English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Tools annual

Global: Zendy Free Plan

Global: Zendy Tools monthly

Global: Zendy Plus monthly

An operator means a bounded linear operator on a Hilbert space. This paper proves the assertion made in its title. Theorem 1 yields the famous result that      A ≥  B ≥  0 A \geq B \geq 0       assures      A α  ≥  B α  {A^\alpha } \geq {B^\alpha }       for each      α  ∈  [ 0 , 1 ] \alpha \in [0,1]       when we put      r = 0 r = 0       in Theorem 1. Also Corollary 1 implies that      A ≥  B ≥  0 A \geq B \geq 0       assures      ( B A p B ) 1 / p ≥  B ( p + 2 ) / p {(B{A^p}B)^{1/p}} \geq {B^{(p + 2)/p}}       for each      p ≥  1 p \geq 1       and this inequality for      p = 2 p = 2       is just an affirmative answer to a conjecture posed by Chan and Kwong. We cite three counterexamples related to Theorem 1 and Corollary 1.

𝐴≥𝐵≥0 assures (𝐵^{𝑟}𝐴^{𝑝}𝐵^{𝑟})^{1/𝑞}≥𝐵^{(𝑝+2𝑟)/𝑞} for 𝑟≥0, 𝑝≥0, 𝑞≥1 with (1+2𝑟)𝑞≥𝑝+2𝑟