A Counterexample to the One-Set Sufficiency Suggested in Maz'ya's Problem 56

1. Introduction

Let $\mathrm{\Omega }$ be an open subset of ${\mathbb{R}}^n$ and $Q_T=\mathrm{\Omega }\times (0,T)$. Problem 56 in Maz’ya’s problem list [6] asks for a complete characterization of the nonnegative measures $\mu$ for which

$${\int }_{\mathrm{\Omega }\times [0,T)}{\left|u\right|}^2\mathrm{d}\mu \le C\left[{\int }_{\mathrm{\Omega }\times [0,T)}|{\nabla }_x{u|}^2\mathrm{d}x\mathrm{d}t+\underset{0\le t<T}{sup}{\int }_{\mathrm{\Omega }}{\left|u(x,t)\right|}^2\mathrm{d}x\right]$$

(1)

holds for every smooth function u vanishing near the lateral boundary $\partial \mathrm{\Omega }\times [0,T]$. The same problem asks for a characterization of measures satisfying

$${\int }_{\mathrm{\Omega }\times [0,T)}{\left|u\right|}^2\mathrm{d}\mu \le C\left[{\int }_{\mathrm{\Omega }\times [0,T)}|{\nabla }_x{u|}^2\mathrm{d}x\mathrm{d}t+{\int }_0^T{\parallel u_t\left(t\right)\parallel }_{H^{-1}\left(\mathrm{\Omega }\right)}^2\mathrm{d}t\right].$$

(2)

Maz’ya observes that a one-set condition analogous to the elliptic capacitary criterion is necessary, with the parabolic capacity studied by Pierre [8], and states that it is “most probably” sufficient.

The elliptic result has no direct parabolic analogue for the energy in (1). The obstruction is multilevel. A single set can be controlled by either its spatial Dirichlet cost or an $L_t^{\infty }{L}_x^2$ cost. A function can nevertheless distribute increasingly high amplitudes over disjoint time intervals. The $L_t^{\infty }{L}_x^2$ term then sees only one amplitude at a time, while a suitable choice of the interval lengths makes the sum of spatial Dirichlet costs finite.

Several notions of parabolic capacity occur in the literature. Pierre proved that the classical heat-equation capacity admits a Hilbert-space formulation based on

$$L^2(0,T;H_0^1\left(\mathrm{\Omega }\right))\cap H^1(0,T;H^{-1}\left(\mathrm{\Omega }\right)).$$

Modern variational formulations and their relation to measure-data or potential-theoretic capacities are developed, for example, in [1,2,3,7]. Since the direction of comparison is decisive here, Section 2 fixes the admissible classes and records the precise equivalences used in the proof.

Our main result is as follows.

Theorem 1.1.  

Let $n\ge 3$. There exist $\mathrm{\Omega }=B(0,2)\subset {\mathbb{R}}^n$, $T=1$, and a finite nonnegative Radon measure $\mu \ll {\mathcal{L}}^{n+1}$ on $Q=\mathrm{\Omega }\times (0,1)$, supported in

$$S=\overline{B(0,1/4)}\times [11/24,13/24]⋐Q,$$

such that:

(i)

for every compact set $K\subset Q$,

$$\mu \left(K\right)\le C{cap}_{\mathrm{heat},Q}\left(K\right);$$

(3)

(ii)

there are functions $u_N\in C_c^{\infty }\left(Q\right)$ satisfying

$$\underset{N}{sup}\left({\int }_Q|{\nabla }_x{u}_N{|}^2\mathrm{d}x\mathrm{d}t+\underset{0<t<1}{ess\; sup}{\parallel u_N\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2\right)<\infty ,$$

(4)

but

$${\int }_Q{\left|u_N\right|}^2\mathrm{d}\mu =N.$$

(5)

Consequently, domination by Pierre’s heat capacity is not sufficient for the first trace inequality in Problem 56 in dimensions $n\ge 3$.

Section 2 gives a separate endpoint-transfer statement that converts (3) into the corresponding estimate for compact subsets of $\mathrm{\Omega }\times [0,1)$ under an explicit half-closed-cylinder normalization of the same capacity.

