Rank-one operator calculus

[Boundedness, norm and linearity] $T_{u,v}$ is linear and $\mathrm{\parallel }{T}_{u,v}x\mathrm{\parallel }=\mathrm{\mid }⟨x,v⟩\mathrm{\mid }\mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }$ by [F4], so $\mathrm{\parallel }{T}_{u,v}\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ [F6]; if $u\mathrm{\ne }0$ and $v\mathrm{\ne }0$, taking $x=v\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }$ gives $T_{u,v}x=\mathrm{\parallel }v\mathrm{\parallel }u$ and hence $\mathrm{\parallel }{T}_{u,v}\mathrm{\parallel }\ge \mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$, while for $u=0$ or $v=0$ the operator vanishes and both sides are $0$. Linearity in $u$ and conjugate-linearity in $v$ are immediate from the definition and the pairing conventions [F4].

[Trace class and trace norm] The range of $T_{u,v}$ is contained in $\mathbb{C}u$, so $T_{u,v}$ has rank at most one, is finite rank, compact [F1]; its one-term nuclear representation is $T_{u,v}={\sum }_j{T}_{v_j,u_j}$ with $v_1=u$, $u_1=v$ and the remaining terms zero, so ${\sum }_j\mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }=\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }$ and [F2] first establishes trace class and gives $\mathrm{\parallel }{T}_{u,v}{\mathrm{\parallel }}_1\le \mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$, while $\mathrm{\parallel }{T}_{u,v}{\mathrm{\parallel }}_1\ge \mathrm{\parallel }{T}_{u,v}\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ by [F5] and step 1.1. Hence $\mathrm{\parallel }{T}_{u,v}{\mathrm{\parallel }}_1=\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$.

[The trace] Applying [F3] to the one-term nuclear representation of step 2.1 gives $\mathrm{tr}(T_{u,v})={\sum }_j⟨v_j,u_j⟩=⟨u,v⟩$. In particular $\mathrm{tr}(T_{x,x})=⟨x,x⟩=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ and, with step 2.1, $\mathrm{\parallel }{T}_{x,x}{\mathrm{\parallel }}_1=\mathrm{\parallel }x{\mathrm{\parallel }}^2$.

[Adjoint, positivity and one-sided products] For all $x,y$, $⟨T_{u,v}x,y⟩=⟨x,v⟩⟨u,y⟩=⟨x,⟨y,u⟩v⟩$ by [F4], so $T_{u,v}^{\ast }=T_{v,u}$; in particular $T_{x,x}$ is self-adjoint, and $⟨T_{x,x}y,y⟩=\mathrm{\mid }⟨y,x⟩{\mathrm{\mid }}^2\ge 0$ for every $y$, so it is positive by [F6]. For bounded $B$: $(B{T}_{u,v})x=B(⟨x,v⟩u)=⟨x,v⟩Bu=T_{Bu,v}x$ and $(T_{u,v}B)x=⟨Bx,v⟩u=⟨x,B^{\ast }v⟩u=T_{u,B^{\ast }v}x$, using the defining property of the adjoint [F4].

[Trace-norm continuity] By the bilinearity of step 1.1, $T_{u,v}-T_{u^{\mathrm{\prime }},v^{\mathrm{\prime }}}=T_{u-u^{\mathrm{\prime }},v}+T_{u^{\mathrm{\prime }},v-v^{\mathrm{\prime }}}$; by [F5] and the estimates of steps 1.1 and 2.1, $\mathrm{\parallel }{T}_{u,v}-T_{u^{\mathrm{\prime }},v^{\mathrm{\prime }}}{\mathrm{\parallel }}_1\le \mathrm{\parallel }u-u^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }{u}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }v-v^{\mathrm{\prime }}\mathrm{\parallel }$, which tends to $0$ whenever $(u^{\mathrm{\prime }},v^{\mathrm{\prime }})\to (u,v)$ in $H\times H$; hence $(u,v)\mapsto T_{u,v}$ is continuous into ${\mathcal{S}}_1(H)$, and fixing one slot gives the stated continuity. This is (f).

[Products with trace-class operators] If $S$ is trace class then $(S{T}_{u,v})x=⟨x,v⟩Su=T_{Su,v}x$ and $(T_{u,v}S)x=⟨Sx,v⟩u=⟨x,S^{\ast }v⟩u=T_{u,S^{\ast }v}x$, so $S{T}_{u,v}=T_{Su,v}$ and $T_{u,v}S=T_{u,S^{\ast }v}$ [F4]; applying step 3.1 to the rank-one operators $T_{Su,v}$ and $T_{u,S^{\ast }v}$ gives $\mathrm{tr}(S{T}_{x,y})=⟨Sx,y⟩$ and $\mathrm{tr}(T_{x,y}S)=⟨x,S^{\ast }y⟩=⟨Sx,y⟩$. This is (e); (a)-(d) are steps 1.1, 2.1, 3.1 and 3.2, and the scope remark records that nothing beyond the rank-one calculus is asserted.