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Let $\mathbb{T} = [0,2\pi]$. It is is well-known that $ L^2(\mathbb{T})$ and $ l_2(\mathbb{Z})$ are isometrically isomorphic via the Fourier transform and the Fourier series is its inverse. Parseval's identity states that $\lVert f \rVert_2^2 = \sum_{n \in \mathbb{Z}} |\hat{f}(n)|^2 $

It is also well-known that the Fréchet spaces $C^\infty(\mathbb{T})$ and $s = \{(c_n)_n \in \mathbb{C}^{\mathbb{Z}} \mid \forall k \in \mathbb{N} : \sum_{n \in \mathbb{Z}} |c_n|n^k < \infty\}$ are isomorphic as Fréchet spaces, again via the Fourier transform and Fourier series.

Now, $C^\infty(\mathbb{T}) \subset L^2(\mathbb{T})$ and $|\widehat{f^{(k)}}(n)|= n^{k}|\hat{f}(n)|$, so all $\sqrt{\sum_{n \in \mathbb{Z}} |\hat{f}(n)|^2n^{2k}}$ are bounded when $f \in C^\infty(\mathbb{T})$.

So, consider $\tilde{s} = \{(c_n)_n \in \mathbb{C}^{\mathbb{Z}} \mid \forall k \in \mathbb{N} : \sqrt{\sum_{n \in \mathbb{Z}} |c_n|^2n^{2k}} < \infty\}$. Are $s$ and $\tilde{s}$ isomorphic as Fréchet spaces?

I can only show that they agree as sets: If $\sum_{n \in \mathbb{Z}} |c_n|$ converges, then $\sum_{n \in \mathbb{Z}} |c_n|^2$ as well, so $s \subset \tilde{s}$. For the converse, note that $\sum_{n \in \mathbb{Z}}n^{-2}$ converges. In general, if both $\sum_{n \in \mathbb{Z}} |a_n|^2$ and $\sum_{n \in \mathbb{Z}} |c_n|^2$ converge, then $\sum_{n \in \mathbb{Z}} |a_n c_n|$ converges as well, so $\sum_{n \in \mathbb{Z}} |c_n|^2n^{2k} < \infty$ implies that $\sum_{n \in \mathbb{Z}} |c_n|n^{k-1} < \infty$, hence $\tilde{s} \subset s$.

mixotrov
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1 Answers1

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Yes, on the circle, the $L^2$ version of Sobolev spaces admits a Sobolev lemma/inequality, simply from Cauchy-Schwarz-Bunyakowsky, that $H^s\subset C^k$ for all $s>k+{1\over 2}$. Thus, the projective limits of the $H^s$'s and the $C^k$'s are isomorphic (as topological vector spaces). The limitand-wise shift by ${1\over 2}$ disappears in the limit.

paul garrett
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