1. The problem and notation
How small is the set where the velocity of a suitable weak solution fails to be locally bounded? We will prove that this singular set has zero parabolic Hausdorff measure for every fixed finite product of iterated logarithmic factors described in Theorem 2. The first theorem isolates a general condition on the masses of dyadic boxes that makes this possible.
We consider the three-dimensional incompressible Navier–Stokes equations, with viscosity equal to one, on the parabolic cylinder
$$\begin{gathered} C_4(0)=B_4(0)\times(-16,16),\\ \partial_tu-\Delta u+(u\cdot\nabla)u+\nabla p=0,\\ \operatorname{div}u=0. \end{gathered}$$
Here the velocity is a vector field and the pressure is a scalar function. For a point and a radius, write
$$\begin{aligned} z&=(x,t),\\ C_r(z)&=B_r(x)\times(t-r^2,t+r^2),\\ P_r(z)&=B_r(x)\times(t-r^2,t). \end{aligned}$$
The first cylinder is symmetric in time; the second is backward in time. Their time lengths are proportional to the square of their spatial radius. This is the geometry respected by the equation's scaling. We write $$B_r(x)$$ only for spatial balls, $$P_r(z)$$ for backward cylinders, and $$Q=X\times I$$ for dyadic spacetime boxes.
On narrow screens, scroll long equations sideways to read the full calculation.
Suitable weak solutions
Definition (suitable weak solution). A pair $$(u,p)$$ is a suitable weak solution on this cylinder if it has the energy and pressure integrability
$$u\in L^\infty\bigl((-16,16);L^2(B_4)\bigr) \cap L^2\bigl((-16,16);H^1(B_4)\bigr), \qquad p\in L^{3/2}(C_4),$$
it solves both equations in distributions, and it satisfies the local energy inequality. Explicitly, the distributional equations mean that, for every smooth vector test function and scalar test function compactly supported in the cylinder,
$$\begin{aligned} 0&=\int_{C_4}\left[-u\cdot\partial_t\varphi +\nabla u:\nabla\varphi -(u\otimes u):\nabla\varphi -p\,\operatorname{div}\varphi\right]\,dx\,dt,\\ 0&=\int_{C_4}u\cdot\nabla\psi\,dx\,dt. \end{aligned}$$
The colon means the sum of entrywise products of two matrices. The local energy inequality means that, for every nonnegative smooth scalar function $$\phi$$ compactly supported in the cylinder and almost every time $$\tau\in(-16,16)$$,
$$\begin{aligned} &\int_{B_4}|u(x,\tau)|^2\phi(x,\tau)\,dx +2\int_{-16}^{\tau}\int_{B_4}|\nabla u|^2\phi\,dx\,dt\\ &\qquad\le \int_{-16}^{\tau}\int_{B_4} |u|^2(\partial_t\phi+\Delta\phi) +(|u|^2+2p)u\cdot\nabla\phi\,dx\,dt. \end{aligned}$$
Thus the kinetic energy and viscous dissipation inside a region are controlled by transport through its boundary. For a smooth solution this follows by multiplying the equation by $$2u\phi$$ and integrating by parts. For a weak solution the inequality is an additional requirement: the word suitable includes precisely this local control. No boundary condition on the lateral boundary of our cylinder is needed, because all test functions are supported in its interior. If one starts with a locally suitable solution, the displayed integrability is required only on compactly contained cylinders; the proof below uses exactly such cylinders.
A point is regular if the velocity is essentially bounded in some open spacetime neighborhood of that point. The singular set $$S$$ consists of the remaining interior points. The problem is to show that this set has zero Hausdorff measure for the gauges in our theorems.
Scale-invariant quantities
For the proof, first restrict attention to $$L=\overline{C_1(0)}$$, and put $$U=C_2(0)=B_2(0)\times(-4,4)$$.
Every cylinder centered in $$L$$ with radius at most $$1/2$$ lies in $$U$$. Throughout, constants denoted by $$C$$ can change from line to line. They depend only on the dimension and on fixed choices of cutoffs, unless another dependence is indicated. For cylinders contained in $$U$$, define
$$\begin{aligned} G(r,z)&=r^{-2}\int_{C_r(z)}\bigl(|u|^3+|p|^{3/2}\bigr)\,dy\,ds,\\ A(r,z)&=r^{-1}\operatorname*{ess\,sup}_{|s-t|<r^2} \int_{B_r(x)}|u(y,s)|^2\,dy,\\ E(r,z)&=r^{-1}\int_{C_r(z)}|\nabla u|^2\,dy\,ds,\\ e(r,z)&=r^{-3/2}\int_{t-r^2}^{t+r^2} \left(\int_{B_r(x)}|\nabla u(y,s)|^{12/7}\,dy\right)^{7/6}ds. \end{aligned}$$
The quantity $$G$$ measures cubic velocity and pressure activity. The quantities $$A$$ and $$E$$ are scaled energy and dissipation. The mixed quantity $$e$$ is chosen to fit a specific Sobolev estimate in Lemma 2. Each is unchanged by the Navier–Stokes rescaling
$$u_r(y,s)=r\,u(x+ry,t+r^2s),\qquad p_r(y,s)=r^2p(x+ry,t+r^2s).$$
All four quantities are finite. In particular, the energy spaces imply local integrability of the cubic velocity: interpolation between spatial $$L^2$$ and $$L^6$$ gives
$$\begin{aligned} \int_U |u|^{10/3} &\le \int_{-4}^4 \|u(\cdot,s)\|_{L^2(B_2)}^{4/3} \|u(\cdot,s)\|_{L^6(B_2)}^2\,ds\\ &\le C\left(\operatorname*{ess\,sup}_{s\in(-4,4)} \|u(\cdot,s)\|_{L^2(B_2)}^2\right)^{2/3} \int_{-4}^4\|u(\cdot,s)\|_{H^1(B_2)}^2\,ds <\infty. \end{aligned}$$
Also, spatial Hölder's inequality yields the useful comparison
$$\begin{aligned} \left(\int_{B_r(x)}|\nabla u|^{12/7}\right)^{7/6} &\le \left[ \left(\int_{B_r(x)}|\nabla u|^2\right)^{6/7} |B_r|^{1/7}\right]^{7/6}\\ &=|B_r|^{1/6}\int_{B_r(x)}|\nabla u|^2\\ &\le Cr^{1/2}\int_{B_r(x)}|\nabla u|^2, \end{aligned} \qquad e(r,z)\le CE(r,z).$$
Gauges and parabolic Hausdorff measure
An increasing continuous function $$h:(0,r_0)\longrightarrow(0,\infty)$$ with $$\lim_{r\downarrow0}h(r)=0$$ is called a gauge. The parabolic Hausdorff measure with this gauge is
$$\mathcal P^h(A) =\lim_{\delta\downarrow0}\inf\left\{ \sum_i h(r_i): A\subset\bigcup_i C_{r_i}(z_i),\quad 0<r_i<\delta \right\}.$$
Dyadic boxes and the dissipation measure
The measure recording the available dissipation is $$\nu(A)=\int_{A\cap U}|\nabla u|^2\,dx\,dt$$, where $$U=C_2(0)$$ is the fixed interior cylinder introduced above. Whenever a box extends outside $$U$$, extend the function $$|\nabla u|$$ by zero outside that region. This makes $$\nu$$ a finite measure on all of spacetime and requires no extension of the velocity itself.
A parabolic dyadic box of level $$n$$ is a product $$Q=X\times I$$ with spatial side $$\ell_n=2^{-n}$$ and time length $$|I|=\ell_n^2$$.
