My understanding is, given a stochastic optimal control problem, one can show that the optimal cost satisfies a Hamilton-Jacobi-Bellman PDE. However, sometimes this PDE has no strong solution, and the profound theory of viscosity solution was invented to make sense of a "solution" in this situation. My question is:
Assuming under reasonable assumptions one can still figure out what the control should be from the viscosity solution, will the control be a function, as opposed to some weak notion like a distribution?
I will use a concrete example to illustrate.
Let $\mathcal{B}(t)$ be a set of controls of the form $b\equiv(b_{s})_{s\in[t,T]}$ such that $b_{s}\in B$. Consider the process defined by $$ X_{s}^{t,x;b}=x+\int_{t}^{T}\mu(s,X_{s}^{t,x;b},b_{s})ds+\int_{t}^{T}\sigma(s,X_{s}^{t,x;b},b_{s})dW_{s}. $$ Now, define the optimal control problem through the value function $V$ given by $$ V(t,x)=\sup_{b\in\mathcal{B}(t)}\mathbb{E}\left[g(X_{T}^{t,x;b})\right]. $$ As you remarked, subject to some technical conditions on the quantities $\mathcal{B},B,\mu,\sigma,g$, it is well-known $V$ satisfies (in the viscosity sense) the Hamilton-Jacobi-Bellman PDE \begin{align*} -V_{t}-\sup_{b\in B}\left\{ \operatorname{tr}(\sigma(t,x,b)\sigma(t,x,b)^{\intercal}D_{x}^{2}V(t,x))+\mu(t,x,b)\cdot D_{x}V(t,x)\right\} & =0\text{ if }t<T\\ V(t,x)-g(x) & =0\text{ if }t=T. \end{align*}
The important distinction here is that the "control" $b$ appearing in the PDE is a scalar, while the control $b$ appearing in the original optimal control problem (involving the supremum and expectation) is a control in the actual sense of the term. This control is not an ordinary function: it is a member of $\mathcal{B}(t)$, which is usually taken to be a subset of the progressively measurable processes on $[t,T]$ (w.r.t., in this case, the filtration generated by the Brownian motion $W$).
As you remark, it is sometimes possible to use properties of $V$ along with the fact that it is a viscosity solution to create an optimal control $b^{*}\in\mathcal{B}(t)$. However, this is not generally possible for all optimal control problems.