<p>We characterize the weights <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13_2025_2124_Article_IEq1.gif" Format="GIF" Height="12" Rendition="HTML" Resolution="72" Type="Linedraw" Width="116" /> </InlineMediaObject> <EquationSource Format="TEX">\(w, v_1, v_2, \dots , v_m \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>w</mi> <mo>,</mo> <msub> <mi>v</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>v</mi> <mn>2</mn> </msub> <mo>,</mo> <mo>⋯</mo> <mo>,</mo> <msub> <mi>v</mi> <mi>m</mi> </msub> </mrow> </math></EquationSource> </InlineEquation> for which the weak-type multilinear gradient inequality <Equation ID="Equ13"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="13_2025_2124_Article_Equ13.gif" Format="GIF" Height="57" Rendition="HTML" Resolution="72" Type="Linedraw" Width="268" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} \left\| \prod _{i=1}^m f_i\right\| _{p,\infty ;w}\le C \prod _{i=1}^m \left\| x \cdot \nabla f_i(x)\right\| _{p_i,v_i} \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <msub> <mfenced close="∥" open="∥"> <munderover> <mo>∏</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>m</mi> </munderover> <msub> <mi>f</mi> <mi>i</mi> </msub> </mfenced> <mrow> <mi>p</mi> <mo>,</mo> <mi>∞</mi> <mo>;</mo> <mi>w</mi> </mrow> </msub> <mo>≤</mo> <mi>C</mi> <munderover> <mo>∏</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>m</mi> </munderover> <msub> <mfenced close="∥" open="∥"> <mi>x</mi> <mo>·</mo> <mi mathvariant="normal">∇</mi> <msub> <mi>f</mi> <mi>i</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> </mfenced> <mrow> <msub> <mi>p</mi> <mi>i</mi> </msub> <mo>,</mo> <msub> <mi>v</mi> <mi>i</mi> </msub> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>holds for all <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13_2025_2124_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="173" /> </InlineMediaObject> <EquationSource Format="TEX">\(f_1, f_2, \dots , f_m \in C_c^{\infty }({\mathbb {R}}^n)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>f</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>f</mi> <mn>2</mn> </msub> <mo>,</mo> <mo>⋯</mo> <mo>,</mo> <msub> <mi>f</mi> <mi>m</mi> </msub> <mo>∈</mo> <msubsup> <mi>C</mi> <mi>c</mi> <mi>∞</mi> </msubsup> <mrow> <mo stretchy="false">(</mo> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mi>n</mi> </msup> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> in the case <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13_2025_2124_Article_IEq3.gif" Format="GIF" Height="25" Rendition="HTML" Resolution="72" Type="Linedraw" Width="157" /> </InlineMediaObject> <EquationSource Format="TEX">\(\frac{1}{p} = \frac{1}{p_1}+\frac{1}{p_2}+ \cdots + \frac{1}{p_m}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfrac> <mn>1</mn> <mi>p</mi> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>p</mi> <mn>1</mn> </msub> </mfrac> <mo>+</mo> <mfrac> <mn>1</mn> <msub> <mi>p</mi> <mn>2</mn> </msub> </mfrac> <mo>+</mo> <mo>⋯</mo> <mo>+</mo> <mfrac> <mn>1</mn> <msub> <mi>p</mi> <mi>m</mi> </msub> </mfrac> </mrow> </math></EquationSource> </InlineEquation>.</p>

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A weighted weak-type multilinear gradient inequality

  • Víctor García García,
  • Pedro Ortega Salvador

摘要

We characterize the weights \(w, v_1, v_2, \dots , v_m \) w , v 1 , v 2 , , v m for which the weak-type multilinear gradient inequality \(\begin{aligned} \left\| \prod _{i=1}^m f_i\right\| _{p,\infty ;w}\le C \prod _{i=1}^m \left\| x \cdot \nabla f_i(x)\right\| _{p_i,v_i} \end{aligned}\) i = 1 m f i p , ; w C i = 1 m x · f i ( x ) p i , v i holds for all \(f_1, f_2, \dots , f_m \in C_c^{\infty }({\mathbb {R}}^n)\) f 1 , f 2 , , f m C c ( R n ) in the case \(\frac{1}{p} = \frac{1}{p_1}+\frac{1}{p_2}+ \cdots + \frac{1}{p_m}\) 1 p = 1 p 1 + 1 p 2 + + 1 p m .