# Preserving continuity of periodic functions under fractional-integration-type transformations

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 $$\pm \infty$$?

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 $$1 (which might be possible), then Sobolev embedding would imply $$g \in C(\mathbb{T})$$ when merely $$a>0$$, but I am not quite sure how to do this.
• 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. – Goulifet Jul 18 '19 at 17:27