Abstract <p>Let <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq1.gif" Format="GIF" Height="12" Rendition="HTML" Resolution="72" Type="Linedraw" Width="16" /> </InlineMediaObject> <EquationSource Format="TEX">\(\varphi \)</EquationSource>  </InlineEquation> be a trace on von Neumann algebra <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq2.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="25" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathcal{M}\)</EquationSource>  </InlineEquation>, <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq3.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="80" /> </InlineMediaObject> <EquationSource Format="TEX">\(A,B \in \mathcal{M}\)</EquationSource>  </InlineEquation>, and <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq4.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="64" /> </InlineMediaObject> <EquationSource Format="TEX">\(\left\| B \right\| < 1\)</EquationSource>  </InlineEquation>, <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq5.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="143" /> </InlineMediaObject> <EquationSource Format="TEX">\([A,B] = AB - BA\)</EquationSource>  </InlineEquation>. Then <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq6.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="160" /> </InlineMediaObject> <EquationSource Format="TEX">\(\varphi \left( {\left| {[A,B]} \right|} \right) \leqslant 2\varphi \left( {\left| A \right|} \right)\)</EquationSource>  </InlineEquation>. Let <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq7.gif" Format="GIF" Height="10" Rendition="HTML" Resolution="72" Type="Linedraw" Width="11" /> </InlineMediaObject> <EquationSource Format="TEX">\(\tau \)</EquationSource>  </InlineEquation> be a faithful normal semifinite trace on <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq8.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="25" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathcal{M}\)</EquationSource>  </InlineEquation>, <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq9.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="63" /> </InlineMediaObject> <EquationSource Format="TEX">\(S(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation> be the <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq10.gif" Format="GIF" Height="8" Rendition="HTML" Resolution="72" Type="Linedraw" Width="9" /> </InlineMediaObject> <EquationSource Format="TEX">\( ^{} \)</EquationSource>  </InlineEquation>-algebra of all <InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq11.gif" Format="GIF" Height="10" Rendition="HTML" Resolution="72" Type="Linedraw" Width="11" /> </InlineMediaObject> <EquationSource Format="TEX">\(\tau \)</EquationSource>  </InlineEquation>-measurable operators. If <InlineEquation ID="IEq12"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq12.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="103" /> </InlineMediaObject> <EquationSource Format="TEX">\(A \in {{L}_{2}}(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation> and <InlineEquation ID="IEq13"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq13.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="96" /> </InlineMediaObject> <EquationSource Format="TEX">\(\operatorname{Re} A = \lambda \left| A \right|\)</EquationSource>  </InlineEquation> with <InlineEquation ID="IEq14"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq14.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="90" /> </InlineMediaObject> <EquationSource Format="TEX">\(\lambda \in \{ - 1,1\} \)</EquationSource>  </InlineEquation>, then <InlineEquation ID="IEq15"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq15.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="74" /> </InlineMediaObject> <EquationSource Format="TEX">\(A = \lambda \left| A \right|\)</EquationSource>  </InlineEquation>. An operator <InlineEquation ID="IEq16"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq16.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="103" /> </InlineMediaObject> <EquationSource Format="TEX">\(A \in {{L}_{2}}(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation> is Hermitian if <InlineEquation ID="IEq17"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq17.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="119" /> </InlineMediaObject> <EquationSource Format="TEX">\(\tau ({{A}^{2}}) = \tau (A{\kern 1pt} ^{}{\kern 1pt} A)\)</EquationSource>  </InlineEquation>. Let positive operators <InlineEquation ID="IEq18"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq18.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="118" /> </InlineMediaObject> <EquationSource Format="TEX">\(A,B \in S(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation> be invertible in <InlineEquation ID="IEq19"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq19.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="63" /> </InlineMediaObject> <EquationSource Format="TEX">\(S(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation> and <InlineEquation ID="IEq20"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq20.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="192" /> </InlineMediaObject> <EquationSource Format="TEX">\(Y: = ({{A}^{{ - 1}}} - {{B}^{{ - 1}}})(A - B)\)</EquationSource>  </InlineEquation>. If <i>Y</i>, <InlineEquation ID="IEq21"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq21.gif" Format="GIF" Height="21" Rendition="HTML" Resolution="72" Type="Linedraw" Width="174" /> </InlineMediaObject> <EquationSource Format="TEX">\({{A}^{{1/2}}}Y{{A}^{{ - 1/2}}} \in {{L}_{1}}(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation>, then <InlineEquation ID="IEq22"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq22.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="68" /> </InlineMediaObject> <EquationSource Format="TEX">\(\tau (Y) \leqslant 0\)</EquationSource>  </InlineEquation>. Let an operator <InlineEquation ID="IEq23"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq23.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="96" /> </InlineMediaObject> <EquationSource Format="TEX">\(A \in S(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation> be hyponormal and <InlineEquation ID="IEq24"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq24.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="90" /> </InlineMediaObject> <EquationSource Format="TEX">\(A = B + {\text{i}}C\)</EquationSource>  </InlineEquation> be its Cartesian decomposition. If (i) <InlineEquation ID="IEq25"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq25.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="117" /> </InlineMediaObject> <EquationSource Format="TEX">\(BC \in {{L}_{1}}(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation> or (ii) <InlineEquation ID="IEq26"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq26.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="101" /> </InlineMediaObject> <EquationSource Format="TEX">\(C = {{C}^{3}} \in \mathcal{M}\)</EquationSource>  </InlineEquation> and <InlineEquation ID="IEq27"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq27.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="133" /> </InlineMediaObject> <EquationSource Format="TEX">\([B,C] \in {{L}_{1}}(\mathcal{M},\tau )\)</EquationSource>  </InlineEquation>, then <InlineEquation ID="IEq28"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11982_2025_10463_Article_IEq28.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(A\)</EquationSource>  </InlineEquation> is normal.</p>

