<p>We find necessary and sufficient conditions on weights <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13324_2025_1101_Article_IEq1.gif" Format="GIF" Height="12" Rendition="HTML" Resolution="72" Type="Linedraw" Width="85" /> </InlineMediaObject> <EquationSource Format="TEX">\(u_1, u_2, v_1, v_2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>u</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <mo>,</mo> <msub> <mi>v</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>v</mi> <mn>2</mn> </msub> </mrow> </math></EquationSource> </InlineEquation>, i.e. measurable, positive, and finite, a.e. on (<i>a</i>,&#xa0;<i>b</i>), for which there exists a positive constant <i>C</i> such that for given <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13324_2025_1101_Article_IEq2.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="154" /> </InlineMediaObject> <EquationSource Format="TEX">\(0&lt; p_1,q_1,p_2,q_2 &lt;\infty \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>0</mn> <mo>&lt;</mo> <msub> <mi>p</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>q</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>p</mi> <mn>2</mn> </msub> <mo>,</mo> <msub> <mi>q</mi> <mn>2</mn> </msub> <mo>&lt;</mo> <mi>∞</mi> </mrow> </math></EquationSource> </InlineEquation> the inequality <Equation ID="Equ127"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="13324_2025_1101_Article_Equ127.gif" Format="GIF" Height="102" Rendition="HTML" Resolution="72" Type="Linedraw" Width="351" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} \begin{aligned}&amp;\bigg (\int _a^b \bigg (\int _a^t f(s)^{p_2} v_2(s)^{p_2} ds\bigg )^{\frac{q_2}{p_2}} u_2(t)^{q_2} dt \bigg )^{\frac{1}{q_2}}\\&amp;\quad \le C \bigg (\int _a^b \bigg (\int _a^t f(s)^{p_1} v_1(s)^{p_1} ds\bigg )^{\frac{q_1}{p_1}} u_1(t)^{q_1} dt \bigg )^{\frac{1}{q_1}} \end{aligned} \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <mtable> <mtr> <mtd /> <mtd columnalign="left"> <mrow> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">(</mo> </mrow> <msubsup> <mo>∫</mo> <mi>a</mi> <mi>b</mi> </msubsup> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">(</mo> </mrow> <msubsup> <mo>∫</mo> <mi>a</mi> <mi>t</mi> </msubsup> <mi>f</mi> <msup> <mrow> <mo stretchy="false">(</mo> <mi>s</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>p</mi> <mn>2</mn> </msub> </msup> <msub> <mi>v</mi> <mn>2</mn> </msub> <msup> <mrow> <mo stretchy="false">(</mo> <mi>s</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>p</mi> <mn>2</mn> </msub> </msup> <mi>d</mi> <mi>s</mi> <msup> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">)</mo> </mrow> <mfrac> <msub> <mi>q</mi> <mn>2</mn> </msub> <msub> <mi>p</mi> <mn>2</mn> </msub> </mfrac> </msup> <msub> <mi>u</mi> <mn>2</mn> </msub> <msup> <mrow> <mo stretchy="false">(</mo> <mi>t</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>q</mi> <mn>2</mn> </msub> </msup> <mi>d</mi> <mi>t</mi> <msup> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">)</mo> </mrow> <mfrac> <mn>1</mn> <msub> <mi>q</mi> <mn>2</mn> </msub> </mfrac> </msup> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="right"> <mrow /> </mtd> <mtd columnalign="left"> <mrow> <mspace width="1em" /> <mo>≤</mo> <mi>C</mi> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">(</mo> </mrow> <msubsup> <mo>∫</mo> <mi>a</mi> <mi>b</mi> </msubsup> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">(</mo> </mrow> <msubsup> <mo>∫</mo> <mi>a</mi> <mi>t</mi> </msubsup> <mi>f</mi> <msup> <mrow> <mo stretchy="false">(</mo> <mi>s</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>p</mi> <mn>1</mn> </msub> </msup> <msub> <mi>v</mi> <mn>1</mn> </msub> <msup> <mrow> <mo stretchy="false">(</mo> <mi>s</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>p</mi> <mn>1</mn> </msub> </msup> <mi>d</mi> <mi>s</mi> <msup> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">)</mo> </mrow> <mfrac> <msub> <mi>q</mi> <mn>1</mn> </msub> <msub> <mi>p</mi> <mn>1</mn> </msub> </mfrac> </msup> <msub> <mi>u</mi> <mn>1</mn> </msub> <msup> <mrow> <mo stretchy="false">(</mo> <mi>t</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>q</mi> <mn>1</mn> </msub> </msup> <mi>d</mi> <mi>t</mi> <msup> <mrow> <mo maxsize="2.047em" minsize="2.047em" stretchy="true">)</mo> </mrow> <mfrac> <mn>1</mn> <msub> <mi>q</mi> <mn>1</mn> </msub> </mfrac> </msup> </mrow> </mtd> </mtr> </mtable> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>holds for every non-negative, measurable function <i>f</i> on (<i>a</i>,&#xa0;<i>b</i>), where <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13324_2025_1101_Article_IEq3.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="113" /> </InlineMediaObject> <EquationSource Format="TEX">\(0 \le a &lt;b \le \infty \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>0</mn> <mo>≤</mo> <mi>a</mi> <mo>&lt;</mo> <mi>b</mi> <mo>≤</mo> <mi>∞</mi> </mrow> </math></EquationSource> </InlineEquation>. The proof is based on a recently developed discretization method that enables us to overcome the restrictions of the earlier results.</p>

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Weighted inequalities involving two Hardy operators

  • Amiran Gogatishvili,
  • Tuǧçe Ünver

摘要

We find necessary and sufficient conditions on weights \(u_1, u_2, v_1, v_2\) u 1 , u 2 , v 1 , v 2 , i.e. measurable, positive, and finite, a.e. on (ab), for which there exists a positive constant C such that for given \(0< p_1,q_1,p_2,q_2 <\infty \) 0 < p 1 , q 1 , p 2 , q 2 < the inequality \(\begin{aligned} \begin{aligned}&\bigg (\int _a^b \bigg (\int _a^t f(s)^{p_2} v_2(s)^{p_2} ds\bigg )^{\frac{q_2}{p_2}} u_2(t)^{q_2} dt \bigg )^{\frac{1}{q_2}}\\&\quad \le C \bigg (\int _a^b \bigg (\int _a^t f(s)^{p_1} v_1(s)^{p_1} ds\bigg )^{\frac{q_1}{p_1}} u_1(t)^{q_1} dt \bigg )^{\frac{1}{q_1}} \end{aligned} \end{aligned}\) ( a b ( a t f ( s ) p 2 v 2 ( s ) p 2 d s ) q 2 p 2 u 2 ( t ) q 2 d t ) 1 q 2 C ( a b ( a t f ( s ) p 1 v 1 ( s ) p 1 d s ) q 1 p 1 u 1 ( t ) q 1 d t ) 1 q 1 holds for every non-negative, measurable function f on (ab), where \(0 \le a <b \le \infty \) 0 a < b . The proof is based on a recently developed discretization method that enables us to overcome the restrictions of the earlier results.