The construction also disproves a natural strong multilevel estimate associated with the energy in (1). We prove separately that this estimate, combined with one-set domination, would imply the trace inequality. The counterexample does not have bounded time-derivative energy, and therefore gives no conclusion about (2). Throughout the paper, C and c denote positive constants that may change from line to line and depend only on the fixed structural parameters unless a different dependence is stated.

2. Capacity Framework and Admissible Classes

Throughout the proof, $\mathrm{\Omega }=B(0,2)$, $T=1$, and $Q=\mathrm{\Omega }\times (0,1)$. Put

$$V=H_0^1\left(\mathrm{\Omega }\right),H=L^2\left(\mathrm{\Omega }\right),V^{\prime }=H^{-1}\left(\mathrm{\Omega }\right),$$

and

$$\mathfrak{W}=\{v\in L^2(0,1;V):{\partial }_{t}v\in L^2(0,1;V^{\prime })\}.$$

(6)

We equip this space with

$${\parallel v\parallel }_{\mathfrak{W}}^2={\int }_0^1\parallel {\nabla }_x{v\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2\mathrm{d}t+{\int }_0^1{\parallel {\partial }_{t}v\left(t\right)\parallel }_{V^{\prime }}^2\mathrm{d}t.$$

(7)

By the Lions–Magenes theorem, $\mathfrak{W}\hookrightarrow C\left(\right[0,1];H)$ continuously. In particular,

$$\underset{0\le t\le 1}{sup}{\parallel v\left(t\right)\parallel }_H^2\le C{\parallel v\parallel }_{\mathfrak{W}}^2(v\in \mathfrak{W}).$$

(8)

Here and below the $H_0^1$ norm is represented by the gradient norm; this is legitimate by Poincare’s inequality.

Let

$$\mathcal{D}=C_c^{\infty }(\mathrm{\Omega }\times \mathbb{R}),$$

with every function restricted to Q when its energy is evaluated. Define

$$\begin{array}{cc}\hfill {\mathcal{E}}_{\infty }\left(v\right)& ={\int }_Q|{\nabla }_x{v|}^2\mathrm{d}x\mathrm{d}t+\underset{0<t<1}{ess\; sup}{\parallel v\left(t\right)\parallel }_H^2,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill {\mathcal{E}}_{\mathrm{var}}\left(v\right)& ={\int }_Q|{\nabla }_x{v|}^2\mathrm{d}x\mathrm{d}t+{\int }_0^1\parallel {\partial }_t{v\left(t\right)\parallel }_{V^{\prime }}^2\mathrm{d}t+\underset{0<t<1}{ess\; sup}{\parallel v\left(t\right)\parallel }_H^2.\hfill \end{array}$$

(10)

For a compact set $K\subset Q$, set

$$\begin{array}{cc}\hfill {cap}_{\infty }\left(K\right)& =inf\{{\mathcal{E}}_{\infty }\left(v\right):v\in \mathcal{D},v\ge 1\mathrm{on}\mathrm{an}\mathrm{open}\mathrm{neighbourhood}\mathrm{of}K\},\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill {cap}_{\mathrm{var}}\left(K\right)& =inf\{{\mathcal{E}}_{\mathrm{var}}\left(v\right):v\in \mathcal{D},v\ge 1\mathrm{on}\mathrm{an}\mathrm{open}\mathrm{neighbourhood}\mathrm{of}K\}.\hfill \end{array}$$

(12)

The time-derivative term in (10) is exactly the $p=2$ instance of the dual term in the functional of Moring–Scheven. Indeed, for $v\in \mathcal{D}$,

$$\parallel {\partial }_t{v\parallel }_{L^2(0,1;V^{\prime })}=\underset{\begin{array}{c}\varphi \in C_c^{\infty }\left(Q\right)\\ \parallel {\nabla }_x{\varphi \parallel }_{L^2\left(Q\right)}\le 1\end{array}}{sup}\left|{\int }_{Q}v{\partial }_t\varphi \mathrm{d}x\mathrm{d}t\right|.$$

(13)

This follows from integration by parts in time and the density of $C_c^{\infty }\left(Q\right)$ in $L^2(0,1;V)$.

We write ${cap}_{\mathrm{heat},Q}$ for Pierre’s classical heat-equation capacity relative to the open cylinder Q. Its normalization is immaterial for the present problem, since all assertions are stable under multiplication of the capacity by a fixed positive constant.