Trace Inequalities for Measurable Operators Affiliated to a von Neumann Algebra

期刊论文
发表日期：

2025年1月1日

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A. M. Bikchentaev

摘要

Abstract

Let \(\varphi \) be a trace on von Neumann algebra \(\mathcal{M}\) , \(A,B \in \mathcal{M}\) , and \(\left\| B \right\| < 1\) , \([A,B] = AB - BA\) . Then \(\varphi \left( {\left| {[A,B]} \right|} \right) \leqslant 2\varphi \left( {\left| A \right|} \right)\) . Let \(\tau \) be a faithful normal semifinite trace on \(\mathcal{M}\) , \(S(\mathcal{M},\tau )\) be the \( ^{*} \) -algebra of all \(\tau \) -measurable operators. If \(A \in {{L}_{2}}(\mathcal{M},\tau )\) and \(\operatorname{Re} A = \lambda \left| A \right|\) with \(\lambda \in \{ - 1,1\} \) , then \(A = \lambda \left| A \right|\) . An operator \(A \in {{L}_{2}}(\mathcal{M},\tau )\) is Hermitian if \(\tau ({{A}^{2}}) = \tau (A{\kern 1pt} ^{*}{\kern 1pt} A)\) . Let positive operators \(A,B \in S(\mathcal{M},\tau )\) be invertible in \(S(\mathcal{M},\tau )\) and \(Y: = ({{A}^{{ - 1}}} - {{B}^{{ - 1}}})(A - B)\) . If Y, \({{A}^{{1/2}}}Y{{A}^{{ - 1/2}}} \in {{L}_{1}}(\mathcal{M},\tau )\) , then \(\tau (Y) \leqslant 0\) . Let an operator \(A \in S(\mathcal{M},\tau )\) be hyponormal and \(A = B + {\text{i}}C\) be its Cartesian decomposition. If (i) \(BC \in {{L}_{1}}(\mathcal{M},\tau )\) or (ii) \(C = {{C}^{3}} \in \mathcal{M}\) and \([B,C] \in {{L}_{1}}(\mathcal{M},\tau )\) , then \(A\) is normal.