[proofplan] We construct a canonical map from $B(H)$ to $M(K(H))$ by letting a bounded operator act on compact operators by left and right composition. To prove that every multiplier arises this way, we fix a unit vector $\eta\in H$ and use the action of the left centralizer on rank-one operators $\theta_{\xi,\eta}$ to recover a bounded operator $T:H\to H$. The double-centralizer identities then force the right centralizer to be right composition by the same $T$, first on rank-one operators, then on finite-rank operators, and finally on all of $K(H)$ by norm density. [/proofplan]

[step:Define the canonical double centralizer associated to a bounded operator] For $\xi,\zeta\in H$, define the rank-one operator \begin{align*} \theta_{\xi,\zeta}:H&\to H \end{align*} \begin{align*} h&\mapsto (h,\zeta)_H\xi. \end{align*} Here the [Hilbert space](/page/Hilbert%20Space) [inner product](/page/Inner%20Product) is linear in the first variable. Let $T\in B(H)$. Define maps \begin{align*} L_T:K(H)&\to K(H), & S&\mapsto TS, \end{align*} \begin{align*} R_T:K(H)&\to K(H), & S&\mapsto ST. \end{align*} If $S\in K(H)$, then $TS$ and $ST$ are compact, because the image of a bounded set under $S$ has compact closure and bounded operators map compact sets to compact sets; for $ST$, compactness follows since $T$ maps bounded sets to bounded sets and $S$ maps bounded sets to relatively compact sets. Thus $L_T$ and $R_T$ are well-defined bounded linear maps, with \begin{align*} \|L_T(S)\|_{B(H)}\le \|T\|_{B(H)}\|S\|_{B(H)} \end{align*} and \begin{align*} \|R_T(S)\|_{B(H)}\le \|S\|_{B(H)}\|T\|_{B(H)}. \end{align*} For $S_1,S_2\in K(H)$, associativity of composition gives \begin{align*} L_T(S_1S_2)=T(S_1S_2)=(TS_1)S_2=L_T(S_1)S_2, \end{align*} \begin{align*} R_T(S_1S_2)=(S_1S_2)T=S_1(S_2T)=S_1R_T(S_2), \end{align*} and \begin{align*} S_1L_T(S_2)=S_1(TS_2)=(S_1T)S_2=R_T(S_1)S_2. \end{align*} Hence $(L_T,R_T)$ is a double centralizer of $K(H)$. The assignment \begin{align*} \Phi:B(H)&\to M(K(H)) \end{align*} \begin{align*} T&\mapsto (L_T,R_T) \end{align*} is linear. With the double-centralizer product convention \begin{align*} (L,R)(L',R')=(L\circ L',R'\circ R), \end{align*} associativity of composition gives $\Phi(TU)=\Phi(T)\Phi(U)$ for all $T,U\in B(H)$. Its unit is preserved because $1_H:H\to H$ is the identity operator and $L_{1_H}$ and $R_{1_H}$ are the identity maps on $K(H)$. The involution on double centralizers is given by $(L,R)^*=(L^\#,R^\#)$, where \begin{align*} L^\#:K(H)&\to K(H), & S&\mapsto R(S^*)^*, \end{align*} and \begin{align*} R^\#:K(H)&\to K(H), & S&\mapsto L(S^*)^*. \end{align*} For $(L_T,R_T)$ this gives $L^\#(S)=(S^*T)^*=T^*S=L_{T^*}(S)$ and $R^\#(S)=(T S^*)^*=ST^*=R_{T^*}(S)$. Hence \begin{align*} \Phi(T)^*=\Phi(T^*), \end{align*} so $\Phi$ is a unital $*$-homomorphism. [/step]