Here $$X$$ is a spatial cube and $$I$$ is a time interval. In one fixed grid, the level-$$n$$ boxes partition spacetime, and each box is partitioned by its boxes at level $$n+1$$. We use half-open cubes and intervals so that every point belongs to exactly one box at each level. Define
$$E_Q=\frac{\nu(Q)}{\ell_n},\qquad e_Q=\ell_n^{-3/2}\int_I \left(\int_X |\nabla u(x,t)|^{12/7}\,dx\right)^{7/6}dt.$$
These are the box versions of the cylindrical quantities. If $$\nu(Q)=0$$, then $$|\nabla u|=0$$ almost everywhere on $$Q$$ and hence $$e_Q=0$$. Every quotient involving $$e_Q/\sqrt{E_Q}$$ is assigned the value zero in this case.
2. Two theorems and the proof strategy
The finite family of nested grids used below is constructed in Lemma 5. The phrase “every box” in the following hypothesis includes every one of those grids.
Abstract Frostman hypothesis. Let $$F(n)>0$$ for $$n\ge n_0$$.
Suppose that every compact set $$T\subset C_4$$ of positive parabolic Hausdorff measure supports a probability measure with the following bound:
$$\mathcal P^h(T)>0 \quad\Longrightarrow\quad \left\{ \begin{aligned} &\operatorname{supp}\mu\subset T,\qquad \mu(T)=1,\\ &\mu(Q)\le C_\mu F(n) &&\text{for every level-}n\text{ box }Q,\quad n\ge n_0. \end{aligned} \right.$$
The constant is allowed to depend on the compact set and its measure. The sequence here bounds the mass itself; it does not have a factor of the box side length hidden in its definition.
Theorem 1 (An abstract Frostman criterion). Let $$(u,p)$$ be a suitable weak solution on $$C_4(0)$$, with interior singular set $$S$$. Assume the abstract Frostman hypothesis for the gauge and sequence above. If, after discarding finitely many indices,
$$2^{n+1}F(n+1)\ge 2^nF(n) \qquad\text{and}\qquad \sum_{n\ge n_0}\frac1{\sqrt{2^nF(n)}}=\infty,$$
then $$\mathcal P^h(S)=0$$.
There is a direct way to check the Frostman hypothesis in applications. Set $$F(n)=h(2^{-n})$$. The monotonicity assumption of Theorem 1 implies $$F(n-1)\le 2F(n)$$. If $$2^{-(n+1)}<r\le2^{-n}$$, then monotonicity of the gauge and two applications of the preceding inequality give
$$\begin{aligned} h(2r) &\le h(2^{-(n-1)})\\ &=F(n-1)\\ &\le4F(n+1)\\ &\le4h(r). \end{aligned}$$
Thus the gauge is doubling at small radii. Positive Hausdorff measure of a compact set implies positive Hausdorff content: if the unrestricted covering cost were zero, then for any $$\delta>0$$ one could choose a cover with cost below both $$h(\delta)$$ and any prescribed positive number. Every radius in that cover would be below $$\delta$$, forcing the Hausdorff measure to vanish. The gauge form of Frostman's lemma in the parabolic metric supplies a probability measure satisfying $$\mu(C_r(z))\le C_\mu h(r)$$.
This is the same Frostman construction used for Euclidean balls, with the parabolic metric $$d_{\mathrm p}((x,t),(y,s))=\max\{|x-y|,|t-s|^{1/2}\}$$.
Its balls are exactly our symmetric cylinders. One can run the usual finite-branching dyadic construction with eight spatial children and four temporal children per box. See Bishop–Peres, Lemma 3.1.1 for the gauge version and its dyadic proof; the change to parabolic boxes changes only the fixed branching and covering constants.
Every parabolic box lies in the cylinder of radius equal to its spatial side length, centered at the midpoint of the box. Indeed, the largest spatial distance from that midpoint is smaller than the side length, and the largest time distance is half the square of the side length. Consequently,
$$\mu(Q)\le\mu(C_{2^{-n}}(z_Q))\le C_\mu h(2^{-n})=C_\mu F(n).$$
This bound works simultaneously for every translated grid. We will use this concrete choice of the mass bound in the next theorem.
Theorem 2 (Iterated logarithmic gauges). Define $$L_1(r)=\log(1/r)$$ and $$L_{m+1}(r)=\log L_m(r)$$.
For each fixed integer $$k\ge2$$, set, for sufficiently small positive radii, $$h_k(r)=r\prod_{m=1}^kL_m(r)^2$$. Then the singular set of every suitable weak solution on the cylinder above satisfies $$\mathcal P^{h_k}(S)=0$$.
The gauge is extended increasingly away from zero. The conclusion is independent of that extension, because Hausdorff measure only uses arbitrarily small covering cylinders.
How the proof fits together
The proof uses the cylindrical activity $$G$$, the mixed dissipation $$e$$, their box counterparts, and a Frostman measure $$\mu$$. The auxiliary coefficients for Carleson embedding will be introduced when needed.
- A singular point has activity at every scale. Epsilon-regularity gives a uniform positive lower bound for $$G(r,z)$$ at every sufficiently small radius. The local energy inequality bounds $$A(r/2,z)+E(r/2,z)$$ by $$C(G(r,z)^{2/3}+G(r,z))$$.
- Persistent activity has an infinite weighted cost. Lemma 2 shows contraction of $$G$$ unless $$e$$ pays for it. Lemma 3 turns this into an infinite sum of mixed dissipation divided by the square root of activity, even after inserting any positive decreasing weights with divergent sum.
- The same cost has a finite trace. From $$\mu(Q)\le C_\mu F(n)$$, Lemma 4 bounds the sum of $$\mu(Q)e_Q/\sqrt{E_Q\,2^nF(n)}$$ over all boxes and all levels by the square root of the two total masses.
- Cylinders and boxes give the same comparison. Lemma 5 constructs finitely many nested grids and compares the cylindrical cost with their box costs. Frostman's lemma then supplies a probability measure on a compact set of persistent points. The weights $$1/\sqrt{2^nF(n)}$$ make the cost infinite at every point of that set, while the trace estimate makes its integral finite. This contradiction proves Theorem 1. Theorem 2 follows by checking its two sequence conditions.
The only Navier–Stokes regularity result taken as an input is the classical velocity–pressure epsilon-regularity criterion. The additional tools are Sobolev–Poincaré, the boundedness of Riesz transforms, dyadic Carleson embedding, and Frostman's lemma. We spell out the embedding argument as well as its Carleson estimate.
3. Local energy and epsilon-regularity
Lemma 1 (Local energy and persistent activity). There is a dimensional constant such that
$$A(r/2,z)+E(r/2,z) \le C\bigl(G(r,z)^{2/3}+G(r,z)\bigr). \tag{1}$$
There is also a fixed positive threshold $$\varepsilon_*$$ such that every singular point satisfies
$$G(r,z)\ge\varepsilon_* \qquad (z\in S\cap L,\ 0<r\le1/2). \tag{2}$$
Proof of the energy estimate. Take a spatial cutoff supported in the large ball and equal to one in the ball of half the radius. Multiply it by a time cutoff which vanishes near the initial time of the large cylinder and equals one on the middle time interval. The resulting function can be chosen so that
$$0\le\phi\le1,\qquad \phi=1\text{ on }C_{r/2}(z),\qquad |\nabla\phi|\le Cr^{-1},\qquad |\partial_t\phi|+|\Delta\phi|\le Cr^{-2}.$$
Apply the local energy inequality up to a time in the middle interval. Taking the essential supremum of its energy term controls $$A(r/2,z)$$. Letting the final time approach the upper end of that interval controls $$E(r/2,z)$$. After division by the radius, both are bounded by
$$C\left[ r^{-3}\int_{C_r(z)}|u|^2 +r^{-2}\int_{C_r(z)}\bigl(|u|^3+|p||u|\bigr) \right].$$
Here is the estimate of each term, including the powers of the radius. Since the cylinder has volume proportional to $$r^5$$,
$$\begin{aligned} r^{-3}\int_{C_r(z)}|u|^2 &\le r^{-3}|C_r|^{1/3} \left(\int_{C_r(z)}|u|^3\right)^{2/3}\\ &\le Cr^{-3}r^{5/3}\bigl(r^2G(r,z)\bigr)^{2/3}\\ &=CG(r,z)^{2/3},\\[2mm] r^{-2}\int_{C_r(z)}|u|^3&\le G(r,z),\\[2mm] r^{-2}\int_{C_r(z)}|p||u| &\le \left(r^{-2}\int_{C_r(z)}|p|^{3/2}\right)^{2/3} \left(r^{-2}\int_{C_r(z)}|u|^3\right)^{1/3}\\ &\le G(r,z)^{2/3}G(r,z)^{1/3}\\ &=G(r,z). \end{aligned}$$
This proves (1).
