Let us choose to postulate (e.g. considering the analogy of the Hamiltonian being a generator of time evolution in classical mechanics)
$$
i\hbar \frac{d\hat{U}}{dt}=\hat{H}\hat{U}\tag{1}
$$
where $\hat{U}$ is the (unitary, linear) evolution operator and $\hat{H}$ the Hamiltonian (the most general version of which; i.e. time-dependent with instances at different times non-commuting).
In S-picture, one can easily show that (1) is equivalent to
$$
i\hbar \frac{d}{dt}|\psi_S(t)\rangle=\hat{H}_S|\psi_S(t)\rangle\tag{2}
$$
where $\psi$ is a state, $|\psi_S(t)\rangle = \hat{U}(t)|\psi_S(0)\rangle \equiv \hat{U}(t)|\psi_H\rangle$ and $\hat{H}_S:=\hat{H}$.
In H-picture, it is straightforward to show that (1) implies
$$
\frac{d\hat{A}_H}{dt} = \frac{\partial\hat{A}_H}{\partial t}+\frac{1}{i\hbar}[\hat{A}_H,\hat{H}_H]\tag{3}
$$
where $A$ is an observable and $\hat{A}_H(t)=\hat{U}(t)^\dagger \hat{A}_H(0)\hat{U}(t)\equiv\hat{U}(t)^\dagger \hat{A}_S\hat{U}(t)$ (and also $[\hat{H}_{H},\hat{H}_H]=0$ implying that the time dependence of $\hat{H}_H$ is purely explicit, i.e. $\hat{H}_H=\hat{H}_S\equiv \hat{H}$ with $[\hat{H},\hat{U}]=0$).
My question: is it possible to obtain (1) from (3), i.e. to show that (1) is equivalent to (3)?
Some thoughts on this: it is extensively mentioned in literature that both pictures yield same answers. Therefore, it should be possible to obtain (1) from (3) since (1) and (2) are equivalent. Assuming (3), the best I can get is that given an observable $A$, the operator
$$
\hat{C}:= \hat{A}_S\left(\frac{d\hat{U}}{dt} \hat{U}^\dagger – \frac{\hat{H}}{i\hbar}\right)
$$
must be skew-Hermitian.
Best Answer
The result can be proved in a general way using, well, math. In particular the theory of semigroups of linear operators on Banach spaces (I know that seems advanced and maybe not physical, but it is an elegant way of proving the result you seek ;-) ).
Define the Banach space $\mathscr{L}^1(\mathscr{H})$ of trace class operators over a separable Hilbert space the set of all bounded operators $u$ such that $\mathrm{Tr}\lvert u\rvert<\infty$. The fact that (3) holds for all observables (I will not discuss about domains here for the sake of simplicity) implies that it holds also for trace class operators that does not depend explicitly on time. In this case (3) is equivalent to the following Cauchy problem on the Banach space $\mathscr{L}^1(\mathscr{H})$: $$\frac{du(t)}{dt}=L u(t)\;, \; u(0)=x$$ where $\mathscr{L}^1(\mathscr{H})\ni x\equiv \hat{A}_0$ that does not depend on time, $u(t)\equiv \hat{A}_H(t)$ and $L$ is the linear operator that acts as $i[\hat{H},\cdot]$ (I am assuming $\hslash=1$). If $x\in D(L)$, i.e. $x$ such that $\mathrm{Tr}\lvert [\hat{H},x]\rvert <\infty$, the solution of the Cauchy problem above is unique, because $H$ is self-adjoint. We know by Stone's theorem an explicit solution, namely $$u(t)=\hat{U}(t)x \hat{U}^\dagger(t)\; ,$$ where $\hat{U}(t)=e^{-it\hat{H}}$ is the unitary group generated by $\hat{H}$, that satisfies (1) on $D(\hat{H})$. That solution is unique, so assuming (3) for all observables implies that the operator $\hat{U}$ you used to define Heisenberg picture operators has to be exactly the group generated by $\hat{H}$, i.e. satisfy (1).
Just for the sake of completeness: once you have solved the Cauchy problem for $x\in D(L)$, you can extend the solution to trace class operators or compact operators or bounded operators; also to unbounded operators (provided $\hat{U}(t)x \hat{U}^\dagger(t)$ makes sense on some dense domain). This is what is called a mild solution of the Cauchy problem, because we don't know a priori if we are allowed to take the derivative. However uniqueness is usually proved under general assumptions and for mild or even weak solutions, so I think it is quite safe to conclude that $\hat{U}(t)x \hat{U}^\dagger(t)$ is the unique solution of (3) in a suitable sense.