[step:Recover a bounded operator from the action on rank-one operators]Let $(L,R)\in M(K(H))$ be a double centralizer. Thus $L,R:K(H)\to K(H)$ are bounded linear maps satisfying \begin{align*} L(AB)=L(A)B,\qquad R(AB)=AR(B),\qquad AL(B)=R(A)B \end{align*} for all $A,B\in K(H)$. Let $\mathcal{L}(K(H))$ denote the [Banach space](/page/Banach%20Space) of bounded linear maps from $K(H)$ to itself, equipped with the operator norm induced by $\|\cdot\|_{B(H)}$ on $K(H)$. Choose a unit vector $\eta\in H$. Define a map \begin{align*} T:H&\to H \end{align*} \begin{align*} \xi&\mapsto L(\theta_{\xi,\eta})\eta. \end{align*} Since $\xi\mapsto \theta_{\xi,\eta}$ is linear and $L$ is linear, the map $T$ is linear. Also \begin{align*} \|T\xi\|_H = \|L(\theta_{\xi,\eta})\eta\|_H \le \|L(\theta_{\xi,\eta})\|_{B(H)} \le \|L\|_{\mathcal{L}(K(H))}\|\theta_{\xi,\eta}\|_{B(H)} = \|L\|_{\mathcal{L}(K(H))}\|\xi\|_H, \end{align*} because $\|\eta\|_H=1$ and $\|\theta_{\xi,\eta}\|_{B(H)}=\|\xi\|_H\|\eta\|_H$. Therefore $T\in B(H)$. We next show that $L(\theta_{\xi,\zeta})=\theta_{T\xi,\zeta}$ for all $\xi,\zeta\in H$. Since \begin{align*} \theta_{\xi,\zeta}=\theta_{\xi,\eta}\theta_{\eta,\zeta}, \end{align*} the left centralizer identity gives \begin{align*} L(\theta_{\xi,\zeta}) = L(\theta_{\xi,\eta}\theta_{\eta,\zeta}) = L(\theta_{\xi,\eta})\theta_{\eta,\zeta}. \end{align*} For $h\in H$, \begin{align*} L(\theta_{\xi,\eta})\theta_{\eta,\zeta}h = L(\theta_{\xi,\eta})((h,\zeta)_H\eta) = (h,\zeta)_H L(\theta_{\xi,\eta})\eta = (h,\zeta)_H T\xi = \theta_{T\xi,\zeta}h. \end{align*} Thus \begin{align*} L(\theta_{\xi,\zeta})=\theta_{T\xi,\zeta}=T\theta_{\xi,\zeta}. \end{align*}[/step]

[guided]The point of fixing $\eta$ is to turn the operator $L(\theta_{\xi,\eta})$ back into a vector: evaluate it at $\eta$. We define \begin{align*} T:H&\to H \end{align*} \begin{align*} \xi&\mapsto L(\theta_{\xi,\eta})\eta. \end{align*} This is linear because $\xi\mapsto \theta_{\xi,\eta}$ is linear, $L$ is linear, and evaluation at $\eta$ is linear. We must also prove that this map is bounded. For every $\xi\in H$, \begin{align*} \|T\xi\|_H = \|L(\theta_{\xi,\eta})\eta\|_H \le \|L(\theta_{\xi,\eta})\|_{B(H)}\|\eta\|_H. \end{align*} Since $\|\eta\|_H=1$ and $L:K(H)\to K(H)$ is bounded, \begin{align*} \|L(\theta_{\xi,\eta})\|_{B(H)} \le \|L\|_{\mathcal{L}(K(H))}\|\theta_{\xi,\eta}\|_{B(H)}. \end{align*} The rank-one norm formula gives \begin{align*} \|\theta_{\xi,\eta}\|_{B(H)}=\|\xi\|_H\|\eta\|_H=\|\xi\|_H. \end{align*} Combining these inequalities, \begin{align*} \|T\xi\|_H\le \|L\|_{\mathcal{L}(K(H))}\|\xi\|_H, \end{align*} so $T\in B(H)$. Now we verify that $T$ really represents the left centralizer on all rank-one operators. For arbitrary $\xi,\zeta\in H$, the rank-one operator $\theta_{\xi,\zeta}$ factors through the fixed vector $\eta$: \begin{align*} \theta_{\xi,\zeta}=\theta_{\xi,\eta}\theta_{\eta,\zeta}. \end{align*} Indeed, for $h\in H$, \begin{align*} \theta_{\xi,\eta}\theta_{\eta,\zeta}h = \theta_{\xi,\eta}((h,\zeta)_H\eta) = ((h,\zeta)_H\eta,\eta)_H\xi = (h,\zeta)_H(\eta,\eta)_H\xi = (h,\zeta)_H\xi = \theta_{\xi,\zeta}h. \end{align*} The left centralizer identity therefore gives \begin{align*} L(\theta_{\xi,\zeta}) = L(\theta_{\xi,\eta}\theta_{\eta,\zeta}) = L(\theta_{\xi,\eta})\theta_{\eta,\zeta}. \end{align*} Evaluating this composition at $h\in H$, \begin{align*} L(\theta_{\xi,\eta})\theta_{\eta,\zeta}h = L(\theta_{\xi,\eta})((h,\zeta)_H\eta) = (h,\zeta)_H L(\theta_{\xi,\eta})\eta = (h,\zeta)_H T\xi = \theta_{T\xi,\zeta}h. \end{align*} Hence \begin{align*} L(\theta_{\xi,\zeta})=\theta_{T\xi,\zeta}=T\theta_{\xi,\zeta}. \end{align*} This proves that the bounded operator recovered from the single vector $\eta$ controls the action of $L$ on every rank-one operator.[/guided]

