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Assume that $f$ is a continuous and periodic function over $\mathbb{T}=[0,1)$, and denote by $f_n$, $n \in \mathbb{Z}$, its Fourier series.

Let $(a_n)_{n\in \mathbb{Z}}$ be a sequence such that $n^a a_n \rightarrow c$ in $\pm \infty$ for some $a > 0$.

Is it always true that the function $g$ with Fourier coefficients $g_n = a_n f_n$ is continuous?

If the answer is yes, then what if $a=0$, that is, $a_n \rightarrow c$ in $\pm \infty$?

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I can provide an affirmative answer if $a>1/2$. Since $f\in C(\mathbb{T})\subset L^2({\mathbb{T}})$ and $(n^a a_n)$ is a bounded sequence by assumption, an application of Parseval's Theorem implies \begin{align*} ||\partial_t^a g||_2 = ||n^a g_n||_{\ell^2} = ||n^a a_n f_n||_{\ell^2} \leq C ||f_n||_{\ell^2} = C ||f||_2<\infty. \end{align*} It follows that $g$ belongs to the Sobolev space $H^a(\mathbb{T})$, which embeds into $C(\mathbb{T})$ wenn $a>1/2$ by Sobolev's Embedding Theorem.

If you can somehow show $g\in W^a_p(\mathbb{T})$ for all $1<p<\infty$ (which might be possible), then Sobolev embedding would imply $g \in C(\mathbb{T})$ when merely $a>0$, but I am not quite sure how to do this.

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  • $\begingroup$ I agree for $a > 1/2$. The idea of working with Sobolev spaces is interesting and deserves to be tried. I will think about it, thanks. $\endgroup$ – Goulifet Jul 18 '19 at 17:27

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