The instantaneous smoothing effect of the heat equation is the property that the solution to $$\begin{cases} \partial_t u= \Delta u, & t>0 \\ u(0, x)=f(x), & x\in \mathbb R^d,\end{cases}$$ is such that $u(t, \cdot)\in C^\infty$ for all $t>0$, regardless of the regularity of $f$; it is enough to suppose $f\in L^\infty(\mathbb R^d)$, for example. This applies to scalar or vector-valued maps; i.e., $u\colon [0, \infty)\times \mathbb R^d\to \mathbb R^d$.
Question. Suppose that $\omega\colon \mathbb R^d\to \mathbb S^{d-1}$ is a measurable map; in particular, $\omega$ can be undefined on a set of measure zero. Does there exist $\omega(t, x)$ such that
- $\omega(t, x)\in \mathbb S^{d-1}$ for all $t>0, x\in\mathbb R^d$;
- $\omega(t, \cdot)\in C^\infty(\mathbb R^d; \mathbb S^{d-1})$;
- $\displaystyle \lim_{t\downarrow 0} \int_{\mathbb R^d}\lvert \omega(t, x)-\omega(x)\rvert^2\, dx=0$ ?
Note. Other modes of convergence for point 3 are welcome. (Calvin points out in comments that $L^\infty$ convergence is unobtainable.)
This question appeared in the context of this answer, step 2. There, I tried to give a solution by considering $\omega$ as a map from $\mathbb R^d$ to $\mathbb R^{d}$ and then solving $$\tag{*}\begin{cases} \partial_t \eta = \Delta \eta, & t>0, \\ \eta(0, x)=\omega(x), & x\in \mathbb R^d, \end{cases} $$ obtaining a map $\eta\colon [0, \infty)\times \mathbb R^d\to \mathbb R^d$ which is indeed smooth for $t>0$$^{[1]}$, but which needs not satisfy $\eta(t, x)\in \mathbb S^{d-1}$. So, I have thought of defining $$ \omega(t,x):=\frac{\eta(t, x)}{\lvert \eta (t, x)\rvert}. $$
However, this is wrong. Nobody guarantees that $\eta(t, x)\ne 0$ for all $t>0$ and $x\in\mathbb R^d$.
The following example illustrates what can go wrong. Let $$ \omega(x)=\frac{x}{\lvert x \rvert}, \quad x\in \mathbb R^d.$$ Since $\omega(-x)=-\omega(x)$, the function $\eta$ must satisfy the same symmetry. But $\eta$ is continuous in $x$ for $t>0$, so $\eta(t, -x)=-\eta(t, x)$ forces $\eta(t,0)=0$.
My intuition is that the initial datum $\omega$ is "too symmetric". To obtain the desired smoothing effect, this symmetry must be broken; but the heat equation, which is isotropic, cannot do that.
A promising idea. Instead of (*), we should consider the general, anisotropic system of heat equations $$ \begin{cases} \partial_t \eta_j = \mathrm{div}(A(x)\nabla \eta_j) + b(x)\cdot \nabla \eta_j, & t>0, j=1,\ldots, d, \\ \eta(0, x)=\omega(x). \end{cases} $$
Here, the matrix-valued $A\colon \mathbb R^d\to \mathbb R^{d\times d}$ and $b\colon \mathbb R^d\to \mathbb R^d$ have to be determined, and they depend on $\omega$. The standard heat equation corresponds to $A$ being the identity matrix, and $b=0$.
[1] See, e.g., Evans "Partial differential equations", 2nd edition, Theorem 8, pag.59.