The Unnormalized Estimate Printed in Maz'ya's Problem 52 Is False

1. Introduction and Precise Scope

Let $B_r=B_r\left(0\right)\subset {\mathbb{R}}^n$. For standard background on p-harmonic functions and nonlinear capacity, see [1]. In the section entitled “Phragmen–Lindelöf Principle for the p-Laplace equation,” Maz’ya considers solutions of

$${\Delta }_{p}u:={div\left(\right|\nabla u|}^{p-2}\nabla u)=0,1<p<n,$$

with zero Dirichlet data near a boundary point $O\in \partial \Omega$. Problem 52 of [2] displays the implication that divergence of the energy integral should force

$${\parallel u\parallel }_{L^p(B_{\delta }\setminus B_r)}\ge A_{\delta }{r}^{-{\displaystyle \frac{n-p}{p-1}}}exp\left(\gamma {\int }_r^{\delta }{\displaystyle \frac{{cap}_p(B_{\rho }\setminus \Omega )}{{cap}_p\left(B_{\rho }\right)}}{\displaystyle \frac{\mathrm{d}\rho }{\rho }}\right),2r<\delta .$$

(1.1)

Here we have renamed the two occurrences of the letter c in the printed formula as $A_{\delta }$ and $\gamma$ in order to distinguish their roles. The p-capacity is

$${cap}_p\left(F\right)=inf\left\{{\int }_{{\mathbb{R}}^n}{|\nabla \phi |}^p\mathrm{d}x:\phi \in C_c^{\infty }\left({\mathbb{R}}^n\right),\phi \ge 1\mathrm{on}F\right\}.$$

(1.2)

The purpose of this note is deliberately limited. We test the displayed inequality (1.1) exactly as written, with its raw annular norm. We neither claim that (1.1) is the intended final formulation nor propose a definitive corrected nonlinear theorem.

Convention 1.1

(Interpretation of the printed statement). Throughout the paper the following natural conventions are used.

(i)

Since u is defined only in Ω, the norm in (1.1) means

$${\parallel u\parallel }_{L^p(\Omega \cap (B_{\delta }\setminus B_r))}.$$

Equivalently, it is the norm of the zero extension of u on the annulus.

(ii)

“The energy integral is divergent” means

$${\int }_{\Omega \cap B_{\delta }}{|\nabla u|}^p\mathrm{d}x=\infty$$

(1.3)

for the boundary neighborhood under consideration.

(iii)

The constants $A_{\delta }>0$ and $\gamma >0$ may depend on any fixed admissible data allowed by the conjecture, but they are independent of the variable radius r. Only positivity and independence from r are used below.

A counterexample for one admissible exponent suffices to disprove a statement asserted for all $1<p<n$. We take $p=2$, which is admissible when $n\ge 3$, and use a classical boundary Poisson-kernel singularity. The main quantitative conclusion is the limit

$$\underset{r\downarrow 0}{lim}{r}^{n-2}{\parallel u\parallel }_{L^2(B_{\delta }^{+}\setminus B_r^{+})}=0.$$

(1.4)

If (1.1) were true, the left-hand side of (1.4) would be bounded below by the positive constant $A_{\delta }$, because the exponential is at least one.

The qualification “unnormalized” is essential. Maz’ya’s later survey [3] states that its Chapter 6 reproduces the relevant 1967 work [4]. In that reproduced linear theory, the growing-solution estimate is expressed using a scale-normalized quadratic mean

$${\mathcal{M}}_r\left(u\right)=r^{-n}{\int }_{K_{{\alpha }^{-1}r,\alpha r}}{u}^2\mathrm{d}x,$$

(1.5)

where $K_{{\alpha }^{-1}r,\alpha r}$ is a fixed-ratio annular region. Thus (1.1) and the linear theorem do not use the same size functional. We return to this distinction in Section 5.