Why singular points satisfy (2). We use the classical velocity–pressure epsilon-regularity criterion: sufficiently small scaled cubic velocity and pressure integral on a backward cylinder implies that the velocity is bounded on the backward cylinder of half its radius. This is the only regularity theorem used here; see Guevara–Phuc, Theorem 1.1.
To apply it to our symmetric cylinders, shift the top time forward:
$$\begin{aligned} P_r(x,t+r^2/8) &=B_r(x)\times(t-7r^2/8,t+r^2/8) \subset C_r(z),\\ P_{r/2}(x,t+r^2/8) &=B_{r/2}(x)\times(t-r^2/8,t+r^2/8). \end{aligned}$$
The second cylinder is an open neighborhood of the original center. Consequently, if $$G(r,z)$$ is below the epsilon-regularity threshold, the scaled integral on the first cylinder is also below that threshold, and the solution is bounded near the original center. Taking the contrapositive proves (2). □
4. The scale recurrence
Lemma 2 (Scale recurrence).
Fix a sufficiently large dyadic constant $$K$$. There are a contraction factor $$q\in(0,1)$$, a dyadic scale ratio $$\theta\in(0,K^{-1})$$, and a constant $$C_*$$ such that
$$G(\theta r,z) \le qG(r,z)+C_*\sqrt{G(r,z)}\,e(r/K,z) \quad\text{whenever }G(r,z)\ge\varepsilon_*. \tag{3}$$
The size of $$K$$ is left open here because it will later be chosen to fit the geometry of dyadic boxes. Once it is fixed, all the other constants are fixed as well. The estimate says that persistent activity requires a payment in mixed dissipation; the square root in that payment will determine the weights in the final argument.
Proof. We first compare two arbitrary nested radii. Suppress the center, and suppose that $$0<s\le\rho/2$$.
We claim that
$$G(s) \le C\frac{s}{\rho}G(\rho) +C\left(\frac{\rho}{s}\right)^2A(\rho)^{1/2}e(\rho). \tag{4}$$
The velocity fluctuation. At almost every time, subtract the spatial mean on the large ball:
$$b(\tau)=\frac1{|B_\rho|}\int_{B_\rho(x)}u(y,\tau)\,dy, \qquad v(y,\tau)=u(y,\tau)-b(\tau).$$
The Sobolev–Poincaré inequality for a mean-zero function gives
$$\|v(\cdot,\tau)\|_{L^4(B_\rho)} \le C\|\nabla u(\cdot,\tau)\|_{L^{12/7}(B_\rho)}. \tag{5}$$
There is no radius factor: scaling the unit-ball inequality produces the factor $$\rho^{3/4+1-3/(12/7)}=\rho^{3/4+1-7/4}=1$$.
This explains our choice of the mixed exponent: the Sobolev conjugate of $$12/7$$ is exactly $$4$$, and Cauchy–Schwarz then connects this spatial fourth power to the cubic velocity. More explicitly,
$$\begin{aligned} \int_{B_\rho}|v|^3 &=\int_{B_\rho}|v|\,|v|^2\\ &\le\left(\int_{B_\rho}|v|^2\right)^{1/2} \left(\int_{B_\rho}|v|^4\right)^{1/2}\\ &=\|v\|_{L^2(B_\rho)}\|v\|_{L^4(B_\rho)}^2\\ &\le C\left(\int_{B_\rho}|u|^2\right)^{1/2} \left(\int_{B_\rho}|\nabla u|^{12/7}\right)^{7/6}. \end{aligned}$$
In the last line we used (5) and the elementary identity
$$\int_{B_\rho}|u-b|^2 =\int_{B_\rho}|u|^2-|B_\rho|\,|b|^2 \le\int_{B_\rho}|u|^2.$$
Now take the first factor uniformly in time and integrate the second factor. By the definitions of the scaled quantities,
$$\begin{aligned} \int_{C_\rho}|v|^3 &\le C\left( \operatorname*{ess\,sup}_{|\tau-t|<\rho^2} \int_{B_\rho}|u(y,\tau)|^2\,dy\right)^{1/2} \int_{t-\rho^2}^{t+\rho^2} \left(\int_{B_\rho}|\nabla u|^{12/7}\right)^{7/6}d\tau\\ &=C\bigl(\rho A(\rho)\bigr)^{1/2}\rho^{3/2}e(\rho)\\ &=C\rho^2A(\rho)^{1/2}e(\rho). \end{aligned} \tag{6}$$
The spatial mean. Jensen's inequality gives, at almost every time,
$$|b(\tau)|^3 =\left|\frac1{|B_\rho|}\int_{B_\rho}u\right|^3 \le\frac1{|B_\rho|}\int_{B_\rho}|u|^3.$$
The mean is constant in space. Therefore its contribution on the small cylinder gains the ratio of the spatial volumes:
$$\begin{aligned} \int_{C_s}|b(\tau)|^3\,dy\,d\tau &=|B_s|\int_{t-s^2}^{t+s^2}|b(\tau)|^3\,d\tau\\ &\le\frac{|B_s|}{|B_\rho|} \int_{t-s^2}^{t+s^2}\int_{B_\rho}|u|^3\,dy\,d\tau\\ &\le\left(\frac{s}{\rho}\right)^3\int_{C_\rho}|u|^3. \end{aligned}$$
Using the pointwise inequality $$|v+b|^3\le4|v|^3+4|b|^3$$ and then (6), we obtain
$$\begin{aligned} s^{-2}\int_{C_s}|u|^3 &\le4s^{-2}\int_{C_\rho}|v|^3 +4s^{-2}\left(\frac{s}{\rho}\right)^3\int_{C_\rho}|u|^3\\ &\le C\left(\frac{\rho}{s}\right)^2A(\rho)^{1/2}e(\rho) +C\frac{s}{\rho}\left(\rho^{-2}\int_{C_\rho}|u|^3\right). \end{aligned} \tag{7}$$
The pressure. Taking the spatial divergence of the equation gives the pressure equation. Because the spatial mean depends only on time, incompressibility also removes every term containing that mean:
$$\begin{aligned} -\Delta p &=\sum_{i,j=1}^3\partial_i\partial_j(u_i u_j),\\ \sum_{i,j=1}^3\partial_i\partial_j(u_i u_j-v_i v_j) &=\sum_{i,j=1}^3 \left[b_i\partial_i\partial_j u_j +b_j\partial_i\partial_j u_i\right]\\ &=\sum_{i=1}^3 b_i\partial_i(\operatorname{div}u) +\sum_{j=1}^3 b_j\partial_j(\operatorname{div}u)\\ &=0. \end{aligned}$$
At each such time, extend the product of the velocity fluctuations by zero outside the large ball and set
$$p_1=\sum_{i,j=1}^3R_iR_j\bigl(\mathbf1_{B_\rho}v_i v_j\bigr), \qquad p_2=p-p_1.$$
Here $$R_i$$ is the spatial Riesz transform with Fourier multiplier $$-\mathrm i\xi_i/|\xi|$$. Thus $$p_1$$ solves the pressure equation generated by this localized product. On the interior of the large ball its right-hand side agrees with that of the actual pressure; consequently $$p_2$$ is harmonic there. The boundedness of Riesz transforms on spatial $$L^{3/2}$$ gives