[step:Identify the right centralizer from the compatibility identity]We prove that $R(\theta_{\alpha,\beta})=\theta_{\alpha,T^*\beta}$ for all $\alpha,\beta\in H$. If $\alpha=0$, then $\theta_{\alpha,\beta}=0$ and the claim follows from the linearity of $R$, so assume $\alpha\ne 0$. Choose $\zeta_0\in H$ with $\zeta_0\ne 0$. The compatibility identity $AL(B)=R(A)B$, applied with $A=\theta_{\alpha,\beta}$ and $B=\theta_{\xi,\zeta_0}$, gives \begin{align*} \theta_{\alpha,\beta}L(\theta_{\xi,\zeta_0}) = R(\theta_{\alpha,\beta})\theta_{\xi,\zeta_0} \end{align*} for every $\xi\in H$. Using the formula already proved for $L$, \begin{align*} \theta_{\alpha,\beta}L(\theta_{\xi,\zeta_0}) = \theta_{\alpha,\beta}\theta_{T\xi,\zeta_0} = (T\xi,\beta)_H\theta_{\alpha,\zeta_0}. \end{align*} Thus \begin{align*} R(\theta_{\alpha,\beta})\theta_{\xi,\zeta_0} = (T\xi,\beta)_H\theta_{\alpha,\zeta_0}. \end{align*} Choose $h_0\in H$ with $(h_0,\zeta_0)_H\ne 0$; for instance $h_0=\zeta_0$. Evaluating the preceding operator identity at $h_0$ gives \begin{align*} (h_0,\zeta_0)_H R(\theta_{\alpha,\beta})\xi = (T\xi,\beta)_H(h_0,\zeta_0)_H\alpha. \end{align*} Since $(h_0,\zeta_0)_H\ne 0$, cancellation gives \begin{align*} R(\theta_{\alpha,\beta})\xi=(T\xi,\beta)_H\alpha. \end{align*} By the defining property of the adjoint $T^*$, \begin{align*} (T\xi,\beta)_H=(\xi,T^*\beta)_H. \end{align*} Therefore \begin{align*} R(\theta_{\alpha,\beta})\xi=(\xi,T^*\beta)_H\alpha=\theta_{\alpha,T^*\beta}\xi \end{align*} for every $\xi\in H$, and hence \begin{align*} R(\theta_{\alpha,\beta})=\theta_{\alpha,T^*\beta}=\theta_{\alpha,\beta}T. \end{align*} Renaming $\alpha,\beta$ as $\xi,\zeta$ gives \begin{align*} R(\theta_{\xi,\zeta})=\theta_{\xi,T^*\zeta}=\theta_{\xi,\zeta}T. \end{align*}[/step]