Proposition 2.1 

(Precise capacity identification). For every compact set $K\subset Q$,

$$c{cap}_{\mathrm{heat},Q}\left(K\right)\le {cap}_{\mathrm{var}}\left(K\right)\le C{cap}_{\mathrm{heat},Q}\left(K\right),$$

(14)

where $c,C>0$ depend only on n, Ω, and the length of the time interval. Moreover, if

$${cap}_{\mathrm{var}}^{\mathrm{ae}}\left(K\right)=inf\{{\mathcal{E}}_{\mathrm{var}}\left(v\right):v\in \mathfrak{W},v\ge 1\mathit{a}.\mathit{e}.\mathrm{on}\mathit{some}\mathit{open}\mathit{U}⋐\mathit{Q}\mathit{with}\mathit{K}\subset \mathit{U}\},$$

(15)

then

$$c{cap}_{\mathrm{var}}^{\mathrm{ae}}\left(K\right)\le {cap}_{\mathrm{var}}\left(K\right)\le C{cap}_{\mathrm{var}}^{\mathrm{ae}}\left(K\right).$$

(16)

Consequently,

$${cap}_{\infty }\left(K\right)\le {cap}_{\mathrm{var}}\left(K\right)\le C{cap}_{\mathrm{heat},Q}\left(K\right).$$

(17)

Proof. 

Pierre proved that the usual heat-equation capacity can be defined, up to equivalence of normalizations, by the Hilbert norm of the parabolic Dirichlet space $\mathfrak{W}$; see [8]. The same identification can be read in the modern variational framework as follows. The functional in [7. Definition 3.5], specialized to $p=2$, is precisely (10), by (13). That definition requires a smooth function to dominate ${\mathbf{1}}_K$ on K. This gives the same infimum as our neighbourhood condition: one direction is immediate, while in the other direction $(1+\epsilon )v>1$ on the compact set K, hence on an open neighbourhood of K by continuity; then let $\epsilon \downarrow 0$. Since $\mathrm{\Omega }=B(0,2)$ has uniformly 2-fat complement, [7, Theorem 1.1] identifies the resulting smooth variational capacity with the parabolic capacity associated with the heat equation. The dependence on a finite reference time is controlled by [7, Lemma 3.14]. In the linear case $p=2$, this is the classical thermal capacity considered by Pierre. This proves (14).

The formulation using functions from the variational space that dominate the obstacle almost everywhere on an open neighbourhood is [7, Definition A.1]. Its equivalence with the smooth formulation of Definition 3.5 is proved in [7, Lemmas A.3 and A.6]; this yields (16). Finally, ${\mathcal{E}}_{\infty }\left(v\right)\le {\mathcal{E}}_{\mathrm{var}}\left(v\right)$ for every smooth v, so taking infima gives (17). □

Remark 2.2 

(Representative and obstacle convention). No pointwise statement is imposed on an arbitrary equivalence class in $\mathfrak{W}$. The variational-space capacity in (15) uses the unambiguous condition $v\ge 1$ almost everywhere on an open neighbourhood. The proof itself uses only smooth admissible functions. Thus no unmentioned choice of a quasi-continuous representative enters the argument.

For the half-closed cylinder in Maz’ya’s formulation, we use the following explicit equivalent normalization. If $K\subset \mathrm{\Omega }\times [0,1)$ is compact, define

$$\begin{array}{cc}\hfill {cap}_{\mathrm{heat},[0,1)}\left(K\right)=inf\{& {\mathcal{E}}_{\mathrm{var}}\left(v\right):v\in \mathcal{D},v\ge 1\mathrm{on}\mathrm{a}\mathrm{relatively}\mathrm{open}\mathrm{neighbourhood}\mathrm{of}K\hfill \\ \hfill & \mathrm{in}\mathrm{\Omega }\times [0,1)\}.\hfill \end{array}$$

(18)

For compact sets contained in Q, Proposition 2.1 shows that this normalization agrees with Pierre’s open-cylinder capacity up to fixed multiplicative constants.