$$\begin{aligned} \|p_1\|_{L^{3/2}(\mathbb R^3)} &\le C\sum_{i,j=1}^3\|\mathbf1_{B_\rho}v_i v_j\|_{L^{3/2}(\mathbb R^3)}\\ &\le C\|v\|_{L^3(B_\rho)}^2, \end{aligned} \qquad \int_{\mathbb R^3}|p_1|^{3/2}\le C\int_{B_\rho}|v|^3. \tag{8}$$
For the harmonic remainder we use the mean-value property and Jensen's inequality. If $$y\in B_{\rho/2}(x)$$, the ball of radius $$\rho/2$$ about $$y$$ is contained in the large ball. Hence
$$\begin{aligned} |p_2(y)|^{3/2} &=\left|\frac1{|B_{\rho/2}|} \int_{B_{\rho/2}(y)}p_2\right|^{3/2}\\ &\le\frac1{|B_{\rho/2}|} \int_{B_{\rho/2}(y)}|p_2|^{3/2}\\ &\le C\rho^{-3}\int_{B_\rho(x)}|p_2|^{3/2}. \end{aligned}$$
Integrate this uniform bound over the small ball and use the pressure decomposition:
$$\begin{aligned} \int_{B_s}|p_2|^{3/2} &\le C\left(\frac{s}{\rho}\right)^3 \int_{B_\rho}|p_2|^{3/2}\\ &\le C\left(\frac{s}{\rho}\right)^3 \left(\int_{B_\rho}|p|^{3/2} +\int_{\mathbb R^3}|p_1|^{3/2}\right)\\ &\le C\left(\frac{s}{\rho}\right)^3 \left(\int_{B_\rho}|p|^{3/2} +\int_{B_\rho}|v|^3\right). \end{aligned}$$
Now integrate over the smaller time interval, enlarge that interval where the integrands are nonnegative, and apply (8). Since the spatial radius ratio is at most one, its factor in front of the fluctuation term can be absorbed:
$$\begin{aligned} \int_{C_s}|p|^{3/2} &\le C\int_{C_s}|p_1|^{3/2}+C\int_{C_s}|p_2|^{3/2}\\ &\le C\int_{C_\rho}|v|^3 +C\left(\frac{s}{\rho}\right)^3\int_{C_\rho}|p|^{3/2},\\[1mm] s^{-2}\int_{C_s}|p|^{3/2} &\le C\left(\frac{\rho}{s}\right)^2A(\rho)^{1/2}e(\rho) +C\frac{s}{\rho}\left(\rho^{-2}\int_{C_\rho}|p|^{3/2}\right). \end{aligned} \tag{9}$$
Adding (7) and (9) proves the claimed two-radius estimate (4).
Choosing the scales. We now use the constants needed in the recurrence. Let $$K\ge2$$ be the fixed dyadic constant that will be chosen in the geometric comparison, and put $$\rho=r/K$$ and $$s=\theta r$$, with $$\theta\le(2K)^{-1}$$.
The containment of cylinders and the definitions give
$$\begin{aligned} G(r/K) &=\frac{K^2}{r^2}\int_{C_{r/K}}(|u|^3+|p|^{3/2})\\ &\le K^2G(r),\\[1mm] A(r/K) &\le\frac Kr\operatorname*{ess\,sup}_{|\tau-t|<(r/2)^2} \int_{B_{r/2}}|u(y,\tau)|^2\,dy\\ &=\frac K2A(r/2)\\ &\le C_K\bigl(G(r)^{2/3}+G(r)\bigr). \end{aligned}$$
Insert these estimates into (4), and use the elementary square-root bound for a sum:
$$\begin{aligned} G(\theta r) &\le C(K\theta)K^2G(r) +C(K\theta)^{-2}A(r/K)^{1/2}e(r/K)\\ &\le CK^3\theta G(r) +C_{K,\theta}\bigl(G(r)^{1/3}+G(r)^{1/2}\bigr)e(r/K). \end{aligned}$$
Choose a sufficiently small dyadic value of $$\theta$$ so that the first coefficient is at most $$q=1/4$$. Whenever the activity exceeds the threshold, its smaller power can be absorbed into its square root:
$$G(r)^{1/3}=G(r)^{1/2}G(r)^{-1/6} \le\varepsilon_*^{-1/6}G(r)^{1/2}.$$
Absorbing the fixed constants into $$C_*$$ proves (3). The numerical contraction factor has no special role; every fixed factor strictly below one would work. □
5. Weighted persistent activity
Lemma 3 (Weighted persistent activity).
Fix a sufficiently small radius and use the geometric scales supplied by Lemma 2, writing $$r_j=R\theta^j$$, $$G_j(z)=G(r_j,z)$$, and $$e_j(z)=e(r_j/K,z)$$.
Let the weights be positive and nonincreasing, and suppose that their series diverges, so $$w_{j+1}\le w_j$$ and $$\sum_{j\ge j_0}w_j=\infty$$.
If activity remains above the regularity threshold at all these scales, then
$$G_j(z)\ge\varepsilon_*\quad(j\ge j_0) \qquad\Longrightarrow\qquad \sum_{j\ge j_0}w_j\frac{e_j(z)}{\sqrt{G_j(z)}}=\infty. \tag{10}$$
Proof. Fix the point and suppress it from the notation. Define the nonnegative logarithmic activity $$a_j=\log\frac{G_j}{\varepsilon_*}\ge0$$.
Divide the recurrence by the activity on the larger scale, take logarithms, and use the elementary inequality $$\log(1+s)\le s$$ for nonnegative inputs:
$$\begin{aligned} a_{j+1}-a_j &=\log\frac{G_{j+1}}{G_j}\\ &\le\log\left(q+C_*\frac{e_j}{\sqrt{G_j}}\right)\\ &=\log q+\log\left(1+\frac{C_*}{q}\frac{e_j}{\sqrt{G_j}}\right)\\ &\le\log q+\frac{C_*}{q}\frac{e_j}{\sqrt{G_j}}. \end{aligned}$$
It remains to check that logarithmic activity cannot pay for an infinite sequence of contractions. This is exactly where monotonicity of the weights enters. Summation by parts gives the full identity
$$\begin{aligned} \sum_{j=j_0}^Nw_j(a_{j+1}-a_j) &=\sum_{j=j_0+1}^{N+1}w_{j-1}a_j -\sum_{j=j_0}^Nw_ja_j\\ &=w_Na_{N+1}-w_{j_0}a_{j_0} +\sum_{j=j_0+1}^N(w_{j-1}-w_j)a_j\\ &\ge-w_{j_0}a_{j_0}. \end{aligned}$$
The final inequality uses both nonnegativity of logarithmic activity and nonincreasing weights. Combining it with the preceding logarithmic bound yields
$$\begin{aligned} \frac{C_*}{q}\sum_{j=j_0}^Nw_j\frac{e_j}{\sqrt{G_j}} &\ge -\log q\sum_{j=j_0}^Nw_j +\sum_{j=j_0}^Nw_j(a_{j+1}-a_j)\\ &\ge -\log q\sum_{j=j_0}^Nw_j-w_{j_0}a_{j_0}\\ &\longrightarrow\infty. \end{aligned}$$
The coefficient of the divergent sum is positive because the contraction factor is less than one. This proves (10). Notice that the proof places no upper bound on activity: even very large activity cannot invalidate the weighted lower bound. □
6. The dissipation trace
Lemma 4 (Dissipation trace). Suppose $$\mu$$ is a finite nonnegative Borel measure, and, for every level-$$n$$ box in a fixed parabolic dyadic grid, $$\mu(Q)\le C_\mu F(n)$$ for $$n\ge n_0$$.
