analysis / Harmonic analysis

Stein’s dimension-free weak-(1,1) Riesz transform problem

The Riesz transforms $R_1,\ldots,R_n$ on $\mathbb{R}^n$ are the Fourier multipliers $-i\xi_j/|\xi|$, the natural higher-dimensional Hilbert transforms. Stein proved in 1983 that their $L^p$ bounds can be taken independent of the dimension for every $1 < p < \infty$. At the 1986 ICM he asked whether the same holds at the endpoint $p=1$: is there an absolute constant $C$, independent of $n$, with $$|\{x : |Rf(x)| > \lambda\}| \le \frac{C}{\lambda}\,\|f\|_{L^1(\mathbb{R}^n)}$$ for every $\lambda > 0$? The Calderon-Zygmund route gives a constant that grows with the dimension, and the best known was Janakiraman's $c\log n$. This paper answers yes, with $C = 2$, for the vector transform $R = (R_1,\ldots,R_n)$ - so the same constant serves every single component $R_j$ uniformly in $n$.

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analysisAug 18, 2026Significance 38/100Registry: unreviewed

Stein’s dimension-free weak-(1,1) Riesz transform problem

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The theorem is the vector-valued endpoint bound $\|Rf\|_{L^{1,\infty}} \le 2\|f\|_{L^1}$ for $R = (R_1,\ldots,R_n)$, so the constant 2 also serves each component $R_j$ uniformly in the dimension; the best previously known component bound grew like $c\log n$. The mechanism is a decomposition theorem stated as Theorem 1.2: for every nonnegative $f \in L^1 \cap L^2$ and every $\lambda > 0$, write $f = \mu + (-\Delta…

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The Riesz transforms $R_1,\ldots,R_n$ on $\mathbb{R}^n$ are the Fourier multipliers $-i\xi_j/|\xi|$, the natural higher-dimensional Hilbert transforms. Stein proved in 1983 that their $L^p$ bounds can be taken independent of the dimension for every $1 < p < \infty$. At the 1986 ICM he asked whether the same holds at the endpoint $p=1$: is there an absolute constant $C$, independent of $n$, with $$|\{x : |Rf(x)| > \lambda\}| \le \frac{C}{\lambda}\,\|f\|_{L^1(\mathbb{R}^n)}$$ for every $\lambda > 0$? The Calderon-Zygmund route gives a constant that grows with the dimension, and the best known was Janakiraman's $c\log n$. This paper answers yes, with $C = 2$, for the vector transform $R = (R_1,\ldots,R_n)$ - so the same constant serves every single component $R_j$ uniformly in $n$.

The theorem is the vector-valued endpoint bound $\|Rf\|_{L^{1,\infty}} \le 2\|f\|_{L^1}$ for $R = (R_1,\ldots,R_n)$, so the constant 2 also serves each component $R_j$ uniformly in the dimension; the best previously known component bound grew like $c\log n$. The mechanism is a decomposition theorem stated as Theorem 1.2: for every nonnegative $f \in L^1 \cap L^2$ and every $\lambda > 0$, write $f = \mu + (-\Delta)^{\alpha/2}u$ with $\mu \le \lambda$ and $u$ in the fractional Sobolev space $H^\alpha$, obtained from an obstacle problem for the fractional Laplacian together with a Lewy-Stampacchia type estimate on an unbounded domain. That replaces the Calderon-Zygmund decomposition, whose cube geometry is where the dimensional loss enters.

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