Proposition 2.3 

(Transfer to the half-closed cylinder). Assume that a finite measure ν is supported in a compact set $S⋐Q$ and satisfies

$$\nu \left(L\right)\le A{cap}_{\mathrm{var}}\left(L\right)\mathit{for}\mathit{every}\mathit{compact}\mathit{L}\subset \mathit{Q}.$$

Then, for every compact $K\subset \mathrm{\Omega }\times [0,1)$,

$$\nu \left(K\right)\le A{cap}_{\mathrm{heat},[0,1)}\left(K\right).$$

(19)

Proof. 

Set $L=K\cap S$. Then L is compact in Q and $\nu \left(K\right)=\nu \left(L\right)$. Every function admissible in (18) for K is, after restriction to Q, admissible for ${cap}_{\mathrm{var}}\left(L\right)$: because $L⋐Q$, a relatively open neighbourhood of K contains an ordinary open neighbourhood of L in Q. Hence ${cap}_{\mathrm{var}}\left(L\right)\le {cap}_{\mathrm{heat},[0,1)}\left(K\right)$, and (19) follows. □

Remark 2.4 

(Terminology). The auxiliary capacity ${cap}_{\infty }$ is generated by the energy in the first trace inequality. It is not the “energy capacity” defined by restricting admissible functions to supercaloric functions; compare [7]. No potential-theoretic assertion is attached to ${cap}_{\infty }$.

3. The Block Measure

Fix $n\ge 3$ and write

$$\alpha ={\displaystyle \frac{n-2}{n}},\beta ={\displaystyle \frac{2}{n}},$$

(20)

so that $\alpha +\beta =1$. Choose $q_1>0$, sufficiently small, and define

$$q_k=q_1{2}^{-{\displaystyle \frac{n}{4}}k(k-1)},k\ge 1.$$

(21)

Then

$${\displaystyle \frac{q_{k+1}}{q_k}}=2^{-kn/2}.$$

(22)

Set

$${\delta }_k=q_{k+1}^{2/n}=2^{-k}{q}_k^{2/n}.$$

(23)

The sequences decrease super-exponentially, and therefore

$${\displaystyle \sum _{k=1}^{\infty }}{q}_k<\infty ,{\displaystyle \sum _{k=1}^{\infty }}{q}_k^{1/n}<\infty ,{\displaystyle \sum _{k=1}^{\infty }}{\delta }_k<\infty .$$

(24)

Let ${\omega }_n=\left|B(0,1)\right|$ and $r_k={(q_k/{\omega }_n)}^{1/n}$. Reducing $q_1$ if necessary, arrange

$$4{\displaystyle \sum _{k=1}^{\infty }}{r}_k<{\displaystyle \frac{1}{4}},2r_1<{\displaystyle \frac{1}{8}},3{\displaystyle \sum _{k=1}^{\infty }}{\delta }_k<{\displaystyle \frac{1}{12}}.$$

(25)

Choose pairwise disjoint closed intervals of lengths $4r_k$ whose union has closure contained in $(-1/8,1/8)$. If $s_k$ is the midpoint of the kth interval, set $x_k=s_k{e}_1$. Then the balls $B(x_k,2r_k)$ are pairwise disjoint and, by (25),

$$\overline{{\bigcup }_{k\ge 1}B(x_k,2r_k)}\subset \overline{B(0,1/4)}⋐\mathrm{\Omega }.$$

(26)

Put

$$A_k=\overline{B(x_k,r_k)},\left|A_k\right|=q_k.$$

(27)

Likewise, choose pairwise disjoint open intervals $J_k$ of length $3{\delta }_k$ such that

$$\overline{{\bigcup }_{k\ge 1}{J}_k}\subset [11/24,13/24]⋐(0,1),$$

(28)

and choose concentric closed intervals $I_k⋐J_k$ with

$$|I_k|={\delta }_k.$$

(29)

Define

$$w(x,t)={\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{1}{{\delta }_k}}{\mathbf{1}}_{A_k}\left(x\right){\mathbf{1}}_{I_k}\left(t\right),\mathrm{d}\mu =w\mathrm{d}x\mathrm{d}t.$$

(30)

Writing $F_k=A_k\times I_k$, we have

$$\mu \left(F_k\right)=q_k,\mu \left(Q\right)={\displaystyle \sum _{k=1}^{\infty }}{q}_k<\infty .$$

(31)

Thus $\mu$ is a finite Radon measure, absolutely continuous with respect to Lebesgue measure, and supported in the fixed compact set S specified in Theorem 1.1, by (26) and (28).

