How do the existence and $L^p$ integrability of weak derivative affect the smoothness of Sobolev functions? In the definition of Sobolev spaces, let us say the space $W^{1,p}(\Omega)$, where $\Omega$ is an open subset of $\mathbb{R}^n$. What contributes more to the smoothness of a function $f$:
1-The fact that $f$ has a weak derivative (i.e. a locally integrable function $v$ such that $\int_{\Omega}u \phi'=-\int_{\Omega}v \phi $ for all $\phi\in C^{\infty}_c(\Omega)$)
Or
2-The fact that the derivative belongs to $L^p(\Omega).$
In other words, if a function has a weak derivative, then it has some sort of smoothness. Now if in addition we assume that this weak derivative belongs to $L^p(\Omega)$, do we gain a huge additional amount of smoothness ?
In other words consider those three spaces:
$$S_1=L^p(\Omega),$$
$$S_2=\{u\in L^p(\Omega),\ D^{\alpha}u \text{ exists for }|\alpha|\leq1 \}$$
and
$$S_3=\{u\in L^p(\Omega),\ D^{\alpha}u \text{ exists and }D^{\alpha}u\in L^p(\Omega) \text{ for } |\alpha|\leq1 \}.$$
There's a big difference between $S_1$ and $S_2$. But I don't know there's also a big difference between $S_2$ and $S_3$.
I know that this question is to soft and maybe you will ask: "what do you mean by bi difference", and I am not sure if we can describe this difference by cardinality. I would appreciate examples of functions which belong to $S_2$ but not $S_3$.
 A: I think smoothness is the wrong term to focus on; the difference concerns the continuity of $f$. To get first-order classical derivative of $f$, we would need $f\in W^{2,p}$ with $p>n$; compare with item 4 below. 


*

*The fact that $f$ has a weak derivative makes it locally absolutely continuous on almost every line. 

*If the derivative is also in $L^1(\Omega)$, we don't gain anything inside the domain (since the weak derivative is locally integrable by definition) but we can push continuity up to the boundary. That is, if $\Omega$ has smooth boundary (imagine a disk), we can assign  some boundary values to $f$ by taking limits along radial segments. The limit will exist for almost every radius. 

*The increase of $p$ from $1$ to $p=n$ brings little visible change. However, there is a slight gradual improvement: $W^{1,p}$ functions have a $p$-quasicontinuous representative, which means a representative continuous on a set with very small complement (not just measure zero, but capacity zero). This is a bit technical, so I refer to page 160 of Measure Theory and Fine Properties of Functions by 
Evans and   Gariepy. 

*Once $p>n$, we gain continuity of $f$: more precisely, it has a Hölder continuous representative. See Morrey's inequality. The   Hölder is $1-n/p$, and in the limit $p\to\infty$ we approach Lipschitz continuity. Although the geometry of domain may  get in the way of direct comparison of Lipschitz and Sobolev classes.


So... I'd say the improvement from an arbitrary $L^p$ function to one with a weak derivative is more dramatic than from there to $W^{1,p}$. 
You also ask for examples. Try $|x|^{-\alpha}$ on the unit ball of $\mathbb R^n$. When $\alpha<n-1$, this function has integrable weak derivative. The derivative is in $L^p$ only for $p<n-\alpha$.