Assume $$F(n)>0$$ and that $$2^nF(n)$$ is nondecreasing. Then
$$\boxed{ \sum_{n\ge n_0}\ \sum_{\operatorname{level}(Q)=n} \frac{\mu(Q)e_Q}{\sqrt{E_Q\,2^nF(n)}} \le C\sqrt{\mu(\mathbb R^3\times\mathbb R)\, \nu(\mathbb R^3\times\mathbb R)}. }$$
The constant depends only on $$C_\mu$$ and the dimension. In particular, it does not depend on the number of scales in the sum.
Proof. We first identify the estimate supplied by Carleson embedding and use it to prove the trace inequality. We then prove the required Carleson estimate.
At a fixed time $$t$$, denote by $$I_n(t)$$ the unique level-$$n$$ time interval containing $$t$$. For a level-$$n$$ spatial cube $$X$$, define $$\alpha_X(t)=\ell_n^3 \sqrt{\frac{\mu(X\times I_n(t))}{F(n)}}$$.
Suppose for the moment that we can prove the following spatial Carleson estimate: for every spatial dyadic cube $$X_0$$ of level at least $$n_0$$,
$$\sum_{X\subseteq X_0}\alpha_X(t)\le C|X_0|, \tag{C}$$
with a constant independent of the time and the cube. The sum includes $$X_0$$ and all its descendants. This assertion controls the sum of coefficients over all scales inside every root cube; a bound at only one scale would not suffice.
The dyadic Carleson embedding theorem says that a family satisfying (C) obeys
$$\sum_{X\subseteq X_0}\alpha_X(t) \left(\frac1{|X|}\int_X f(x)\,dx\right)^p \le C_p\int_{X_0}f(x)^p\,dx, \qquad f\ge0,\quad 1<p<\infty.$$
Apply this with $$p=\frac76$$ and $$f(x)=|\nabla u(x,t)|^{12/7}$$. The choice of exponent gives $$f^p=|\nabla u|^2$$, so the conclusion is
$$\sum_{X\subseteq X_0}\alpha_X(t) \left(\frac1{|X|}\int_X|\nabla u(x,t)|^{12/7}\,dx\right)^{7/6} \le C\int_{X_0}|\nabla u(x,t)|^2\,dx. \tag{CE}$$
We next use (CE) to prove the lemma. Fix a spacetime root box $$Q_0=X_0\times I_0$$. Since $$|X|=\ell_n^3$$, the exact conversion from spatial integrals to spatial averages is
$$\begin{aligned} e_Q &=\ell_n^{-3/2}\int_I \left(\int_X|\nabla u|^{12/7}\,dx\right)^{7/6}dt\\ &=\ell_n^{-3/2}|X|^{7/6}\int_I \left(\frac1{|X|}\int_X|\nabla u|^{12/7}\,dx\right)^{7/6}dt\\ &=\ell_n^{-3/2}\ell_n^{7/2}\int_I \left(\frac1{|X|}\int_X|\nabla u|^{12/7}\,dx\right)^{7/6}dt\\ &=\ell_n^2\int_I \left(\frac1{|X|}\int_X|\nabla u|^{12/7}\,dx\right)^{7/6}dt. \end{aligned}$$
The coefficient that occurs after integrating (CE) will be used repeatedly, so now define $$\beta_Q=\ell_n\sqrt{\frac{\mu(Q)}{F(n)}}\,e_Q$$.
For a fixed time in $$I_0$$ and a fixed spatial descendant $$X$$, exactly one descendant time interval at the same level contains that time. Consequently, Tonelli's theorem, the preceding identity, and (CE) give
$$\begin{aligned} \sum_{Q\subseteq Q_0}\beta_Q &=\sum_{\substack{Q=X\times I\subseteq Q_0\\\operatorname{level}(Q)=n,\ n\ge\operatorname{level}(Q_0)}} \int_I\ell_n^3\sqrt{\frac{\mu(Q)}{F(n)}} \left(\frac1{|X|}\int_X|\nabla u|^{12/7}\,dx\right)^{7/6}dt\\ &=\int_{I_0}\sum_{X\subseteq X_0}\alpha_X(t) \left(\frac1{|X|}\int_X|\nabla u(x,t)|^{12/7}\,dx\right)^{7/6}dt\\ &\le C\int_{I_0}\int_{X_0}|\nabla u(x,t)|^2\,dx\,dt\\ &=C\nu(Q_0). \end{aligned} \tag{R}$$
All terms are nonnegative, which is why exchanging the integrals and infinite sums requires no prior convergence assertion. The estimate holds inside every root box $$Q_0$$ of level at least $$n_0$$.
To finish, we must introduce the extra factor that converts $$\beta_Q$$ into the summand in the lemma. The algebra is
$$\begin{aligned} \beta_Q\sqrt{\frac{\mu(Q)}{\nu(Q)}} &=\ell_n\sqrt{\frac{\mu(Q)}{F(n)}}\,e_Q \sqrt{\frac{\mu(Q)}{\nu(Q)}}\\ &=\frac{\mu(Q)\ell_ne_Q}{\sqrt{F(n)\nu(Q)}}\\ &=\frac{\mu(Q)e_Q} {\sqrt{(\nu(Q)/\ell_n)(F(n)/\ell_n)}}\\ &=\frac{\mu(Q)e_Q}{\sqrt{E_Q\,2^nF(n)}}. \end{aligned}$$
This identity is used only when $$\nu(Q)>0$$. The terms with zero dissipation have already been assigned the value zero.
Fix $$\lambda>0$$ and consider the boxes for which $$\nu(Q)>0$$ and $$\mu(Q)>\lambda\nu(Q)$$.
Take the maximal such boxes among levels at least $$n_0$$. Every box under consideration is contained in one of them: there are only finitely many ancestors between its level and $$n_0$$, so one can take the highest ancestor still satisfying these conditions. Distinct maximal boxes are disjoint, because two boxes of one nested grid are either disjoint or one contains the other.
Apply (R) inside each maximal box. Disjointness bounds their total $$\nu$$-mass by the total dissipation; their defining inequality also bounds it by the total $$\mu$$-mass divided by $$\lambda$$. Writing the maximal boxes as $$Q_i$$, we get both estimates in the chain
$$\begin{aligned} \sum_{\substack{\nu(Q)>0\\\mu(Q)>\lambda\nu(Q)}}\beta_Q &\le\sum_i\sum_{Q\subseteq Q_i}\beta_Q\\ &\le C\sum_i\nu(Q_i)\\ &\le C\min\left\{\nu(\mathbb R^3\times\mathbb R), \frac{\mu(\mathbb R^3\times\mathbb R)}{\lambda}\right\}. \end{aligned}$$
The layer-cake representation expresses a square root through its level sets:
$$\sqrt a=\frac12\int_0^\infty\lambda^{-1/2}\mathbf1_{\{a>\lambda\}}\,d\lambda, \qquad a\ge0.$$
Apply this to the mass ratio and use Tonelli again. For brevity, write the two total masses only in this calculation as $$m=\mu(\mathbb R^3\times\mathbb R)$$ and $$v=\nu(\mathbb R^3\times\mathbb R)$$.
If either is zero, the desired trace sum is zero. Otherwise split the integral at $$\lambda=m/v$$:
$$\begin{aligned} \sum_{\nu(Q)>0}\beta_Q\sqrt{\frac{\mu(Q)}{\nu(Q)}} &=\frac12\int_0^\infty\lambda^{-1/2} \sum_{\substack{\nu(Q)>0\\\mu(Q)>\lambda\nu(Q)}}\beta_Q\,d\lambda\\ &\le\frac C2\int_0^\infty\lambda^{-1/2}\min\{v,m/\lambda\}\,d\lambda\\ &=\frac C2\left( v\int_0^{m/v}\lambda^{-1/2}\,d\lambda +m\int_{m/v}^\infty\lambda^{-3/2}\,d\lambda\right)\\ &=\frac C2\left(2v\sqrt{m/v}+2m\sqrt{v/m}\right)\\ &=2C\sqrt{mv}. \end{aligned}$$
This proves the trace inequality, subject to (C).
