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In the first answer to this question the users states that if we define a norm $$\|f\|_{k,p} = \left( \sum_{|\alpha| \leq k} \|D^{\alpha} f\|_p^p \right)^{1/p},$$ and write $$\widetilde{C}^k(\Omega) = \{f \in C^k(\Omega) : \|f\|_{k,p} < \infty\},$$ then $W^{k,p}(\Omega)$ is the completion of $(\widetilde{C}^k(\Omega), \|\cdot\|_{k,p})$ as long as $p \in [1, \infty)$.

Why is $W^{k,p}(\Omega)$ the completion of $(\widetilde{C}^k(\Omega), \|\cdot\|_{k,p})$? I don't see the difference between them. What functions are missing from $(\widetilde{C}^k(\Omega), \|\cdot\|_{k,p})$ that are in $W^{k,p}(\Omega)$?

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csss
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    There is a big difference between them. In general, the functions in $W^{k,p}(\Omega)$ are not $k$ continuously differentiable. Consider for simplicity the case $k=0$: the space $L^p$ is much bigger than the space of (integrable) continuous functions. – guestDiego Jul 20 '16 at 15:07
  • So is the poster wrong? It seems you are saying that as the functions in $W^{k,p}(\Omega)$ are not $k$ continuously differentiable then we can't have $(\tilde{C}^k(\Omega), \Vert \cdot \Vert_{k,p}) \subseteq W^{k,p}(\Omega)$? So how could $W^{k,p}(\Omega)$ be its completion in that case? – csss Jul 20 '16 at 15:15
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    I am absolutely not saying that! I was only pointing out that the statement "I don't see the difference between them." is unmotivated: the difference exists. Therefore the completion is not trivial: it gives a bigger set. – guestDiego Jul 20 '16 at 16:48

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The space $\tilde{C}^k$ is not complete with respect to the norm $||.||_{k,p}$ -- assuming the norm $||.||_p$ is the $L^p$ norm $$||f||_p^p := \int_\Omega |f|^p dx$$ The reason is that convergence in this kind of norm does not preserve differentiability.

Edit: provide simple example to reply to a comment: look, for example, at the functions $$f_n(x) = |x|^{1+\frac{1}{n}}$$ defined on the interval $(-1,1)$. These funktions are (once) continously differentible, and it is not difficult to see that they all belong to what you called $\tilde{C}^1((-1,1))$ It's also not difficult to see that they converge (pointwise and uniformly, first of all) to the function $f(x) = |x|$ They do form a Cauchy Sequence in that space with respect to the $||.||_{1,p}$-norm, too (for each $p$, which I leave as an exercise to you to show), but the limit (which can only be $f$) does not belong to that space. So convergence in that norm does not preserve differentiability.

It is in fact known that $W^{1,1} $ is the space of absolutely continuous functions, which is much larger thant the space of continuous differentiable funcions.

Thomas
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  • The norm is defined as $$|f|{k,p} = \left( \sum{|\alpha| \leq k} |D^{\alpha} f|_p^p \right)^{1/p},$$. I don't see how that doesn't preserve differentiability? In fact it looks to be exactly the same as the Sobolev norm to me? Hence why I say that the two spaces seem to be equal? – csss Jul 21 '16 at 08:00
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    @csss I added an example to my answer. – Thomas Jul 21 '16 at 16:00
  • @csss (and the definition of $||.||_{k,p}$ involves the norms $||.||_p $ which you did not specify. The remark in my answer just makes clear what I assumed about the definition of this latter norm) – Thomas Jul 21 '16 at 16:04
  • Thanks for providing the example! – csss Jul 22 '16 at 10:20