2. Admissible Boundary Singularity

Fix $n\ge 3$ and put

$$\Omega =B_1^{+}:=B_1\cap \{x_n>0\},B_r^{+}:=B_r\cap \{x_n>0\},O=0.$$

The boundary is flat and smooth in a neighborhood of O; the nonsmooth edge of the half-ball is separated from O. Fix $0<\delta <1/2$ and define

$$u\left(x\right)={\displaystyle \frac{x_n}{{\left|x\right|}^n}},x\in \Omega .$$

(2.1)

This is, up to a positive dimensional constant, the Poisson kernel of the upper half-space with pole at the origin, restricted to $B_1^{+}$; see [5].

Proposition 2.1.

The function in (2.1) satisfies the following properties.

(i)

$u\in C^{\infty }(\Omega )\cap W_{\mathrm{loc}}^{1,2}(\Omega )$, $u>0$ in Ω, and $\Delta u=0$ there. In particular, it is a local weak solution:

$${\int }_{\Omega }\nabla u·\nabla \phi \mathrm{d}x=0(\phi \in C_c^{\infty }(\Omega )).$$

(ii)

u extends continuously with value zero to

$$(\partial \Omega \cap B_{\delta })\setminus \left\{0\right\}.$$

(iii)

The local Dirichlet energy diverges:

$${\int }_{B_{\delta }^{+}}{|\nabla u|}^2\mathrm{d}x=\infty .$$

Proof. 

Let $\Phi \left(x\right)={\left|x\right|}^{2-n}$. Since $\Phi$ is harmonic in ${\mathbb{R}}^n\setminus \left\{0\right\}$ and

$${\partial }_{x_n}\Phi \left(x\right)=(2-n)x_n{\left|x\right|}^{-n},$$

we have

$$u=-{\displaystyle \frac{1}{n-2}}{\partial }_{x_n}\Phi .$$

Hence u is smooth and harmonic in $\Omega$. Positivity follows from $x_n>0$. Every compact subset of $\Omega$ has positive distance from the origin, so $u\in W_{\mathrm{loc}}^{1,2}(\Omega )$, and the weak formulation follows from classical harmonicity. If $x\in (\partial \Omega \cap B_{\delta })\setminus \left\{0\right\}$, then $x_n=0$, proving the local boundary condition.

For the energy, write $x=s\theta$ with $s=\left|x\right|$ and

$$\theta \in {\mathbb{S}}_{+}^{n-1}:={\mathbb{S}}^{n-1}\cap \{{\theta }_n>0\}.$$

Direct differentiation gives

$$\nabla u\left(x\right)={\left|x\right|}^{-n}\left(e_n-n{\theta }_n\theta \right),$$

(2.2)

and therefore

$${|\nabla u\left(x\right)|}^2={\left|x\right|}^{-2n}\left(1+n(n-2){\theta }_n^2\right).$$

(2.3)

By symmetry,

$${\int }_{{\mathbb{S}}_{+}^{n-1}}{\theta }_n^2\mathrm{d}\sigma ={\displaystyle \frac{|{\mathbb{S}}^{n-1}|}{2n}}.$$

(2.4)

Consequently, for $0<\epsilon <\delta$,

$$\begin{array}{cc}\hfill {\int }_{B_{\delta }^{+}\setminus B_{\epsilon }^{+}}{|\nabla u|}^2\mathrm{d}x& =\left({\int }_{{\mathbb{S}}_{+}^{n-1}}\left(1+n(n-2){\theta }_n^2\right)\mathrm{d}\sigma \right){\int }_{\epsilon }^{\delta }{s}^{-n-1}\mathrm{d}s\hfill \\ \hfill & ={\displaystyle \frac{(n-1)|{\mathbb{S}}^{n-1}|}{2n}}\left({\epsilon }^{-n}-{\delta }^{-n}\right).\hfill \end{array}$$

(2.5)

The right-hand side tends to $+\infty$ as $\epsilon \downarrow 0$. □

Remark 2.2.

The example requires no irregular boundary geometry. The distinguished point is a locally flat $C^{\infty }$ boundary point. The global edge of $B_1^{+}$ does not enter any calculation because $\delta <1/2$.

3. Exact Annular Norm and Failure of the Printed Bound

Lemma 3.1.

