}\nu$ means $R^{s}\nu$ with $s=2$.

Let $\Delta^{2}:=\Delta\circ\Delta$, and let

[TABLE]

Note that $\Delta(\frac{1}{|x|^{3}})=0$ and $\Delta^{2}(\frac{1}{|x|})=0$ in ${\mathbb{R}}^{5}\setminus\{0\}$. Hence, $\Delta^{2}u(x)=0,|x|\neq 1$. Moreover, $\nabla u$ is continuous in ${\mathbb{R}}^{5}$ and $\nabla u(x)=0$, $|x|=1$.

For $\delta\in(0,\varepsilon)$, let $\varphi_{\delta}(x)$ be a $C^{\infty}$-function in ${\mathbb{R}}^{5}$ such that $\varphi_{\delta}>0$, $\operatorname{supp}\varphi_{\delta}=\{x\in{\mathbb{R}}^{5}:|x|\leq\delta\}$, and $\int\varphi_{\delta}(x)\,dm_{5}(x)=1$ (for example, a bell-like function on $|x|\leq\delta\}$). Let $U_{\delta}:=u\ast\varphi_{\delta}$. Then $\Delta^{2}U_{\delta}(x)=0$ as $x\not\in D_{\delta}$. Also, $\Delta U_{\delta}(x)\to 0$ as $|x|\to\infty$. Hence, the function $\Delta U_{\delta}$ can be represented in the form $\Delta U_{\delta}=c(\frac{1}{|x|^{3}}\ast\Delta^{2}U_{\delta})$ (here and in the sequel by $c$ we denote various absolute constants). Set $d\nu_{\delta}=\Delta^{2}U_{\delta}\,dm_{5}$. Then $\operatorname{supp}\nu_{\delta}\in D_{\delta}$, and (b) is satisfied. Since $\Delta(\frac{1}{|x|})=\frac{c}{|x|^{3}}$, we have $\Delta U_{\delta}=c\Delta(\frac{1}{|x|}\ast\Delta^{2}U_{\delta})$, that is $U_{\delta}=c(\frac{1}{|x|}\ast\Delta^{2}U_{\delta})+h$, where $h$ is a harmonic function in ${\mathbb{R}}^{5}$. Since both $U_{\delta}$ and $\frac{1}{|x|}\ast\Delta^{2}U_{\delta}$ tend to 0 as $x\to\infty$, we have

[TABLE]

Thus,

[TABLE]

Obviously, $\nabla U_{\delta}=\nabla(u\ast\varphi_{\delta})=(\nabla u)\ast\varphi_{\delta}$, and hence $\max_{x\in D_{\delta}}|\nabla U_{\delta}(x)|\to 0$ as $\delta\to 0$. On the other hand, for fixed $x$ with $|x|>1$ (say, $|x|=2$) we have $\lim_{\delta\to 0}|\nabla u\ast\varphi_{\delta}|=|\nabla u(x)|>0$. Thus, (c) is satisfied if $\delta$ is chosen sufficiently small.

Remark. It is well-known that the maximum principle (with a constant $C$) holds for potentials $\int K(|x-y|)\,d\mu(y)$ with non-negative kernels $K(t)$ decreasing on $(0,\infty)$, and non-negative finite Borel measures $\mu$. Our arguments show that for non-positive measures the analog of (1.1) fails even for potentials with positive Riesz kernels. In fact we have proved that for every $\varepsilon>0$, there exists a signed measure $\eta=\eta(\varepsilon)$ in ${\mathbb{R}}^{5}$ with $C^{\infty}$-density and such that $\operatorname{supp}\eta\in D_{\varepsilon}:=\{1-\varepsilon\leq|x|\leq 1+\varepsilon\}$, $|u_{\eta}(x)|<\varepsilon$ for $x\in\operatorname{supp}\eta$, but $|u_{\eta}((2,0,\dots,0))|>b>0$, where $u_{\eta}(x):=\int\frac{d\eta(y)}{|y-x|}$, and $b$ is an absolute constant.

Indeed, for the first component $R_{1}^{2}\nu$ of $R^{2}\nu$ we have

[TABLE]