4. The One-Set Estimate

Proposition 4.1.  

For every compact $K\subset Q$,

$$\mu \left(K\right)\le C{cap}_{\infty }\left(K\right).$$

(32)

Consequently,

$$\mu \left(K\right)\le C{cap}_{\mathrm{heat},Q}\left(K\right)\mathit{for}\mathit{every}\mathit{compact}K\subset Q.$$

(33)

Proof. 

Fix a compact $K\subset Q$ and an admissible $v\in \mathcal{D}$ with $v\ge 1$ on an open neighbourhood of K. For $t\in I_k$, let

$$E_k\left(t\right)=\{x\in A_k:(x,t)\in K\},e_k\left(t\right)=\left|E_k\left(t\right)\right|.$$

(34)

Then $0\le e_k\left(t\right)\le q_k$. Put

$$M=\underset{k\ge 1}{sup}\underset{t\in I_k}{ess\; sup}{e}_k\left(t\right).$$

(35)

Since $v\ge 1$ on $E_k\left(t\right)$,

$$M\le \underset{0<t<1}{ess\; sup}{\parallel v\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2.$$

(36)

Let $2^{*}=2n/(n-2)$. The Sobolev inequality gives, for almost every $t\in I_k$,

$$\begin{array}{cc}\hfill e_k{\left(t\right)}^{\alpha }& \le {\left({\int }_{E_k\left(t\right)}{\left|v(x,t)\right|}^{2^{*}}\mathrm{d}x\right)}^{2/2^{*}}\le C{\int }_{\mathrm{\Omega }}{\left|{\nabla }_{x}v(x,t)\right|}^2\mathrm{d}x.\hfill \end{array}$$

(37)

Because the intervals $I_k$ are disjoint,

$${\displaystyle \sum _{k=1}^{\infty }}{\int }_{I_k}{e}_k{\left(t\right)}^{\alpha }\mathrm{d}t\le C{\int }_Q{\left|{\nabla }_{x}v\right|}^2\mathrm{d}x\mathrm{d}t.$$

(38)

By Fubini’s theorem,

$$\mu \left(K\right)={\displaystyle \sum _{k=1}^{\infty }}{s}_k,s_k={\displaystyle \frac{1}{{\delta }_k}}{\int }_{I_k}{e}_k\left(t\right)\mathrm{d}t.$$

(39)

If $M=0$, then $\mu \left(K\right)=0$. Assume $M>0$. Since $M\le q_1$ and $q_k\downarrow 0$, there is $N\ge 1$ such that

$$q_{N+1}<M\le q_N.$$

(40)

For $k\le N-1$, $q_{k+1}\ge q_N\ge M$, and hence

$${\delta }_k=q_{k+1}^{\beta }\ge M^{\beta }.$$

Since $e_k\left(t\right)\le M$ and $\alpha +\beta =1$,

$$e_k\left(t\right)\le M^{\beta }{e}_k{\left(t\right)}^{\alpha }.$$

Therefore

$$s_k\le {\int }_{I_k}{e}_k{\left(t\right)}^{\alpha }\mathrm{d}t(k\le N-1).$$

(41)

For the critical index, $s_N\le M$. For $k\ge N+1$, the trivial estimate gives $s_k\le q_k$. If $\rho =2^{-n/2}<1$, then (22) yields

$${\displaystyle \sum _{k=N+1}^{\infty }}{q}_k\le {\displaystyle \frac{q_{N+1}}{1-\rho }}<{\displaystyle \frac{M}{1-\rho }}.$$

(42)

Combining these estimates with (36) and (38),

$$\begin{array}{cc}\hfill \mu \left(K\right)& \le {\displaystyle \sum _{k=1}^{N-1}}{\int }_{I_k}{e}_k{\left(t\right)}^{\alpha }\mathrm{d}t+CM\hfill \\ \hfill & \le C\left({\int }_Q|{\nabla }_x{v|}^2\mathrm{d}x\mathrm{d}t+\underset{0<t<1}{ess\; sup}{\parallel v\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2\right)=C{\mathcal{E}}_{\infty }\left(v\right).\hfill \end{array}$$

Taking the infimum proves (32). Proposition 2.1 gives (33). □

Corollary 4.2.  