For every $0<r<\delta <1/2$,

$${\parallel u\parallel }_{L^2(B_{\delta }^{+}\setminus B_r^{+})}^2={\displaystyle \frac{|{\mathbb{S}}^{n-1}|}{2n(n-2)}}\left(r^{2-n}-{\delta }^{2-n}\right).$$

(3.1)

In particular,

$${\parallel u\parallel }_{L^2(B_{\delta }^{+}\setminus B_r^{+})}\sim {\left({\displaystyle \frac{|{\mathbb{S}}^{n-1}|}{2n(n-2)}}\right)}^{1/2}{r}^{-(n-2)/2}(r\downarrow 0).$$

(3.2)

Proof. 

In polar coordinates, $u\left(s\theta \right)={\theta }_n{s}^{1-n}$. Using (2.4),

$$\begin{array}{cc}\hfill {\int }_{B_{\delta }^{+}\setminus B_r^{+}}{\left|u\right|}^2\mathrm{d}x& ={\int }_{{\mathbb{S}}_{+}^{n-1}}{\theta }_n^2\mathrm{d}\sigma {\int }_r^{\delta }{s}^{2-2n}{s}^{n-1}\mathrm{d}s\hfill \\ \hfill & ={\displaystyle \frac{|{\mathbb{S}}^{n-1}|}{2n}}{\int }_r^{\delta }{s}^{1-n}\mathrm{d}s,\hfill \end{array}$$

which is (3.1). □

Theorem 3.2

(Counterexample to the unnormalized display). For every $n\ge 3$ and every $0<\delta <1/2$, the domain $\Omega =B_1^{+}$ and the function (2.1) satisfy all conditions in Convention 1.1, but the lower bound (1.1) fails for $p=2$. More strongly,

$$\underset{r\downarrow 0}{lim}{r}^{n-2}{\parallel u\parallel }_{L^2(B_{\delta }^{+}\setminus B_r^{+})}=0.$$

(3.3)

Thus no positive constant $A_{\delta }$ independent of r can make even the weaker estimate

$${\parallel u\parallel }_{L^2(B_{\delta }^{+}\setminus B_r^{+})}\ge A_{\delta }{r}^{-(n-2)}$$

(3.4)

hold for all sufficiently small r.

Proof. 

Admissibility and energy divergence follow from Proposition 2.1. By Lemma 3.1,

$$r^{n-2}{\parallel u\parallel }_{L^2(B_{\delta }^{+}\setminus B_r^{+})}\sim {\left({\displaystyle \frac{|{\mathbb{S}}^{n-1}|}{2n(n-2)}}\right)}^{1/2}{r}^{(n-2)/2},$$

which proves (3.3).

Now suppose that the printed estimate (1.1) holds for $p=2$ with $A_{\delta }>0$ and $\gamma >0$ independent of r. Since capacities are nonnegative, the exponential factor is at least one. Hence (1.1) implies (3.4). Multiplication by $r^{n-2}$ and passage to the limit $r\downarrow 0$ give $0\ge A_{\delta }$, a contradiction. □

Corollary 3.3.

The theorem disproves the lower bound printed in Problem 52 when its left-hand side is interpreted as the unnormalized annular $L^p$-norm. It does not disprove a statement formulated with a scale-normalized local mean or another size functional.

4. The Capacity Factor Strengthens the Contradiction

The proof of Theorem 3.2 uses only that the exponential in (1.1) is at least one. For completeness, the capacity ratio in the example can be bounded from below uniformly in scale.

Proposition 4.1.

For $0<\rho <\delta <1/2$,

$${\displaystyle \frac{{cap}_2(B_{\rho }\setminus \Omega )}{{cap}_2\left(B_{\rho }\right)}}\ge 4^{2-n}.$$

(4.1)

Consequently,

$${\int }_r^{\delta }{\displaystyle \frac{{cap}_2(B_{\rho }\setminus \Omega )}{{cap}_2\left(B_{\rho }\right)}}{\displaystyle \frac{\mathrm{d}\rho }{\rho }}\ge 4^{2-n}log{\displaystyle \frac{\delta }{r}}.$$

(4.2)

Proof. 