[guided]The left centralizer has already produced the operator $T$. The compatibility identity is what forces the right centralizer to use the same operator, and the safest way to extract vectors from the rank-one identity is to evaluate at a vector whose inner product with the chosen second vector is nonzero. Fix $\alpha,\beta\in H$. If $\alpha=0$, then $\theta_{\alpha,\beta}=0$, so $R(\theta_{\alpha,\beta})=0=\theta_{\alpha,T^*\beta}$ by linearity of $R$. Assume $\alpha\ne 0$. Choose a nonzero vector $\zeta_0\in H$. For every $\xi\in H$, apply the compatibility identity $AL(B)=R(A)B$ with $A=\theta_{\alpha,\beta}$ and $B=\theta_{\xi,\zeta_0}$: \begin{align*} \theta_{\alpha,\beta}L(\theta_{\xi,\zeta_0}) = R(\theta_{\alpha,\beta})\theta_{\xi,\zeta_0}. \end{align*} The operator $T$ was defined by $T\xi=L(\theta_{\xi,\eta})\eta$, and the preceding left-centralizer computation proves, for every $u,v\in H$, that $L(\theta_{u,v})=\theta_{Tu,v}$. Applying this with $u=\xi$ and $v=\zeta_0$ gives $L(\theta_{\xi,\zeta_0})=\theta_{T\xi,\zeta_0}$. Therefore \begin{align*} R(\theta_{\alpha,\beta})\theta_{\xi,\zeta_0} = \theta_{\alpha,\beta}\theta_{T\xi,\zeta_0} = (T\xi,\beta)_H\theta_{\alpha,\zeta_0}. \end{align*} Now choose $h_0\in H$ with $(h_0,\zeta_0)_H\ne 0$; taking $h_0=\zeta_0$ works because $\zeta_0\ne 0$. Evaluating the operator identity at $h_0$ gives \begin{align*} R(\theta_{\alpha,\beta})\theta_{\xi,\zeta_0}h_0 = (T\xi,\beta)_H\theta_{\alpha,\zeta_0}h_0. \end{align*} Using the definition of the rank-one operator on both sides, \begin{align*} (h_0,\zeta_0)_H R(\theta_{\alpha,\beta})\xi = (T\xi,\beta)_H(h_0,\zeta_0)_H\alpha. \end{align*} The scalar $(h_0,\zeta_0)_H$ is nonzero, so cancellation is valid and yields \begin{align*} R(\theta_{\alpha,\beta})\xi=(T\xi,\beta)_H\alpha. \end{align*} Because $T^*$ is the Hilbert-space adjoint of $T$ and the inner product is linear in the first variable, \begin{align*} (T\xi,\beta)_H=(\xi,T^*\beta)_H. \end{align*} Thus \begin{align*} R(\theta_{\alpha,\beta})\xi=(\xi,T^*\beta)_H\alpha=\theta_{\alpha,T^*\beta}\xi \end{align*} for every $\xi\in H$. Therefore $R(\theta_{\alpha,\beta})=\theta_{\alpha,T^*\beta}$. Finally, for every $h\in H$, \begin{align*} \theta_{\alpha,\beta}Th=(Th,\beta)_H\alpha=(h,T^*\beta)_H\alpha=\theta_{\alpha,T^*\beta}h, \end{align*} so $R(\theta_{\alpha,\beta})=\theta_{\alpha,\beta}T$.[/guided]

[step:Extend the rank-one formulas to every compact operator] Let $\mathcal{F}(H)$ denote the linear subspace of $K(H)$ consisting of finite-rank operators. Every $F\in \mathcal{F}(H)$ is a finite linear combination of rank-one operators. By linearity and the rank-one formulas proved above, \begin{align*} L(F)=TF,\qquad R(F)=FT \end{align*} for every $F\in \mathcal{F}(H)$. By the definition of $K(H)$ in the theorem statement as the operator-norm closure of the finite-rank operators, $\mathcal{F}(H)$ is norm dense in $K(H)$. Let $S\in K(H)$, and choose a net $(F_i)_{i\in I}$ in $\mathcal{F}(H)$ such that \begin{align*} \|F_i-S\|_{B(H)}\to 0. \end{align*} Since $L$, $R$, left composition by $T$, and right composition by $T$ are bounded maps on $K(H)$, passing to the norm limit gives \begin{align*} L(S)=\lim_i L(F_i)=\lim_i TF_i=TS \end{align*} and \begin{align*} R(S)=\lim_i R(F_i)=\lim_i F_iT=ST. \end{align*} Therefore $(L,R)=(L_T,R_T)=\Phi(T)$. [/step]

[step:Prove injectivity and conclude the canonical identification] Suppose $T\in B(H)$ satisfies $\Phi(T)=0$. Then $T\theta_{\xi,\zeta}=0$ for all $\xi,\zeta\in H$. Choose any unit vector $\zeta\in H$. For every $\xi\in H$, \begin{align*} 0=T\theta_{\xi,\zeta}\zeta=T((\zeta,\zeta)_H\xi)=T\xi. \end{align*} Hence $T=0$, so $\Phi$ is injective. The preceding step proves that every double centralizer of $K(H)$ is equal to $(L_T,R_T)$ for some $T\in B(H)$, so $\Phi$ is surjective. Since $\Phi$ is a bijective unital $*$-homomorphism between C*-algebras, it is a unital $*$-isomorphism. Thus \begin{align*} M(K(H))\cong B(H) \end{align*} as unital C*-algebras, with $T\in B(H)$ acting on $K(H)$ by left and right composition. [/step]