For every compact $K\subset \mathrm{\Omega }\times [0,1)$,

$$\mu \left(K\right)\le C{cap}_{\mathrm{heat},[0,1)}\left(K\right).$$

(43)

Proof. 

By (32) and (17), the measure satisfies $\mu \left(L\right)\le C{cap}_{\mathrm{var}}\left(L\right)$ for every compact $L\subset Q$. Apply Proposition 2.3, using the compact support S from Theorem 1.1. □

5. Failure of the Trace Inequality

Choose fixed cutoff profiles so that, for every k, there are functions

$${\varphi }_k\in C_c^{\infty }\left(B(x_k,2r_k)\right),{\psi }_k\in C_c^{\infty }\left(J_k\right)$$

with

$$0\le {\varphi }_k\le 1,{\varphi }_k=1\mathrm{near}{A}_k,{\int }_{\mathrm{\Omega }}{\varphi }_k^2\mathrm{d}x\le C{q}_k,{\int }_{\mathrm{\Omega }}{|\nabla {\varphi }_k|}^2\mathrm{d}x\le C{q}_k^{\alpha },$$

(44)

and

$$0\le {\psi }_k\le 1,{\psi }_k=1\mathrm{near}{I}_k,{\int }_0^1{\psi }_k^2\mathrm{d}t\le C{\delta }_k.$$

(45)

The time supports are pairwise disjoint. Define

$$u_N(x,t)={\displaystyle \sum _{k=1}^N}{q}_k^{-1/2}{\varphi }_k\left(x\right){\psi }_k\left(t\right).$$

(46)

Then $u_N\in C_c^{\infty }\left(Q\right)$.

Proposition 5.1.  

The sequence in (46) satisfies

$$\underset{N}{sup}{\mathcal{E}}_{\infty }\left(u_N\right)<\infty ,$$

(47)

but

$${\int }_Q{\left|u_N\right|}^2\mathrm{d}\mu =N.$$

(48)

Hence (1) fails for μ.

Proof. 

At each time, at most one ${\psi }_k$ is nonzero. Thus

$$\underset{0<t<1}{ess\; sup}{\parallel u_N\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2\le \underset{1\le k\le N}{max}{q}_k^{-1}{\int }_{\mathrm{\Omega }}{\varphi }_k^2\mathrm{d}x\le C.$$

(49)

The same disjointness eliminates the cross terms in the spatial gradient energy. By (44), (45), and (23),

$$\begin{array}{cc}\hfill {\int }_Q{\left|{\nabla }_x{u}_N\right|}^2\mathrm{d}x\mathrm{d}t& \le C{\displaystyle \sum _{k=1}^N}{q}_k^{-1}{\delta }_k{q}_k^{\alpha }=C{\displaystyle \sum _{k=1}^N}{\delta }_k{q}_k^{-2/n}=C{\displaystyle \sum _{k=1}^N}{2}^{-k}\le C.\hfill \end{array}$$

(50)

This proves (47).

On $F_k=A_k\times I_k$, one has $u_N=q_k^{-1/2}$ for $k\le N$. Therefore

$${\int }_Q{\left|u_N\right|}^2\mathrm{d}\mu ={\displaystyle \sum _{k=1}^N}{q}_k^{-1}\mu \left(F_k\right)={\displaystyle \sum _{k=1}^N}1=N.$$

□

Propositions 4.1 and 5.1 prove Theorem 1.1.

6. The Multilevel Obstruction

Let

$$K_1\supset K_2\supset \dots \supset K_m$$

be compact subsets of Q, and let $0=a_0<a_1<\dots <a_m$. Define

$${cap}_{\infty }^{\left(m\right)}(K_{•};a_{•})=inf\{{\mathcal{E}}_{\infty }\left(v\right):v\in \mathcal{D},v\ge a_j\mathrm{near}{K}_j,1\le j\le m\}.$$

(51)

The natural strong multilevel estimate is

$${\displaystyle \sum _{j=1}^m}(a_j^2-a_{j-1}^2){cap}_{\infty }\left(K_j\right)\le C{cap}_{\infty }^{\left(m\right)}(K_{•};a_{•}).$$

(52)

The next proposition supplies the implication that is sometimes left implicit in discussions of one-set criteria.

