Strong bounded interaction picture

[The correct conjugation and strong continuity] For $t\to u$ and fixed $\psi \in H$, subtract $E(-u)V(u)E(u)\psi$ from $E(-t)V(t)E(t)\psi$ and write the difference as $E(-t)V(t)[E(t)-E(u)]\psi +E(-t)[V(t)-V(u)]E(u)\psi +[E(-t)-E(-u)]V(u)E(u)\psi$. Its three norms tend to zero respectively by the bound $v$ and group continuity, strong continuity of $V$ on the fixed vector $E(u)\psi$, and group continuity on the fixed vector $V(u)E(u)\psi$. Unitary conjugation gives self-adjointness and the exact norm identity. Thus $B$ is strongly continuous with $\mathrm{\parallel }B(t)\mathrm{\parallel }\le K$ and $B(t{)}^{\ast }=-B(t)$. If $x(t)$ is any continuous $H$-valued curve, $B(t)x(t)$ is continuous: subtract $B(u)x(u)$, use the uniform bound on $B(t)[x(t)-x(u)]$, and use strong continuity on the fixed $x(u)$. Iterating this observation proves joint vector continuity of every finite time-ordered product.

[Vector integrals and all-time oriented recursion] For all $t,s\in I$, define $D_0(t,s)=I$ and $D_{n+1}(t,s)\psi ={\int }_s^{t}B(r)D_n(r,s)\psi dr$, with the usual signed orientation when $t<s$. Induction using [F2] proves existence, linearity and joint strong continuity; one may also change variables to fixed unit-interval simplexes to prove continuity in both endpoints. The bound $\mathrm{\parallel }{D}_n(t,s)\mathrm{\parallel }\le K^n\mathrm{\mid }t-s{\mathrm{\mid }}^n\mathrm{/}n!$ follows inductively from the integral inequality. For $t\ge s$, expanding the iterated integral puts the newest factor $B(t_n)$ on the left, giving precisely the increasing-simplex expression in the Statement. For $t<s$, the signed recursion gives the genuinely backward construction; it is not postulated to be the adjoint of a forward series. The operator norm bounds are uniform on $I\times I$, despite each integral being defined vectorwise.

[Norm convergence, tails and actual strong differential equation] The factorial bounds make ${\sum }_n{D}_n(t,s)$ uniformly convergent in operator norm, with norm at most $e^{K\mathrm{\mid }t-s\mathrm{\mid }}$. For $x=K\mathrm{\mid }t-s\mathrm{\mid }$, $(N+1+j)!\ge (N+1)!j!$ gives ${\sum }_{n>N}{x}^n\mathrm{/}n!\le e^x{x}^{N+1}\mathrm{/}(N+1)!$, including the zero case. Uniform convergence on each vector permits summation through the recursive integral and gives $U_I(t,s)\psi =\psi +{\int }_s^{t}B(r)U_I(r,s)\psi dr$. The integrand is continuous by step 1.1 and joint strong continuity of the uniformly convergent series. [F2] therefore proves the strong first-endpoint derivative on every $H$ vector, with continuous derivative. If two continuous solutions of this integral identity have the same initial vector, their difference satisfies $\mathrm{\parallel }w(t)\mathrm{\parallel }\le M{K}^n\mathrm{\mid }t-s{\mathrm{\mid }}^n\mathrm{/}n!$ for all $n$, where $M$ is its finite maximum norm on the compact interval; repeated integration proves the inequality and its limit gives zero. This is vector uniqueness in either time orientation.

