inner automorphisms of $B({\cal H})$, reference It is a well-known fact that if $\phi : B({\cal H})\to B({\cal H})$, with ${\cal H}$ complex Hilbert space, is a $^*$-automorphism, then there exists a unitary operator $U: {\cal H} \to {\cal H}$ such that $$\phi(A)= UAU^{-1}\quad  \mbox{for every $A \in B({\cal H})$.}$$
Could you give me a precise reference where this statement is proved?
 A: Recall that the pure positive functionals on the C*-algebra of compact operators on a Hilbert space, $K(H)$, are exactly the vector functionals $\omega_x$ for $x\in H$; these functionals are defined by $\omega_x(T)=\langle Tx, x \rangle$. Furthermore, the map $\omega_x \rightarrow \omega_x\circ \phi$ clearly defines a bijection between the set of pure positive functionals; if $\omega_x\circ \phi=\psi_1+ \psi_2$ is a non-trivial dcecomposition into positive functionals, then $\psi_1\circ \phi^{-1}+\psi_2\circ \phi^{-1}$ is a nontrivial decomposition of $\omega_x$.
Hence, for any $x\in H$ there exists a unique vector we will suggestively denote by $Ux$, such that $\langle \phi(T)x,x\rangle=\langle T(Ux), Ux \rangle$ $\forall T\in K(H)$. A priori, since $\phi$ is ultraweakly continuous, $\langle \phi(\cdot)x,x\rangle$, when seen as a functional on $B(H)$, must also be ultrweakly continuous. Hence there exists a sequence of square summable $\{x_n\}_n$ vectors such that $\langle \phi(T)x,x\rangle=\sum\limits_n\langle Tx_n,x_n\rangle$ for all $T\in B(H)$. In particular $\langle T(Ux), Ux \rangle=\sum\limits_n\langle Tx_n,x_n\rangle$ for compact $T$. Letting $T$ range over all orthogonal projections onto one dimensional subspaces of $\{Ux\}^{\perp}$ this equation implies that $\langle Tx_n, x_n \rangle=\langle Tx_n, Tx_n \rangle=0$ for such $T$, so we are left to conclude that all $x_n$s are scalar multiples of $Ux$. This then implies that $\langle \phi(T)x, x \rangle=\langle T(Ux), Ux \rangle$ must hold for ALL operators $T\in B(H)$ (1).
We now claim that the mapping $U: x \mapsto Ux$ is a unitary operator. It is isometric since, by (1), we have $\langle \phi(I)x, x \rangle=\langle I(Ux), Ux \rangle$, i.e. $||x||^2=||Ux||^2$. From the bijection in the first paragraph it is also immediate that $U$ is surjective. We finally claim that is also linear. That it is homogenous is obvious so we need to show that $U(x+y)=Ux+Uy$ when $x$ and $y$ are linearly independent (not parallell). For this, start by defining a linear operator $S_0: \text{Span}\{x, y\}\rightarrow \text{Span}\{x, y\}$ by $S_0(x)=y$ and $S_0(y)$ (this is possible by linear independence). Extend $S_0$ to an operator $S: H \rightarrow H$ by declaring $S_{|\text{Span}\{x, y\}^{\perp}}=0$. We claim first that $\phi^{-1}(S)(Ux)=Uy$. Indeed, this is because $\langle \phi(T)y,y\rangle=\langle \phi(T) Sx,Sx\rangle=\langle S^*\phi(T)Sx,x \rangle=\langle \phi(\phi^{-1}(S^*)T\phi^{-1}(S))x, x\rangle=\langle \phi^{-1}(S^*)T\phi^{-1}(S)(Ux), Ux\rangle=\langle T\phi^{-1}(S)(Ux), \phi^{-1}(S)(Ux)\rangle$ holds for all $T\in K(H)$ and (1) then shows that $\phi^{-1}(S)(Ux)$ satisfies the property that uniquely determines $Uy$, i.e. they are equal.
To show that $U(x+y)=Ux+Uy$ it will suffice to show that $\langle \phi(T)(x+y),x+y\rangle=\langle T(Ux+Uy),Ux+Uy\rangle$ for all compact $T$, but since the LHS can be written as $\langle \phi(T)x,x\rangle+\langle \phi(T)x,y \rangle+\langle \phi(T)y,x\rangle+\langle \phi(T)y, y\rangle$ and the RHS equals $\langle T(Ux),Ux\rangle+\langle T(Ux),Uy \rangle+\langle T(Uy),Ux\rangle+\langle T(Uy),Uy\rangle$ it will of course suffice to show that the second and third terms of the LHS and RHS coincide, respectively. By symmetry it suffices to show that $\langle \phi(T)x,y \rangle=\langle T(Ux), Uy\rangle$. For this, note that $\langle \phi(T)x,y \rangle=\langle \phi(T)x,Sx \rangle=\langle \phi(\phi^{-1}(S^*)T)x,x \rangle=\langle \phi^{-1}(S^*)T(Ux),Ux \rangle=\langle T(Ux),\phi^{-1}(S)Ux \rangle$, but by the previous paragraph this equals $\langle T(Ux),Uy\rangle$ as required. 
So $U$ is a unitary operator and by (1) we see that for fixed $T\in B(H)$ we have $\langle \phi(T)x,x\rangle=\langle U^*TUx,x\rangle$ for all $x\in B(H)$. By the polarization identity, it follows that $\phi(T)=U^*TU$. This holds for all $T\in B(H)$ so we are done.


