David–Semmes Riesz-transform problem in intermediate dimensions

David–Semmes Riesz-transform problem in intermediate dimensions

OpenAI's claim

Higher-codimension Riesz rectifiability: The claimed result turns uniform $L^2$ control of all positive hard truncations of the Riesz transform into quantitative uniform rectifiability. It concerns $n$-regular measures in $\mathbb R^d$ with $d\ge4$ and $2\le n\le d-2$.

Mathematical statement of the claimed result

Mathematical claim 1.

Let $d\ge4$ and $2\le n\le d-2$. Suppose a Radon measure $\mu$ on $\mathbb R^d$ satisfies $C^{-1}r^n\le\mu(B(x,r))\le Cr^n$ for $x\in\operatorname{supp}\mu$ and $0<r\le\operatorname{diam}(\operatorname{supp}\mu)$. If \[R_{\mu,\epsilon}f(x)=\int_{|x-y|>\epsilon}\frac{x-y}{|x-y|^{n+1}}f(y)\,d\mu(y),\qquad\sup_{\epsilon>0}\|R_{\mu,\epsilon}\|_{L^2(\mu)\to L^2(\mu;\mathbb R^d)}\le C_R,\] then there are $\theta>0$ and $M<\infty$, depending only on $d,n,C,C_R$, such that each admissible $B(x,r)$ contains at least $\theta r^n$ of $\mu$ on an $M$-Lipschitz image of the radius-$r$ ball in $\mathbb R^n$. This is uniform $n$-rectifiability; pointwise principal values are not assumed.

Outline of the argument

Higher-codimension Riesz rectifiability: After a flat blowup, the normal height is renormalized by the initial excess so that relative improvement can be detected. A fractional-energy test yields compactness and a limiting equation whose integrability forces the height to be affine. Dyadic packing then counts failures of excess improvement, leading to large Lipschitz pieces at every support scale.

Formal verification

Family 081 describes Lean coverage for the quantitative Lipschitz-piece conclusion in the stated higher-codimension Riesz setting.

Higher-codimension Riesz rectifiability: The manuscript links RieszQuantitative.lean (declaration OAI.RieszRectifiability.quantitative_higher_codimension_riesz_rectifiability).

The documented selection includes this manuscript’s main assertion; that assertion retains the restrictions stated above.

This site has not run the Lean development. These links describe OpenAI’s selected formal statements and should be read with their stated scope.

Status caveat

OpenAI reports that an internal model produced the manuscripts. Their arguments have not been independently verified as proofs by this site. Any portion outside the documented formal scope may contain mistakes, and the catalogue label reports the announced mathematical claim.

Sources

• OpenAI, Riesz transforms and uniform rectifiability in higher codimension, 2026-09-24; citation key OAI:Riesz-transforms-and-uniform-rectifiability-in-higher-codimension-September-24-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main.
• Family 081: Lean scope.
• Comparator RieszQuantitative.json.
• Comparator RieszQuantitative.lean.
• OpenAI catalogue entry.

David–Semmes Riesz-transform problem in intermediate dimensions

Definitions and mathematical statement

A Radon measure $\mu$ on ${\mathbb{R}}^d$ is $n$-Ahlfors--David regular if some $C\ge1$ satisfies \[ C^{-1}r^n\le\mu(B(x,r))\le Cr^n \] for $x\in\operatorname{supp}\mu$ and $0<r<\operatorname{diam}(\operatorname{supp}\mu)$. Uniform $n$-rectifiability can be defined by big pieces of Lipschitz images: fixed $\theta,L>0$ work in every such ball, giving an $L$-Lipschitz map from an $n$-ball of radius $r$ whose image meets that ball in $\mu$-measure at least $\theta r^n$.

For $2\le n\le d-2$, suppose the truncated vector Riesz transforms \[ R_{\mu,{\varepsilon}}f(x)=\int_{|x-y|>{\varepsilon}} \frac{x-y}{|x-y|^{n+1}}f(y){\,\mathrm{d}}\mu(y) \] have uniformly bounded norms from $L^2(\mu)$ to $L^2(\mu;{\mathbb{R}}^d)$ over all ${\varepsilon}>0$. Must $\mu$ be uniformly $n$-rectifiable? The constants need not be numerically prescribed; they must be uniform across location and scale.

Formulation and concept references: [S1].

Short English statement

Does square-integrable boundedness of the Riesz transform force quantitative rectifiability for regular measures in the intermediate dimensions?

Research attempt: cancellation, flat models, and the missing geometric estimate

Status. The intermediate-dimensional implication is not proved here. The results below are an exact normalization of the analytic hypothesis, a local testing consequence, a complete geometric subcase, and an explicit obstruction to importing the three-point positivity argument from dimension one. The obstruction concerns a proposed proof, not the conjecture.

Source and scope audit. The introduction of [S1], arXiv:2509.26547v1, identifies the cases $n=1$ and $n=d-1$ as solved and the other integer dimensions as open. Its Theorem 1.1 assumes codimension one and an additional bilateral approximation condition involving unions of parallel planes. It cannot be applied to the present range merely by retaining its Riesz-transform assumption. This audit uses the displayed primary manuscript; it is not a certification that every subsequent announcement has been checked.

