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I have the following question when I was reading a lecture note, (which can be found here: https://www.math.ucdavis.edu/~hunter/book/ch7.pdf)

If $f$ is a function on the 1-dimension torus, which is in the form of $\sum_{n\in\mathbb{Z}}c_n e^{inx}$ with $c_n\in\mathbb{R}$ and $\sum_{n\in \mathbb{Z}}n^2 c_n^2<\infty.$ Is the weak derivative of $f$ always $\sum_{n\in\mathbb{Z}}inc_n e^{inx}$?

From the lecture note, it defines the Sobolev space on the torus to be $H^1(\mathbb{T})=\left\{f\in L^2: f(x)=\sum_{n\in\mathbb{Z}}c_n e^{inx}, \sum_{n\in \mathbb{Z}}n^2 c_n^2<\infty\right\}$. And it also "defines" the weak derivative of a function in the space to be $\sum_{n\in\mathbb{Z}}inc_n e^{inx}$. However I want to make sure it fits with the definition of weak derivative I learned before, so I was trying to show that $$ \int_{\mathbb{T}}(\sum_{n\in\mathbb{Z}}inc_n e^{inx})\phi=-\int_{\mathbb{T}}(\sum_{n\in\mathbb{Z}}c_n e^{inx})\phi', $$ by integration by parts and using Dominated Convergence theorem to pass the limit, but I couldn't find the integrable function to bound $|\sum_{n\in\mathbb{Z}}c_n e^{inx}|$ and $|\sum_{n\in\mathbb{Z}}inc_n e^{inx}|$ almost everywhere under the assumption that $\sum_{n\in \mathbb{Z}}n^2 c_n^2<\infty.$ Could I ask is it the right way to see it? If it's not, how can we explain that the derivative of the infinite sum can be done term by term? Thanks for any help.

suomynonA
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Vera
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    If $g(x) = \sum_n c_n in e^{inx} \in L^2(\mathbb{T})$, then $\int_a^x g(t)dt =C+ \sum_n c_n e^{inx}$ – reuns Jan 09 '17 at 05:07
  • @user1952009: I understand that on torus $L^2\subset L^1$, so it's legal to write $\int_a^x g(t)dt$ if we have $g(x)\in L^2$. But could you give me more hints about why it's equal to $C+\sum_n c_n e^{inx}?$ It looks to me that we can integrate it term by term and I still don't know why we can do it... – Vera Jan 09 '17 at 17:14
  • As usual in functional analysis : Let $g_N(x) = \sum_{n=-N}^N c_n in e^{inx}$. It converges to $g(x)$ in $L^1$ and so $\int_a ^x g_N(t)dt \to \int_a ^x g(t)dt $ – reuns Jan 09 '17 at 17:15
  • @user1952009: Thank you for this quick reply. I got it now. Just another question. It seems to me that this argument still works in higher dimension? I just thought $\partial_i (\sum_n c_n e^{in\cdot x})=\sum_n c_n in_i e^{in\cdot x}$ should still be true in higher dimension. Is that right? – Vera Jan 09 '17 at 17:34

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