The closed ball

$$D_{\rho }:=\overline{B_{\rho /4}\left(-{\displaystyle \frac{\rho }{2}}{e}_n\right)}$$

is contained in $B_{\rho }\cap \{x_n<0\}$ and therefore in $B_{\rho }\setminus \Omega$. By monotonicity and homogeneity of Newtonian capacity,

$${cap}_2(B_{\rho }\setminus \Omega )\ge {cap}_2\left(D_{\rho }\right)=4^{2-n}{cap}_2\left(B_{\rho }\right).$$

Integrating (4.1) over $(r,\delta )$ yields (4.2). □

For any fixed $\gamma >0$, (4.2) gives

$$exp\left(\gamma {\int }_r^{\delta }{\displaystyle \frac{{cap}_2(B_{\rho }\setminus \Omega )}{{cap}_2\left(B_{\rho }\right)}}{\displaystyle \frac{\mathrm{d}\rho }{\rho }}\right)\ge {\left({\displaystyle \frac{\delta }{r}}\right)}^{\gamma 4^{2-n}}.$$

(4.3)

Thus the right-hand side of the printed inequality is bounded below by a positive constant times

$$r^{-(n-2)-\gamma 4^{2-n}},$$

which diverges still faster than the already impossible power $r^{-(n-2)}$.

5. Homogeneity and the Missing Scale Normalization

The failure can be understood before any capacity calculation. The Poisson kernel (2.1) is homogeneous of degree $1-n$:

$$u\left(\lambda x\right)={\lambda }^{1-n}u\left(x\right).$$

(5.1)

On a fixed-ratio half-annulus $B_{\kappa r}^{+}\setminus B_r^{+}$, where $\kappa >1$ is fixed, scaling therefore gives

$${\parallel u\parallel }_{L^2(B_{\kappa r}^{+}\setminus B_r^{+})}\asymp r^{1-n+n/2}=r^{-(n-2)/2}.$$

(5.2)

The factor $r^{n/2}$ in this exponent is the volume contribution carried by an unnormalized $L^2$-norm. Dividing by the square root of the volume scale removes that contribution:

$$r^{-n/2}{\parallel u\parallel }_{L^2(B_{\kappa r}^{+}\setminus B_r^{+})}\asymp r^{1-n}.$$

(5.3)

The normalized expression has the same homogeneity as the pointwise size of u.

This is precisely the type of normalization present in the linear theory. Maz’ya’s 2015 survey ([3], Chapter 6), which explicitly reproduces the relevant earlier paper [4], introduces

$${\mathcal{M}}_r\left(u\right)=r^{-n}{\int }_{K_{{\alpha }^{-1}r,\alpha r}}{u}^2\mathrm{d}x$$

(5.4)

(see ([3], Equation (6.1.3))). Its growing-solution estimate is stated in terms of ${\mathcal{M}}_r{\left(u\right)}^{1/2}$; see ([3], Theorem 6.1.3). Formula (5.4) is a fixed-ratio annular quadratic mean, whereas (1.1) uses the raw norm over $B_{\delta }\setminus B_r$.

Remark 5.1.

The comparison above diagnoses a scale mismatch; it is not a proof that simply replacing the left-hand side of (1.1) by $r^{-n/p}$ times a local $L^p$-norm yields the correct nonlinear theorem. A valid nonlinear analogue may also require a fixed-ratio annulus, a different capacity exponent, additional solution hypotheses, or a different normalization. Determining that statement remains open here.

6. Conclusions

The upper-half-space Poisson kernel restricted to a half-ball gives a complete counterexample to the unnormalized lower bound printed in Maz’ya’s Problem 52. At a locally flat smooth boundary point, the function is positive and harmonic, satisfies the local zero Dirichlet condition away from the singular point, and has infinite Dirichlet energy. Nevertheless,

$${\parallel u\parallel }_{L^2(B_{\delta }^{+}\setminus B_r^{+})}\asymp r^{-(n-2)/2},$$

so it cannot dominate any positive multiple of $r^{-(n-2)}$. The capacitary exponential only enlarges the incompatible right-hand side.

The conclusion is therefore exact but limited: the literal raw-norm display is false. The correctly normalized nonlinear Phragmén–Lindelöf lower-growth principle is a separate question and is not resolved by this counterexample.