Write $K_n(z)=z/|z|^{n+1}$ and let $A$ be a common bound for the truncated operators. All vector norms below are Euclidean. The integral defining a truncation is first interpreted on bounded, compactly supported functions and then by its bounded extension. An infinite measure need not have the constant function $1$ in $L^2$.

1. A scale-invariant formulation

Fix $x_0\in\operatorname{supp}\mu$ and $r>0$, and put \[ T(y)=(y-x_0)/r,\qquad \mu_{x_0,r}=r^{-n}T_\#\mu. \] If $f'(T(y))=f(y)$, homogeneity of the kernel gives \[ R_{\mu_{x_0,r},\varepsilon}f'(T(x)) =R_{\mu,r\varepsilon}f(x),\qquad \|f'\|_{L^2(\mu_{x_0,r})}=r^{-n/2}\|f\|_{L^2(\mu)}. \] The same scaling holds for the output norm. Thus the AD constant and the operator bound $A$ survive every translation and dilation. Any useful geometric conclusion must therefore have constants depending on these quantities and on $n,d$, rather than on the radius under inspection.

If $E$ is measurable and $\nu=\mu|_E$, zero extension gives \[ \|R_{\nu,\varepsilon}f\|_{L^2(\nu)} \leq A\|f\|_{L^2(\nu)}. \tag{1} \] Indeed, on $E$ the integral is $R_{\mu,\varepsilon}(f\mathbf1_E)$, and restriction only decreases its output norm. Restricting a measure preserves this analytic estimate but need not preserve its AD lower bound. Consequently $1$ does not permit arbitrary deletion of geometrically inconvenient pieces followed by another application of the conjecture.

For a ball $B=B(x,r)$ with $x\in\operatorname{supp}\mu$ and admissible radius, testing the hypothesis on $\mathbf1_B$ gives \[ \int_B|R_{\mu,\varepsilon}\mathbf1_B|^2\,d\mu \leq A^2\mu(B)\leq A^2 C r^n. \tag{2} \] It also gives, for every $t>0$, \[ \mu\{z\in B:|R_{\mu,\varepsilon}\mathbf1_B(z)|>t\} \leq A^2\mu(B)/t^2. \tag{3} \] These estimates are uniform in the truncation. They control a local singular integral in measure, but make no reference to a plane. Identifying the missing plane from $2$ is a genuine further step.

There is an exact cancellation which can otherwise be misleading. Since $K_n$ is odd and the truncated kernel is bounded on $B\times B$, \[ \int_B R_{\mu,\varepsilon}\mathbf1_B\,d\mu=0. \tag{4} \] Fubini and interchanging the two variables prove $4$. It holds for every finite measure on $B$, whether or not that measure is rectifiable. Thus smallness, or even vanishing, of this vector average is not evidence for flatness. An argument based only on the average in $4$ has discarded the relevant information before geometry is considered.

2. A fully geometric subcase

Suppose in addition that the support is contained in a union of at most $N$ affine $n$-planes, \[ \operatorname{supp}\mu\subset P_1\cup\cdots\cup P_N. \] Then $\mu$ is uniformly $n$-rectifiable, with constants depending only on its lower AD constant and $N$. No Riesz-transform assumption is needed for this subcase.

To prove this, fix $B(x,r)$. At least one plane, say $P_j$, satisfies \[ \mu(B(x,r)\cap P_j)\geq \mu(B(x,r))/N \geq (CN)^{-1}r^n. \] Choose an isometric linear parametrization $U:\mathbb R^n\to P_j-P_j$ and let $p_j$ be the orthogonal projection of $x$ onto $P_j$. The map \[ F:B_{\mathbb R^n}(0,r)\longrightarrow\mathbb R^d, \qquad F(t)=p_j+Ut \] is $1$-Lipschitz and its image contains $B(x,r)\cap P_j$: orthogonal projection gives $|z-p_j|\leq |z-x|<r$ for $z\in P_j\cap B(x,r)$. This proves the big-pieces condition with $\theta=(CN)^{-1}$ and $L=1$. Boundaries of open balls cause no issue, since only their open intersections are used.

The proof remains valid if the planes depend on $B$, provided the number $N$ is uniform and their union contains the support inside $B$. The substantial assumption is uniform control of the number of planes at every scale. A cover by countably many planes, or a finite cover whose size increases as $r\downarrow0$, does not yield the fixed positive $\theta$ required by the problem.

This isolates a plausible target for an analytic argument: produce, at sufficiently many locations and scales, a bounded-complexity planar description carrying a fixed fraction of the mass. Estimate $2$ alone has not been shown here to produce such a description.

3. Why the one-dimensional curvature calculation does not transfer

For three distinct points define the symmetrized scalar kernel \[ p_n(x,y,z)=K_n(x-y)\cdot K_n(x-z) +K_n(y-x)\cdot K_n(y-z) +K_n(z-x)\cdot K_n(z-y). \tag{5} \] When $n=1$, the corresponding planar identity is tied to a nonnegative curvature quantity. The sign needed for that strategy already fails for the kernel in this problem, even on a straight line.