[Cocycle and genuine unitary reverse] For every fixed $\psi$, $t\mapsto U_I(t,s)U_I(s,r)\psi$ and $t\mapsto U_I(t,r)\psi$ solve the same strong equation with the same value $U_I(s,r)\psi$ at $t=s$. Step 3.1's vector uniqueness gives the full cocycle for all orientations. In particular, the independently constructed reverse family satisfies $U_I(t,s)U_I(s,t)=U_I(s,t)U_I(t,s)=I$. For every vector, differentiation gives $\frac{d}{dt}\mathrm{\parallel }{U}_I(t,s)\psi {\mathrm{\parallel }}^2=2\mathrm{Re}⟨B(t)U_I(t,s)\psi ,U_I(t,s)\psi ⟩=0$ by skew-adjointness. Thus both orientations preserve norms. The two-sided inverse implies onto, so the evolution is unitary with $U_I(t,s{)}^{\ast }=U_I(s,t)$. This proves both unitary identities without illegitimate differentiation of an operator adjoint from strong convergence alone.

[Endpoint continuity and the reverse derivative] The integral identity and unitary bound give $\mathrm{\parallel }{U}_I(t,s)-I\mathrm{\parallel }\le K\mathrm{\mid }t-s\mathrm{\mid }$. Cocycle and unitarity then imply $\mathrm{\parallel }{U}_I(t,s)-U_I(t^{\mathrm{\prime }},s^{\mathrm{\prime }})\mathrm{\parallel }\le K(\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid })$, so the interaction evolution is even jointly norm-continuous; this conclusion does not imply norm continuity of $V_I$. The cocycle identity $U_I(t,s+h)\psi -U_I(t,s)\psi =U_I(t,s+h)[\psi -U_I(s+h,s)\psi ]$ and the diagonal first-endpoint derivative yield ${\mathrm{\partial }}_s{U}_I(t,s)\psi =-U_I(t,s)B(s)\psi$. Strong continuity of $B$ and joint continuity of $U_I$ make this derivative continuous. A first-time derivative of the reverse operator $U_I(s,t)$ instead obeys ${\mathrm{\partial }}_s{U}_I(s,t)\psi =B(s)U_I(s,t)\psi$.

[Physical propagator and vector mild identity] Define $U(t,s)=E(t)U_I(t,s)E(-s)$. Products of uniformly bounded strongly continuous families are strongly continuous on each vector, as in step 1.1. The group law of $E$ and the cocycle and unitarity of $U_I$ prove the physical cocycle, unitarity and inverse. Apply the integral identity of step 3.1 to the fixed vector $E(-s)\psi$, then multiply by $E(t)$. In its integrand, $E(t)V_I(r)U_I(r,s)E(-s)\psi =E(t-r)V(r)U(r,s)\psi$. This proves the mild formula on every vector. Its integrand is continuous in $(t,r,s)$ on vectors by uniform boundedness and strong continuity, so it is a well-defined vector strong integral defining a bounded operator. No claim of a norm-Bochner integral into $\mathcal{B}(H)$ follows or is needed.

[Weak equation with no hidden domain assertion] Fix initial $\psi \in H$ and test vector $\phi \in D(H_0)$, and write $x(t)=U_I(t,s)E(-s)\psi$, so $x\in C^1(I;H)$ by step 3.1. The scalar pairing is $⟨U(t,s)\psi ,\phi ⟩=⟨x(t),E(-t)\phi ⟩$. The second curve belongs to the autonomous domain and has derivative $(i\mathrm{/}\mathrm{\hslash })H_{0}E(-t)\phi$ by [F1]. The ordinary scalar product rule, with conjugate linearity in the second argument, gives derivative $⟨B(t)x(t),E(-t)\phi ⟩-(i\mathrm{/}\mathrm{\hslash })⟨x(t),H_{0}E(-t)\phi ⟩$. The first term equals $-(i\mathrm{/}\mathrm{\hslash })⟨V(t)U(t,s)\psi ,\phi ⟩$ by the exact conjugation; autonomous commutation gives the second term $-(i\mathrm{/}\mathrm{\hslash })⟨U(t,s)\psi ,H_0\phi ⟩$. Both are continuous by strong continuity and uniform bounds. This proves the stated weak equation, including its original domain-restricted initial-vector clause, without applying $H_0$ to $U(t,s)\psi$ or assuming this vector lies in its domain. All retained assertions and the explicit ordering/integral corrections follow.