Finite Hayashi–Nagaoka operator inequality

If $H=\{0\}$, every endomorphism, including $I_H$, is zero, so the asserted inequality is $0⪯0$. Henceforth assume $H\mathrm{\ne }\{0\}$, as required by [F2]. Set $X=S+T$, $R=\sqrt{X}$, and $U=I-R$. For $c>0$ and endomorphisms $A,B$, expanding $(A-cB{)}^{\ast }(A-cB)⪰0$ and dividing by $c$ gives $A^{\ast }B+B^{\ast }A⪯c^{-1}{A}^{\ast }A+c{B}^{\ast }B$. Apply this with $A=\sqrt{T}R$ and $B=\sqrt{T}U$. Since $R+U=I$, expansion of $(R+U)T(R+U)$ and the preceding inequality give $T⪯(1+c^{-1})RTR+(1+c)UTU\mathrm{.}$ Now $T⪯X=R^2$, so congruence by $U$ gives $UTU⪯U{R}^{2}U=R{U}^{2}R$; the last equality follows because $U=I-R$ commutes with $R$. Since $U^2=I-2R+R^2=I+S+T-2R$, substitution and collection of the two $T$ terms yield $T⪯R((2+c+c^{-1})T+(1+c)(I+S-2R))R\mathrm{.}$ By [F2], $R⪰\sqrt{S}$; [F3] and $0⪯S⪯I$ give $\sqrt{S}⪰S$ by diagonalizing $S$. Therefore $I+S-2R⪯I-S$. Taking $c=1$ in the last display proves $T⪯R(4T+2(I-S))R\mathrm{.}$

Let $P$ be the orthogonal projection onto $\mathrm{supp}X$. If $v\in \ker X$, then $0=⟨v,Xv⟩=⟨v,Sv⟩+⟨v,Tv⟩$; both terms are nonnegative, so each is zero. By [F3], these terms are $\mathrm{\parallel }\sqrt{S}v{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }\sqrt{T}v{\mathrm{\parallel }}^2$, hence $Sv=Tv=0$. Since $S,T$ are self-adjoint, they preserve the orthogonal complement $PH$ of $\ker X$ as well. Thus both sides of the desired inequality are block diagonal for $H=PH\oplus \ker X$. On $PH$, congruence of step 1.1 by the inverse of the restriction of $R$ gives $X^{-1\mathrm{/}2}T{X}^{-1\mathrm{/}2}⪯4T+2(I-S)$. On this block, $X^{-1\mathrm{/}2}(S+T)X^{-1\mathrm{/}2}=I_{PH}$, so its left side is $I_{PH}-X^{-1\mathrm{/}2}S{X}^{-1\mathrm{/}2}$. On $\ker X$, the desired inequality is $I⪯2I$, since $S=T=X^{-1\mathrm{/}2}=0$ there. Combining these two block inequalities proves the Statement, including when either block is zero-dimensional.