Proposition 6.1 

(Multilevel estimate implies the trace bound). Let ν be a finite nonnegative measure on Q. Suppose that

$$\nu \left(K\right)\le A{cap}_{\infty }\left(K\right)$$

(53)

for every compact $K\subset Q$, and that (52) holds with constant B. Then

$${\int }_Q{\left|u\right|}^2\mathrm{d}\nu \le 2AB{\mathcal{E}}_{\infty }\left(u\right)$$

(54)

for every real-valued $u\in C_c^{\infty }\left(Q\right)$.

Proof. 

We first treat $u_{+}=max\{u,0\}$. Fix $\eta >0$, $\lambda >1$, and $\epsilon >0$, and set $a_j=\eta {\lambda }^{j-1}$, $a_0=0$. Choose m so large that

$$\parallel u_{+}{\parallel }_{L^{\infty }\left(Q\right)}<(1+\epsilon )a_{m+1},$$

and put

$$K_j^{+}=\{(x,t)\in Q:u(x,t)\ge (1+\epsilon )a_j\},1\le j\le m.$$

These are compact because $a_j>0$ and u has compact support. Moreover, compactness and continuity imply that $u\ge a_j$ on an open neighbourhood of $K_j^{+}$. Thus u is admissible in the multilevel capacity associated with $(K_j^{+};a_j)$.

For every point of Q,

$$u_{+}^2\le {(1+\epsilon )}^2{\eta }^2+{\lambda }^2{(1+\epsilon )}^2{\displaystyle \sum _{j=1}^m}(a_j^2-a_{j-1}^2){\mathbf{1}}_{K_j^{+}}.$$

(55)

Indeed, if $u_{+}<(1+\epsilon )a_1$, the first term suffices. Otherwise, let j be the largest index for which the point lies in $K_j^{+}$. Then

$$u_{+}<(1+\epsilon )a_{j+1}=\lambda (1+\epsilon )a_j,$$

while the sum in (55) telescopes to $a_j^2$.

Integrating, using (53), (52), and the admissibility of u, gives

$${\int }_Q{u}_{+}^2\mathrm{d}\nu \le {(1+\epsilon )}^2{\eta }^2\nu \left(Q\right)+{\lambda }^2{(1+\epsilon )}^{2}AB{\mathcal{E}}_{\infty }\left(u\right).$$

Applying the same argument to $-u$ gives the corresponding estimate for $u_{-}$. Since $u^2=u_{+}^2+u_{-}^2$, letting $\eta \downarrow 0$, $\lambda \downarrow 1$, and $\epsilon \downarrow 0$ proves (54). □

Corollary 6.2.  

The strong multilevel estimate (52) is false.

Proof. 

For $N\ge 1$, put

$$K_j^{\left(N\right)}={\displaystyle \bigcup _{k=j}^N}{F}_k,a_0=0,a_j=q_j^{-1/2}(1\le j\le N).$$

(56)

The sets are nested and compact, and the levels increase. On a neighbourhood of $F_k$, the function $u_N$ equals $a_k$. Hence $u_N\ge a_j$ near $K_j^{\left(N\right)}$ whenever $k\ge j$, so

$${cap}_{\infty }^{\left(N\right)}(K_{•}^{\left(N\right)};a_{•})\le {\mathcal{E}}_{\infty }\left(u_N\right)\le C.$$

(57)

On the other hand,

$$\begin{array}{cc}\hfill {\displaystyle \sum _{j=1}^N}(a_j^2-a_{j-1}^2)\mu \left(K_j^{\left(N\right)}\right)& ={\displaystyle \sum _{k=1}^N}{q}_k{\displaystyle \sum _{j=1}^k}(a_j^2-a_{j-1}^2)={\displaystyle \sum _{k=1}^N}{q}_k{a}_k^2=N.\hfill \end{array}$$

(58)

If (52) held, Proposition 4.1 would imply

$$N\le C{\displaystyle \sum _{j=1}^N}(a_j^2-a_{j-1}^2){cap}_{\infty }\left(K_j^{\left(N\right)}\right)\le C{cap}_{\infty }^{\left(N\right)}(K_{•}^{\left(N\right)};a_{•}),$$

contradicting (57) as $N\to \infty$. □

7. Why the Second Inequality Is Not Affected

The cutoffs may be chosen by scaling fixed profiles so that, in addition to (45),

$$c{\delta }_k^{-1}\le {\int }_0^1{\left|{\psi }_k^{\prime }\left(t\right)\right|}^2\mathrm{d}t\le C{\delta }_k^{-1}.$$

(59)

The lower bound follows already from the transition from 0 to 1 across a time interval of length at most ${\delta }_k$, by Cauchy–Schwarz.