Take a unit vector $e$ and the collinear triple $x=0$, $y=e$, $z=2e$. The contributions in $5$ are respectively $2^{-n}$, $-1$, and $2^{-n}$. Hence \[ p_n(0,e,2e)=2^{1-n}-1<0\qquad(n>1). \tag{6} \] At scale $a$ the value is multiplied by $a^{-2n}$. Thus this is not a defect caused by a particular length normalization. In particular, $p_n$ cannot equal a nonnegative squared curvature expression multiplied by a positive constant for all triples.

The negative value persists for noncollinear triples. Put $x=0$, $y=e_1$, $z=2e_1+h e_2$. Direct dot products give \[ p_n(x,y,z)=\frac{2}{(4+h^2)^{(n+1)/2}} -\frac{1}{(1+h^2)^{(n+1)/2}} +\frac{2+h^2}{[(4+h^2)(1+h^2)]^{(n+1)/2}}. \tag{7} \] Continuity at $h=0$ and $6$ show negativity for all sufficiently small nonzero $h$. This avoids a possible objection that exact collinearity is a degenerate exceptional configuration.

Neither $6$ nor $7$ constructs an AD-regular counterexample. Three atoms are not an $n$-AD-regular measure with $n\geq2$. Their role is narrower and rigorous: they disprove the pointwise positivity assertion that would be needed to convert a triple-integral expansion directly into a positive curvature energy. The failure is at the algebraic step before any measure is selected.

One might hope that averaging over an AD-regular measure repairs the sign. That is a new inequality, not a consequence of pointwise symmetrization. Negative and positive triples could cancel, and controlling the norm of the resulting singular integral would not by itself control either part separately. A proof using this idea must state and establish an averaged lower bound which survives that cancellation.

4. A precise multiscale bottleneck

Here is one way to formulate the missing geometry without equating it with the elementary estimates above. For $E=\operatorname{supp}\mu$, set \[ b\beta_E(x,r)=\inf_P\left\{ \sup_{y\in E\cap B(x,r)}\frac{\operatorname{dist}(y,P)}r +\sup_{z\in P\cap B(x,r)}\frac{\operatorname{dist}(z,E)}r \right\}, \tag{8} \] where $P$ runs over affine $n$-planes; an empty second intersection contributes zero. The first term detects mass far from a plane; the second detects holes in a putative plane. Bounding only the first term is not the same geometric condition.

The bilateral weak geometric lemma discussed in [S1] is a Carleson packing condition for the balls or dyadic cells where this bilateral deviation exceeds a fixed threshold. In a ball formulation the desired kind of estimate is \[ \int_0^R\int_{B(x_0,R)} \mathbf1_{\{b\beta_E(x,r)>\delta\}}\,d\mu(x)\,\frac{dr}{r} \leq C_\delta R^n, \tag{9} \] with the usual harmless changes of ball aperture and admissible scales. The known geometric theory relates this condition to uniform rectifiability for AD-regular sets. No proof of that deep equivalence is supplied here; $9$ is being used to locate the analytic-to-geometric gap, not as an elementary replacement for it.

Applying $2$ separately at each of $m$ scales would at best give a bound proportional to $mR^n$. The right side of $9$ must be independent of the number of scales. Consequently a levelwise $L^2$ estimate, repeated infinitely often, is insufficient. One needs a mechanism charging geometrically bad scales to a summable analytic energy, with overlap controlled across scales. The three-point formula does not supply such a positive energy in the intermediate dimensions by $6$.

An ambient-space reduction also stops at a clear point. An arbitrary $n$-AD-regular subset of $\mathbb R^d$ is not contained in an affine $\mathbb R^{n+1}$. Projecting it onto such a space can identify separated pieces, can destroy its lower regularity at relevant scales, and changes the kernel. In particular, \[ \pi\!\left(\frac{x-y}{|x-y|^{n+1}}\right) \ne \frac{\pi x-\pi y}{|\pi x-\pi y|^{n+1}} \] in general. The known codimension-one theorem therefore cannot be used after an arbitrary projection without proving additional estimates that are absent here.

5. Verification and outcome

The finite diagnostic accompanying this attempt checks $6$ exactly for $2\leq n\leq8$, and checks the three dot products behind $7$ numerically at several small nonzero heights. This is an algebraic stress test of a proposed proof mechanism. It does not test the conjecture by sampling finitely many atoms. The scaling identities, restriction estimate, and finite-plane lemma have proofs independent of the diagnostic.

What is established is uniform local testing, exact antisymmetric cancellation, uniform rectifiability for uniformly bounded unions of planes, and the failure of pointwise three-point positivity in the target range. The first unresolved step is an estimate converting Riesz boundedness and AD regularity into a quantitative, summable control of bad geometric scales. Neither a codimension reduction nor the signed symmetrization above supplies it. The full intermediate-dimensional problem remains unresolved by this attempt.

Sources

• S1 Xavier Tolsa, Riesz transforms and the BAUPP and BWGL criteria for uniform rectifiability $2025$. https://arxiv.org/html/2509.26547. Formulation source.