Verification of the Carleson estimate. Let $$X_0$$ have level $$k$$. At a fixed finer level $$n\ge k$$, its spatial descendants are disjoint and number $$\bigl(\ell_k/\ell_n\bigr)^3$$.
Their products with the one time interval $$I_n(t)$$ are disjoint as well. Moreover, the nesting of the time intervals gives
$$\bigcup_{\substack{X\subseteq X_0\\\operatorname{level}(X)=n}} \bigl(X\times I_n(t)\bigr) =X_0\times I_n(t) \subseteq X_0\times I_k(t).$$
Apply Cauchy–Schwarz to the finite sum of square roots of their masses. Disjointness and the last inclusion then give
$$\begin{aligned} \sum_{\substack{X\subseteq X_0\\\operatorname{level}(X)=n}}\alpha_X(t) &=\frac{\ell_n^3}{\sqrt{F(n)}} \sum_{\substack{X\subseteq X_0\\\operatorname{level}(X)=n}} \sqrt{\mu(X\times I_n(t))}\\ &\le\frac{\ell_n^3}{\sqrt{F(n)}} \left(\frac{\ell_k}{\ell_n}\right)^{3/2} \left(\sum_{\substack{X\subseteq X_0\\\operatorname{level}(X)=n}} \mu(X\times I_n(t))\right)^{1/2}\\ &\le\ell_k^{3/2}\sqrt{\mu(X_0\times I_k(t))} \frac{\ell_n^{3/2}}{\sqrt{F(n)}}. \end{aligned}$$
This is the only place we need the number of spatial descendants. The monotonicity assumption on $$2^nF(n)=F(n)/\ell_n$$ yields
$$\frac{F(n)}{\ell_n}\ge\frac{F(k)}{\ell_k}, \qquad \frac{\ell_n^{3/2}}{\sqrt{F(n)}} \le\ell_n\sqrt{\frac{\ell_k}{F(k)}}.$$
Sum the preceding level estimate and use the geometric series, followed by the mass bound at the root:
$$\begin{aligned} \sum_{X\subseteq X_0}\alpha_X(t) &\le\ell_k^{3/2}\sqrt{\mu(X_0\times I_k(t))} \sum_{n\ge k}\frac{\ell_n^{3/2}}{\sqrt{F(n)}}\\ &\le\ell_k^2 \sqrt{\frac{\mu(X_0\times I_k(t))}{F(k)}} \sum_{n\ge k}\ell_n\\ &=2\ell_k^3 \sqrt{\frac{\mu(X_0\times I_k(t))}{F(k)}}\\ &\le2\sqrt{C_\mu}\,\ell_k^3\\ &=2\sqrt{C_\mu}\,|X_0|. \end{aligned}$$
This proves (C), and completes the proof. $$\square$$
Proof of the Carleson embedding used above
Here is the short argument behind the embedding used above. Fix the time and the root spatial cube. Let $$\alpha_X\ge0$$ satisfy (C), and define the localized dyadic maximal function
$$M f(x)=\sup_{\substack{x\in X\subseteq X_0\\X\text{ dyadic}}} \frac1{|X|}\int_X f.$$
For each $$s>0$$, the maximal dyadic cubes whose average of $$f$$ exceeds $$s$$ are disjoint and their union is $$\{Mf>s\}$$. Every cube with such an average lies inside one of those maximal cubes. Thus (C), applied separately inside each maximal cube, gives
$$\sum_{\substack{X\subseteq X_0\\ |X|^{-1}\int_X f>s}}\alpha_X \le C|\{Mf>s\}|.$$
Using the layer-cake formula for a power gives
$$\begin{aligned} \sum_{X\subseteq X_0}\alpha_X \left(\frac1{|X|}\int_X f\right)^p &=p\int_0^\infty s^{p-1} \sum_{\substack{X\subseteq X_0\\ |X|^{-1}\int_X f>s}}\alpha_X\,ds\\ &\le Cp\int_0^\infty s^{p-1}|\{Mf>s\}|\,ds\\ &=C\int_{X_0}(Mf)^p\\ &\le C\left(\frac p{p-1}\right)^p\int_{X_0}f^p. \end{aligned}$$
For completeness, the last maximal-function inequality follows from those same disjoint maximal cubes. On each such cube, its measure is less than its integral of $$f$$ divided by $$s$$. Summing gives $$|\{Mf>s\}|\le\frac1s\int_{\{Mf>s\}}f$$.
For the maximal function restricted to finitely many levels, layer cake and Hölder therefore imply
$$\begin{aligned} \|Mf\|_p^p &=p\int_0^\infty s^{p-1}|\{Mf>s\}|\,ds\\ &\le p\int_0^\infty s^{p-2}\int_{\{Mf>s\}}f(x)\,dx\,ds\\ &=p\int_{X_0}f(x)\int_0^{Mf(x)}s^{p-2}\,ds\,dx\\ &=\frac p{p-1}\int_{X_0}f(x)(Mf(x))^{p-1}\,dx\\ &\le\frac p{p-1}\|f\|_p\|Mf\|_p^{p-1}. \end{aligned}$$
Divide by the last factor when it is nonzero, and then let the number of levels tend to infinity by monotone convergence. This proves the asserted bound without assuming in advance that the infinite-level maximal function belongs to the required space.
7. From cylinders to dyadic boxes
The trace estimate concerns boxes in one nested grid, whereas the recurrence concerns cylinders centered at arbitrary points. A single grid cannot compare all those cylinders: a small cylinder may straddle a grid boundary at every large scale. We now construct a finite collection of grids so that at least one of them has a box containing any given cylinder at a comparable scale. The familiar name “one-third trick” refers to the spacing of the three grids used in each coordinate; the construction below is the whole trick needed here.
Lemma 5 (Cylinder-to-box comparison). There are $$81$$ fixed nested parabolic dyadic grids with the following property. Choose $$K=128$$, and let $$r$$ be a dyadic radius. For every center $$z=(x,t)$$, there is a box in one of these grids such that
$$C_{r/K}(z)\subset Q\subset C_{r/2}(z), \qquad \operatorname{side}(X)=16r/K.$$
If $$G(r,z)\ge\varepsilon_*$$ and the cylinders lie in the region where the local energy estimate holds, then $$\frac{e(r/K,z)}{\sqrt{G(r,z)}} \le C\frac{e_Q}{\sqrt{E_Q}}$$.
If $$E_Q=0$$, both sides are zero. The constant is independent of the center and the scale.
Proof. We construct the grids, prove the containment, and then compare the integrals.
The three grids on a spatial axis. At level $$n$$, put $$\ell=2^{-n}$$. For each of the three choices $$a\in\{0,1,2\}$$, use the intervals
$$\ell\left[m+\frac{(-1)^na}{3},\ m+1+\frac{(-1)^na}{3}\right), \qquad m\in\mathbb Z. \tag{S}$$
The alternating sign ensures nesting. Indeed, a level-$$n$$ boundary, when expressed in units of the finer length $$\ell/2$$, is
$$\begin{aligned} 2\left(m+\frac{(-1)^na}{3}\right) &=2m+\frac{2(-1)^na}{3}\\ &=\bigl(2m+(-1)^na\bigr)+\frac{(-1)^{n+1}a}{3}. \end{aligned}$$
The expression in parentheses is an integer. Thus every boundary at level $$n$$ is also a boundary at level $$n+1$$, and every interval is the union of two intervals at the next level. This is why simply choosing the same fractional shift $$a/3$$ at every spatial scale would not be enough.
At any one level, the union of the boundaries from the three grids is exactly $$\left\{\frac{j\ell}{3}:j\in\mathbb Z\right\}$$.
These boundary points are spaced $$\ell/3$$ apart. Any closed interval of length less than $$\ell/3$$ therefore meets boundaries from at most one of the three grids. In either of the other two grids, that interval lies entirely in the interior of one level-$$n$$ interval.
