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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 $S_3$.

user50618
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  • What do you mean by "amount of smoothness"? – tomasz Aug 07 '14 at 22:22
  • The definition of weak derivative is local in nature, whilst the integrability is global. It is the local weak derivative that gives us smoothness, a local property. – jxnh Aug 07 '14 at 22:23
  • @tomasz I will add some notes to make the question clear. – user50618 Aug 07 '14 at 22:25
  • By a weak derivative do you mean a locally integrable function with certain relation to $u$? Sometimes this term is used for distributional derivatives, which can be any distribution whatsoever. The answer depends on this distinction. –  Aug 07 '14 at 22:43
  • @900sit-upsaday I mean by the weak derivative of a locally integrable function $u$ the function $v$ such that $\int u \phi'=-\int v \phi$, for all $\phi$ smooth with compact support. – user50618 Aug 07 '14 at 22:49

1 Answers1

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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.

  1. The fact that $f$ has a weak derivative makes it locally absolutely continuous on almost every line.
  2. 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.
  3. 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.
  4. 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 $p<n-\alpha$.

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