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I apologize for repeating the same question from ME, but it seems more subtle than I expected.

Let me fix the notations here first: \begin{equation} C^\infty_c(0,1):= \{ f : (0,1) \to \mathbb{C} \mid f \text{ is smooth and compactly supported} \} \end{equation}

\begin{equation} C^\infty(S^1):= \{ f : S^1 \to \mathbb{C} \mid f \text{ is smooth } \} \end{equation} where $S^1 := \mathbb{R}/ \bigl(2\pi\mathbb{Z} \bigr)$. Note that we identify each $f \in C^\infty(S^1)$ with a unique smooth function $F : \mathbb{R} \to \mathbb{C}$ such that $F(n)=F(0)$ for all $n \in \mathbb{Z}$. With this convention, we may define the integral on $S^1$ like \begin{equation} \int_{S^1} f := \int_0^1 F(x) dx \end{equation} and somehow extend this formula continuously to define $L^p(S^1)$ and so on. Also, $C^\infty_c(0,1) \subset C^\infty(S^1)$ through such identification.

Now, let us fix any $k \in \mathbb{N} \cup \{0\}$ and $p \in [1,\infty]$ and think of the two Sobolev spaces: $W^{k,p}\bigl([0,1] \bigr)$ and $W^{k,p}(S^1)$.

Following the answer here, weak derivatives for functions in $W^{k,p}\bigl([0,1] \bigr)$ can be defined just by using test functions in $C^\infty_c(0,1)$.

Now, I run into two confusions:

How does one actually define $W^{k,p}(S^1)$? Do we use $C^\infty(S^1)$ or $C^\infty_c(0,1)$ to define weak derivatives?

Can we somehow identify $W^{k,p}\bigl([0,1] \bigr)$ with $W^{k,p}(S^1)$?

For the second question, it seems trivial for $k=0$. Also, $C^\infty_c(0,1)$ can be identified as a dense in $C^\infty(S^1)$, so that I feel positive about making some sort of identification.

However, I cannot find a precise mapping between the two spaces that would qualify as "identification". Moreover, de Rham cohomology and the function $f(x)=x$ mentioned in the original ME post confuse me still..

I tried to avoid topological concepts like covering space or lifting and keep everything explicitly written down. Maybe I need those concepts for a complete answer? Could anyone please help me?

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    $\begingroup$ The space $C^\infty_c(0,1)$ is obviously not dense in $C^\infty(S^1)$ . $\endgroup$
    – Oleg Eroshkin
    Commented Oct 8 at 18:20
  • $\begingroup$ @OlegEroshkin OK, I didn't think carefully about values at the endpoints of $(0,1)$. Thanks for pointing this out. $\endgroup$
    – Isaac
    Commented Oct 8 at 21:18

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Well...the function $f(x)=x$ on $[0,1]$ indeed is a counterexample showing that $W^{k,p}(S^1) \subsetneq W^{k,p}\bigl([0,1] \bigr)$.

Let us "artificially modfiy" such $f$ to a periodic function $F$ by setting $$ F(x)=x \text{ for } 0 \leq x <1, F(1)=0 \text{ and }F(x+n)=F(x) \text{ for }x \in [0,1] \text{ and }n \in \mathbb{Z} $$ Then, $F$ has a jump discontinuity at $x=1$ if we choose a representation of $S^1$ as $[1/2, 3/2]$, which results in the Dirac-delta function as a weak derivative.

I am not sure how to make all these perfectly rigorous, but it is at least certain that $f \in W^{k,p}\bigl([0,1] \bigr)$ while $F \notin W^{k,p}(S^1)$ for any choice of $k \in \mathbb{N}$ and $p \in [1,\infty]$.

I believe that we can somehow find a connection between this counterexample and de Rham cohomology as well.

I deeply appreciate any comment or feedback.

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    $\begingroup$ One connection is that you can obtain $F$ by integrating the closed one-form $d\theta$ from a base point identified with $0$, and the fact the total integral is not zero means that the one-form is not exact. $\endgroup$
    – Neal
    Commented Oct 8 at 16:29
  • $\begingroup$ What exactly is the problem of $d \theta$ not being exact? I am sorry that I have forgotten most of the details I learned before. $\endgroup$
    – Isaac
    Commented Oct 8 at 16:33
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    $\begingroup$ It is not a problem, just a connection with de Rham cohomology. $\endgroup$
    – Neal
    Commented Oct 8 at 16:44
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    $\begingroup$ Sorry Isaac, but you need to review some basic things. Every exact form is closed but not every closed is exact. That failure is measured by de Rham cohomology. The form $d\theta$ is closed but not exact. That's the relation. $\endgroup$
    – Oleg Eroshkin
    Commented Oct 8 at 19:28
  • $\begingroup$ @OlegEroshkin Well, I am aware of the definitions. And de Rham cohomology doesn't seem "critically" relevant for my question in the first place, as it only pertains to smooth functions. As you can see in my answer above, it is more about appearance of jump discontinuity (and the Dirac delta distribution) that causes the problem.. $\endgroup$
    – Isaac
    Commented Oct 8 at 21:04

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