The three grids on the time axis. The time length at level $$n$$ is $$\ell^2=4^{-n}$$. For each $$b\in\{0,1,2\}$$, use
$$\ell^2\left[m+\frac b3,\ m+1+\frac b3\right), \qquad m\in\mathbb Z. \tag{T}$$
Here no alternating sign is needed, because subdivision is into four intervals and $$4\left(m+\frac b3\right)=(4m+b)+\frac b3$$.
Thus every coarse boundary is again a fine boundary. At a fixed level, the three time grids have combined boundaries spaced $$\ell^2/3$$ apart.
Choose one of the three spatial grids independently on each of the three spatial axes, and one of the three time grids. Their products give $$3\cdot3\cdot3\cdot3=81$$ parabolic grids. Each is nested: a box has eight spatial children and four temporal children, hence $$32$$ children in spacetime.
Finding a box for the cylinder. Write $$\rho=r/K$$ and $$\ell=16\rho$$.
Both are dyadic, so $$\ell$$ is the spatial length at an integer level. The projection of the closed spatial ball onto the $$i$$th coordinate axis is contained in $$[x_i-\rho,x_i+\rho]$$, whose length is $$2\rho=\ell/8<\ell/3$$.
Choose a grid on each spatial axis with no boundary in that closed interval. The product of the resulting three containing intervals is a spatial cube $$X$$ containing the ball. Independently, the time projection satisfies
$$|[t-\rho^2,t+\rho^2]|=2\rho^2=\ell^2/128<\ell^2/3,$$
so one time grid has a level-$$n$$ interval $$I$$ containing that whole closed interval. Consequently $$C_\rho(z)\subset X\times I=Q$$.
In fact the cylinder closure lies in the interior of this box. In particular, $$Q$$ is the unique level-$$n$$ box of its chosen grid containing $$z$$; there is no ambiguity at a half-open boundary.
Next verify the outer containment. Since both $$y$$ and $$x$$ belong to the same spatial cube of side $$\ell$$, and both $$s$$ and $$t$$ belong to the same time interval of length $$\ell^2$$, every point $$(y,s)\in Q$$ satisfies
$$\begin{aligned} |y-x| &\le\sqrt{|y_1-x_1|^2+|y_2-x_2|^2+|y_3-x_3|^2}\\ &\le\sqrt3\,\ell =16\sqrt3\,\rho <64\rho=r/2, \end{aligned}$$
and
$$|s-t|\le \ell^2=256\rho^2<4096\rho^2=(r/2)^2.$$
This proves $$Q\subset C_{r/2}(z)$$.
Comparing the dissipation integrals. The inner containment and nonnegativity give, with all changes of normalization written out,
$$\begin{aligned} e(\rho,z) &=\rho^{-3/2}\int_{t-\rho^2}^{t+\rho^2} \left(\int_{B_\rho(x)}|\nabla u(y,s)|^{12/7}\,dy\right)^{7/6}ds\\ &\le\rho^{-3/2}\int_I \left(\int_X|\nabla u(y,s)|^{12/7}\,dy\right)^{7/6}ds\\ &=\rho^{-3/2}\ell^{3/2}e_Q\\ &=16^{3/2}e_Q. \end{aligned}$$
The outer containment gives the estimate in the other direction that we need for the denominators:
$$\begin{aligned} E_Q &=\ell^{-1}\int_Q|\nabla u|^2\\ &\le \ell^{-1}\int_{C_{r/2}(z)}|\nabla u|^2\\ &=\frac{r/2}{\ell}\,E(r/2,z)\\ &=\frac K{32}E(r/2,z)\\ &=4E(r/2,z)\\ &\le C\bigl(G(r,z)^{2/3}+G(r,z)\bigr)\\ &\le C\bigl(\varepsilon_*^{-1/3}+1\bigr)G(r,z). \end{aligned}$$
The penultimate line is the local energy estimate, and the last line uses $$G(r,z)\ge\varepsilon_*$$. Absorb the fixed factor in the constant. If $$E_Q>0$$, this implies
$$\begin{aligned} \frac{e(r/K,z)}{\sqrt{G(r,z)}} &\le16^{3/2}\frac{e_Q}{\sqrt{G(r,z)}}\\ &\le C\frac{e_Q}{\sqrt{E_Q}}. \end{aligned}$$
If $$E_Q=0$$, the gradient vanishes almost everywhere on $$Q$$; the inner containment then gives $$e(r/K,z)=0$$, so the zero convention gives the same result. $$\square$$
One useful consequence concerns measurability. Denote by $$Q_n^{(a)}(z)$$ the unique level-$$n$$ box containing $$z$$ in the $$a$$th grid. We need not choose a successful grid as a measurable function of the center. Because all terms are nonnegative, Lemma 5 immediately implies
$$\frac{e(r/K,z)}{\sqrt{G(r,z)}} \le C\sum_{a=1}^{81} \frac{e_{Q_n^{(a)}(z)}}{\sqrt{E_{Q_n^{(a)}(z)}}}, \qquad 2^{-n}=16r/K.$$
Each summand on the right is constant on a half-open box and is therefore measurable. Integrating it against $$\mu$$ gives exactly the box sum in Lemma 4:
$$\int \frac{e_{Q_n^{(a)}(z)}}{\sqrt{E_{Q_n^{(a)}(z)}}}\,d\mu(z) =\sum_{\operatorname{level}(Q)=n} \frac{\mu(Q)e_Q}{\sqrt{E_Q}}.$$
Thus the geometric comparison has produced precisely the expression controlled by the dissipation trace.
8. Proof of Theorem 1
Use the localization $$L=\overline{C_1(0)}$$, the interior cylinder $$U=C_2(0)$$, and the dissipation measure $$\nu$$ introduced in Section 1.
The proof works on every compact subcylinder of the original cylinder, by choosing all radii below its distance from the boundary. No rescaling of the gauge is needed.
Choose the dyadic constants from Lemmas 2 and 5, and then a sufficiently small dyadic starting radius. Write $$r_j=R\theta^j$$, $$G_j(z)=G(r_j,z)$$, and $$e_j(z)=e(r_j/K,z)$$.
All the cylinders below then lie inside the region where the finite dissipation measure is defined. Suppose, toward a contradiction, that $$\mathcal P^h(S\cap L)>0$$.
For a fixed radius, the activity is a continuous function of its center. To see this, extend its locally integrable density by zero after restricting it to a slightly larger compact region. Translation is continuous in the space of integrable functions, and hence
$$\begin{aligned} |G(r,z+a)-G(r,z)| &\le r^{-2}\int_{C_r(0)} \big|f(z+a+w)-f(z+w)\big|\,dw\\ &\le r^{-2}\|f(\,\cdot+a)-f\|_{L^1(\mathbb R^4)}\\ &\longrightarrow0, \end{aligned}$$
where the density agrees with $$f=|u|^3+|p|^{3/2}$$ on all cylinders in question. Therefore the persistent set $$T=L\cap\bigcap_{j\ge0}\{z:G_j(z)\ge\varepsilon_*\}$$ is compact. The epsilon-regularity criterion shows that it contains every singular point in the localization. In particular, $$\mathcal P^h(T)\ge\mathcal P^h(S\cap L)>0$$.
Apply the abstract Frostman hypothesis to obtain a probability measure supported on the persistent set, with $$\mu(Q)\le C_\mu F(n)$$ for every sufficiently fine level in every grid.
Let the box level corresponding to the radius in Lemma 5 be $$n_j=Mj+c$$, where $$\theta=2^{-M}$$. More explicitly, if $$R=2^{-b}$$ and $$K=128$$, then $$16r_j/K=2^{-(Mj+b+3)}$$, so $$c=b+3$$ is fixed independently of the center and scale index. Choose the starting index large enough that all these levels satisfy the hypotheses. The weight we need is exactly $$w_j=\frac1{\sqrt{2^{n_j}F(n_j)}}$$. These weights decrease. They also have divergent sum. For completeness, define only in this verification $$w(n)=\frac1{\sqrt{2^nF(n)}}$$.
