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4.How to prove that there is a continuous periodic function $f$ (with period $2\pi$), such that

$$\hat{f}(n) = \log(n)/(n^{3/2}).$$

$n\neq 0$ and $\hat{f}(0) = -1$. I know only the basics of fourier series.

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That is f cap . – ewew23 May 6 '12 at 8:34
I edited your post, please check if everythings ok. BTW, welcome to MathStackExchange. – draks ... May 6 '12 at 22:16

The key here is that $\sum_{n=0}^{\infty} |\hat{f}(n)| < \infty$.

Let $S_N(t) = \sum_{n=0}^N \hat{f}(n) e^{i w t}$. Each $S_N$ is clearly continuous.

For any $t$, the sequence $S_N$ is Cauchy, since $|S_N(t)-S_M(t)| \leq \sum_{k=\min(N,M)}^{\infty} |\hat{f}(n)| $. This defines a function $t \rightarrow \phi(t)$ pointwise, ie, $\phi(t) = \lim_{N \rightarrow \infty} S_N(t)$.

Then we have $|\phi(t) - S_N(t) | \leq \sum_{k=N+1}^{\infty} |\hat{f}(n)| $, so $\phi$ is the limit of a uniformly convergent sequence of continuous functions, hence $\phi$ is continuous.

Since each $S_N$ is $2 \pi$-periodic, we have $S_N(t+2\pi) = S_N(t)$. We have $$\phi(t+2\pi) = \lim_{N \rightarrow \infty} S_N(t+ 2\pi) = \lim_{N \rightarrow \infty} S_N(t) = \phi(t),$$ hence $\phi$ is $2 \pi$-periodic as well.

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