Skip to main content
QUICK REVIEW

[Paper Review] Continuity of weighted estimates for sublinear operators

Michael Papadimitrakis, Nikolaos Pattakos|arXiv (Cornell University)|Jun 20, 2012
Advanced Harmonic Analysis Research4 references3 citations
TL;DR

This paper establishes the continuity of operator norms for sublinear operators on weighted $L^p$ spaces with respect to the $d_*$ metric on $A_p$ weights. It proves that if a sublinear operator $T$ satisfies a weighted norm inequality depending only on the $A_p$ characteristic, then its operator norm is continuous at $w_0$ as $w \to w_0$ in the $d_*$ topology.

ABSTRACT

In this note we prove that if a sublinear operator T satisfies a certain weighted estimate in the $L^{p}(w)$ space for all $w\in A_{p}$, $1

Motivation & Objective

  • To extend the continuity of operator norms from linear to sublinear operators in weighted $L^p$ spaces.
  • To investigate whether the operator norm $\|T\|_{L^p(w)\to L^p(w)}$ varies continuously with respect to the weight $w$ in the $d_*$ metric.
  • To establish that the $A_p$ characteristic remains bounded under small $d_*$-perturbations of a fixed weight $w_0$.
  • To show that the constant $c_{[w]_{A_p}}$ in the key inequality depends continuously on $[w]_{A_p}$, ensuring boundedness near $w_0$.

Proposed method

  • Uses a key inequality from [3]: $\|T\|_{L^p(u)\to L^p(u)} \leq \|T\|_{L^p(v)\to L^p(v)}(1 + c_{[v]_{A_p}} d_*(u,v))$ for sublinear operators and $u,v$ close in $d_*$.
  • Applies this inequality with $u = w$, $v = w_0$ to obtain an upper bound on $\|T\|_{L^p(w)\to L^p(w)}$ that tends to $\|T\|_{L^p(w_0)\to L^p(w_0)}$ as $d_*(w,w_0) \to 0$.
  • Establishes a reverse inequality by swapping roles: $u = w_0$, $v = w$, requiring control of $c_{[w]_{A_p}}$ as $w \to w_0$.
  • Uses Hölder's inequality and conjugate exponents $R, R' = 1+\epsilon$ to bound $[w]_{A_p}$ uniformly for $w$ near $w_0$ in $d_*$.
  • Leverages the fact that $\left(\frac{w}{w_0}\right)^R \in A_p$ with uniformly bounded characteristic for large $R$, and $w_0^{1+\epsilon} \in A_p$ for small $\epsilon$.
  • Uses the continuity of $c_{[w]_{A_p}}$ in $[w]_{A_p}$—derived from Riesz-Thorin interpolation with measure change—to conclude boundedness of the constant.

Experimental results

Research questions

  • RQ1Does the operator norm $\|T\|_{L^p(w)\to L^p(w)}$ depend continuously on the weight $w$ in the $d_*$ metric for sublinear operators?
  • RQ2Can the continuity result for linear operators in [3] be extended to sublinear operators under the same $A_p$-norm assumptions?
  • RQ3Is the $A_p$ characteristic $[w]_{A_p}$ continuous with respect to the $d_*$ metric near a fixed $w_0 \in A_p$?
  • RQ4Does the constant $c_{[w]_{A_p}}$ in the key operator norm inequality remain bounded as $w \to w_0$ in $d_*$?
  • RQ5Can the $A_p$ characteristic of $w$ be uniformly bounded for all $w$ sufficiently close to $w_0$ in the $d_*$ metric?

Key findings

  • The operator norm $\|T\|_{L^p(w)\to L^p(w)}$ is continuous at $w_0$ with respect to the $d_*$ metric: $\lim_{d_*(w,w_0)\to 0}\|T\|_{L^p(w)\to L^p(w)} = \|T\|_{L^p(w_0)\to L^p(w_0)}$.
  • For $w$ sufficiently close to $w_0$ in $d_*$, the $A_p$ characteristic $[w]_{A_p}$ is uniformly bounded by a constant $C$ depending only on $[w_0]_{A_p}$ and the distance $\delta = d_*(w,w_0)$.
  • The $A_p$ characteristic satisfies $\limsup_{d_*(w,w_0)\to 0}[w]_{A_p} \leq [w_0]_{A_p}$, and dually $[w_0]_{A_p} \leq \liminf_{d_*(w,w_0)\to 0}[w]_{A_p}$, proving continuity of $[w]_{A_p}$ in $d_*$.
  • The constant $c_{[w]_{A_p}}$ in the key inequality is continuous in $[w]_{A_p}$, and thus remains bounded as $w \to w_0$ in $d_*$.
  • The proof relies on Hölder’s inequality with conjugate exponents $R$ and $R' = 1+\epsilon$, and the fact that $\left(\frac{w}{w_0}\right)^R \in A_p$ with uniformly bounded characteristic for large $R$.
  • The result generalizes the continuity result from linear operators to sublinear operators, resolving a gap where previous methods failed.

Better researchstarts right now

From reading papers to final review, dramatically reduce your research time.

No credit card · Free plan available

This review was created by AI and reviewed by human editors.