Lemma 7.1 

(Negative Sobolev scaling). For the spatial cutoffs in (44),

$$c{r}_k^{n+2}\le {\parallel {\varphi }_k\parallel }_{H^{-1}\left(\mathrm{\Omega }\right)}^2\le C{r}_k^{n+2}.$$

(60)

Equivalently,

$$\parallel {\varphi }_k{\parallel }_{H^{-1}\left(\mathrm{\Omega }\right)}^2\asymp q_k^{1+2/n}.$$

Proof. 

For the upper estimate, Sobolev duality gives

$$\parallel {\varphi }_k{\parallel }_{H^{-1}\left(\mathrm{\Omega }\right)}\le C{\parallel {\varphi }_k\parallel }_{L^{2n/(n+2)}\left(\mathrm{\Omega }\right)}\le C{r}_k^{(n+2)/2}.$$

For the lower estimate, choose ${\zeta }_k\in C_c^{\infty }\left(B(x_k,2r_k)\right)$ equal to 1 on $A_k$, obtained by scaling a fixed cutoff. Then

$${\int }_{\mathrm{\Omega }}{\varphi }_k{\zeta }_k\mathrm{d}x\ge |A_k|\asymp r_k^n,{\parallel \nabla {\zeta }_k\parallel }_{L^2\left(\mathrm{\Omega }\right)}\asymp r_k^{(n-2)/2}.$$

Testing the $H^{-1}$ norm against ${\zeta }_k$ yields the lower bound. □

Proposition 7.2.  

For the sequence $u_N$,

$${\int }_0^1{\parallel {\left(u_N\right)}_t\left(t\right)\parallel }_{H^{-1}\left(\mathrm{\Omega }\right)}^2\mathrm{d}t\asymp {\displaystyle \sum _{k=1}^N}{2}^k,$$

(61)

and in particular this quantity is unbounded.

Proof. 

The supports of the ${\psi }_k^{\prime }$ are pairwise disjoint, so

$$\begin{array}{cc}\hfill {\int }_0^1{\parallel {\left(u_N\right)}_t\left(t\right)\parallel }_{H^{-1}}^2\mathrm{d}t& ={\displaystyle \sum _{k=1}^N}{q}_k^{-1}\parallel {\varphi }_k{\parallel }_{H^{-1}}^2{\int }_0^1{\left|{\psi }_k^{\prime }\left(t\right)\right|}^2\mathrm{d}t\hfill \\ \hfill & \asymp {\displaystyle \sum _{k=1}^N}{q}_k^{-1}{q}_k^{1+2/n}{\delta }_k^{-1}={\displaystyle \sum _{k=1}^N}{\displaystyle \frac{q_k^{2/n}}{{\delta }_k}}={\displaystyle \sum _{k=1}^N}{2}^k.\hfill \end{array}$$

□

Thus the test sequence is deliberately adapted to the first trace energy and cannot disprove (2). The second part of Problem 56 remains open within the scope of this construction. Likewise, the slice Sobolev estimate used in Section 4 is specific to $n\ge 3$; no conclusion is claimed for $n=1$ or $n=2$.

8. Conclusion

For every $n\ge 3$, the block measure (30) obeys uniform one-set domination by Pierre’s classical heat capacity on the open cylinder and, by Proposition 2.3, under the explicit half-closed-cylinder normalization (18). Nevertheless, it fails the trace embedding governed by spatial Dirichlet energy and the $L_t^{\infty }{L}_x^2$ norm. The example identifies the missing mechanism: one-set domination does not provide the strong multilevel control required to pass from capacities of individual superlevel sets to the full quadratic trace integral.

Accordingly, a characterization of the measures satisfying the first inequality in Maz’ya’s Problem 56 cannot consist solely of one-set domination by Pierre’s capacity. It must contain additional information that detects the multilevel accumulation exhibited by the nested sets and levels above, or an equivalent structural condition of the same strength.