Laplacian Smoothing Irregular Initial Data Apparently for many parabolic and elliptic PDEs the (ir-)regularity of initial data does not have any significant impact on the regularity of (weak) solutions. 
Very often people when people talk about this fact they mention so-called smoothing property of Laplace operator term present in these equations. 
However, it seems to me that this expression is not a part of the formal PDE & analysis terminology, but rather something people tend to use informally. 

Is there an actual definition of the "smoothing property of second-order derivative term"?
What is the (physical?) mechanism behind this elusive property? 

I would be glad if someone could provide an intuitive and/or visual justification of this effect.
 A: Not every appearance of Laplacian in a PDE yields smoothing effect. For example, in the backwards heat equation $u_t=-\Delta u$ it has the opposite effect of reducing the smoothness of solution as time increases. And in the Scrödinger equation $iu_t =\Delta u$ the smoothness of $u$ neither improves nor gets worse. 
One way of seeing this is to imagine $u$ decomposed into harmonics, $$u(x,t)=\sum_n A_n(t)e^{i \omega_n x}$$
The smoothness of $u$ with respect to $x$ is measured by how quickly $|A_n(t)|$ decays as $n\to\infty$. (E.g.,  $|A_n(t)|=O(n^{-k-1-\epsilon})$ gives you $k$ continuous derivatives, because the series remains uniformly convergent after being differentiated $k$ times.)
Example. For the heat equation  $u_t=\Delta u$ we have
$$ \sum_n A_n'(t)e^{i \omega_n x} = - \sum_n \omega_n^2 A_n (t)e^{i \omega_n x}$$
hence $A_n(t) = A_n(0)\exp(-\omega_n^2 t)$. Since $|\omega_n|$ is typically $\Omega(n^\alpha)$ for some $\alpha>0$, it follows that for every $t>0$ the decay of $A_n(t)$ is very  fast (superexponential), which implies $u$ is infinitely differentiable in $x$.
Repeating the above for $iu_t =\Delta u$ you will see that no decay of $A_n(t)$ takes place, and therefore no gain of smoothness is to be expected. 

Summary:


*

*Laplacian is a diagonal operator in the basis of complex exponentials, which  gives large negative coefficients to  high-frequency harmonics.

*Thus, evolving a function in the direction of Laplacian reduces said  high-frequency harmonics, smoothing the functions.

