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In the paper https://projecteuclid.org/download/pdf_1/euclid.aos/1176348252 on page 1351 it is written: "Let $W_2^r$ be the Sobolev space of functions on $[0,1]$ for which f^{(r-1)} is absolutely continuouse and $\int_0^1 |f(x)|^2 dx$ is finite." where the $L^2$-norm is computed on the compact interval $[0,1]$.

How can I understand this sentence?

I thought Sobolev spaces can only be defined on open intervals.

wzell
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1 Answers1

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Actually, $W^{r,2}(0,1)$ is a space of equivalence classes of function, i.e, they are only defined a.e. and you can simply change the function on a set of measure zero but it identifies the same function in $W^{r,2}(0,1)$.

Now, note that $W^{1,1}(a,b)$ is embedded in the space $AC([a,b])$, e.g., see here. That means you can identify the function with its continuous representation on the closed interval, which makes the function in $W^{1,1}(a,b)$ really identifiable.

Since $f \in W^{r,2}(a,b)$, it means $f \in C^{r-1}([a,b])$.

Cahn
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  • Thank you very much for this helpful answer! If I understood you correctly, by your answer and the Rellich–Kondrachov theorem, the following can motivate the interpretation: $W^{r,2}(0,1)\hookrightarrow W^{1,1}(0,1)\hookrightarrow AC([0,1])$. – wzell Nov 30 '20 at 07:19
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    @wzell Yes absolutely. Also be aware that people are sloppy with it. When they write $u \in W^{1,1}(0,1)$, they mostly automatically choose the continuous representation, i.e the function $v \in A([0,1])$ such that $u=v$ almost everywhere. – Cahn Nov 30 '20 at 07:53