For each block of consecutive indices, monotonicity gives
$$\sum_{n=n_j}^{n_{j+1}-1}w(n) \le\sum_{n=n_j}^{n_{j+1}-1}w(n_j) =Mw(n_j)=Mw_j.$$
Summing the blocks yields
$$\sum_{n=n_{j_0}}^{n_{N+1}-1}w(n) \le M\sum_{j=j_0}^{N}w_j.$$
The left side tends to infinity by the hypothesis of Theorem 1, so the sum on the right also diverges. This argument explains why no additional comparison between the sequence at two affine-related indices is needed.
Apply Lemma 3 with these weights. At every point of the persistent set,
$$\sum_{j\ge j_0}w_j\frac{e_j(z)}{\sqrt{G_j(z)}}=\infty.$$
We next integrate the same expression and obtain a finite bound. Index the finitely many grids by $$\mathscr D^1,\ldots,\mathscr D^{81}$$, and denote the unique level box containing the center by $$Q_n^a(z)\in\mathscr D^a$$.
The cylinder-to-box comparison of Lemma 5 says that at least one of these grids gives the estimate needed at each point and scale. All summands are nonnegative, so we may sum over every grid and obtain the pointwise bound
$$\frac{e_j(z)}{\sqrt{G_j(z)}} \le C\sum_{a=1}^{81}\frac{e_{Q_{n_j}^a(z)}}{\sqrt{E_{Q_{n_j}^a(z)}}}.$$
A term with zero dissipation is defined to be zero. Notice that we have not chosen a grid as a measurable function of the center. The finite sum avoids that issue entirely.
Since each level is a disjoint partition, integration of a box-constant function simply multiplies its value by the mass of that box. Tonelli's theorem and Lemma 4 therefore give the complete chain
$$\begin{aligned} \int_T\sum_{j\ge j_0}w_j \frac{e_j(z)}{\sqrt{G_j(z)}}\,d\mu(z) &\le C\sum_{a=1}^{81}\sum_{j\ge j_0} \int_T\frac{e_{Q_{n_j}^a(z)}} {\sqrt{E_{Q_{n_j}^a(z)}\,2^{n_j}F(n_j)}}\,d\mu(z)\\ &=C\sum_{a=1}^{81}\sum_{j\ge j_0} \sum_{\substack{Q\in\mathscr D^a\\\operatorname{level}(Q)=n_j}} \frac{\mu(Q)e_Q}{\sqrt{E_Q\,2^{n_j}F(n_j)}}\\ &\le C\sum_{a=1}^{81}\sum_{n\ge n_0} \sum_{\substack{Q\in\mathscr D^a\\\operatorname{level}(Q)=n}} \frac{\mu(Q)e_Q}{\sqrt{E_Q\,2^nF(n)}}\\ &\le C\cdot81\sqrt{\mu(\mathbb R^4)\nu(\mathbb R^4)}\\ &<\infty. \end{aligned}$$
However, the integrand on the left is infinite at every point of the persistent set, and that set has measure one. This is a contradiction. Hence $$\mathcal P^h(S\cap L)=0$$.
To cover the whole original cylinder, repeat the same proof on compact subcylinders whose interiors exhaust it. The energy is finite on a slightly larger compact cylinder at each stage, and the starting radius is chosen small enough for that stage. Countable subadditivity then gives $$\mathcal P^h(S)=0$$.
9. Proof of Theorem 2
Fix the number of logarithmic factors and take the radius sufficiently small that all the logarithms below are positive. The relevant dyadic mass bound is
$$F(n)=h_k(2^{-n}) =2^{-n}\prod_{m=1}^kL_m(2^{-n})^2.$$
This is the actual mass bound furnished by Frostman's lemma. In particular,
$$2^nF(n)=\prod_{m=1}^kL_m(2^{-n})^2, \qquad \frac1{\sqrt{2^nF(n)}} =\frac1{\displaystyle\prod_{m=1}^kL_m(2^{-n})}.$$
Every factor in the first product increases with the scale index, so the normalized mass bound is increasing and the reciprocal square-root weights are decreasing.
We check that these weights have divergent sum without dropping the constant introduced by the dyadic scale. For real positive scale parameter sufficiently large, $$L_1(2^{-x})=x\log2$$.
Repeated application of the chain rule gives
$$\begin{aligned} \frac{d}{dx}L_{k+1}(2^{-x}) &=\frac1{L_k(2^{-x})}\frac{d}{dx}L_k(2^{-x})\\ &=\frac1{L_k(2^{-x})L_{k-1}(2^{-x})} \frac{d}{dx}L_{k-1}(2^{-x})\\ &=\cdots\\ &=\frac1{\displaystyle\prod_{m=1}^kL_m(2^{-x})} \frac{d}{dx}L_1(2^{-x})\\ &=\frac{\log2}{\displaystyle\prod_{m=1}^kL_m(2^{-x})}. \end{aligned}$$
The integrand decreases, so comparison of each term with the integral over the following unit interval gives
$$\begin{aligned} \sum_{n=n_0}^{N}\frac1{\sqrt{2^nF(n)}} &=\sum_{n=n_0}^{N}\frac1{\displaystyle\prod_{m=1}^kL_m(2^{-n})}\\ &\ge\int_{n_0}^{N+1}\frac{dx}{\displaystyle\prod_{m=1}^kL_m(2^{-x})}\\ &=\frac{L_{k+1}(2^{-(N+1)})-L_{k+1}(2^{-n_0})}{\log2}\\ &\longrightarrow\infty. \end{aligned}$$
It remains to check that the function used to measure covers is indeed a gauge near zero. Put $$y=\log(1/r)$$.
Then
$$h_k(r)=e^{-y}\prod_{m=1}^kL_m(e^{-y})^2,\qquad L_1(e^{-y})=y.$$
For sufficiently large parameter, each logarithmic factor is at most the first factor; hence $$0<h_k(r)\le e^{-y}y^{2k}\longrightarrow0$$.
Furthermore,
$$\begin{aligned} \frac{d}{dy}\log h_k(e^{-y}) &=-1+2\sum_{j=1}^k\frac1{\displaystyle\prod_{m=1}^jL_m(e^{-y})}\\ &\longrightarrow-1. \end{aligned}$$
Thus the gauge decreases as the logarithmic parameter increases, which means that it increases as the radius increases. It is continuous and tends to zero at the origin. The preceding dyadic monotonicity also supplies its doubling property, as explained after Theorem 1. Frostman's lemma therefore verifies the abstract hypothesis. Both sequence conditions of Theorem 1 have now been checked, so that theorem proves $$\mathcal P^{h_k}(S)=0$$.
References
- Yongxi Lin, A logarithmic refinement of CKN partial regularity, original manuscript. This webpage expands and reorganizes its proof; Theorem 2 is Proposition 5 of the manuscript. The source manuscript describes the proof as AI-generated under the author's guidance.
- L. Caffarelli, R. Kohn, and L. Nirenberg, Partial regularity of suitable weak solutions of the Navier–Stokes equations, Communications on Pure and Applied Mathematics 35 (1982), 771–831. DOI.
- C. Guevara and N. C. Phuc, Local energy bounds and epsilon-regularity criteria for the 3D Navier–Stokes system, Calculus of Variations and Partial Differential Equations 56 (2017), article 68. Author-hosted manuscript. Theorem 1.1 is the velocity–pressure epsilon-regularity input used in Lemma 1.
- C. J. Bishop and Y. Peres, Fractals in Probability and Analysis, Cambridge University Press, 2017. Author-hosted manuscript, Lemma 3.1.1 and its dyadic proof.
- T. Hytönen and A. Kairema, Systems of dyadic cubes in a doubling metric space, Colloquium Mathematicum 126 (2012), 1–33. DOI. Lemma 5 above gives the explicit construction needed here, so no adjacent-grid theorem is left implicit.