Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f$ is absolutely continuous and the Fourier coefficients of $f$ satisfy $$ \hat f(n)= \frac{a_n sgn(n)}{n} \ge 0 $$
where the $a_n$ are postive, even, and decreasing to zero as $|n| \to \infty$, so that the Fourier coefficients of $f$ are even and positive. Show $$ \sum_{n=-\infty}^{\infty} \hat f(n)<\infty $$

This was answered in this post. The post says:

A standard theorem says that the Fourier series of an absolutely continuous function converges to it uniformly so taking $x=0$ you get the result.

I suspect the theorem holds almost everywhere, so I don't see why taking $x=0$ is justified.
If possible, can you point me to a proof for this theorem?
Or explain why the choice is possible.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Since $f$ is absolutely continuous, we can write $$ f(x)=f(0)+\int_0^x g(t)\,\mathrm{d}t $$ where $g\in L^1$.

Since $f$ is even, it's Fourier coefficients are real. Since $\hat{f}(n)\downarrow0$, we have $\hat{f}(n)\ge0$

Therefore, $$ \begin{align} \sum_{n\in\mathbb{Z}}\hat{f}(n) &=f(0)+\sum_{n\in\mathbb{Z}}\int_0^1\int_0^x g(t)\,\mathrm{d}t\,e^{-2\pi inx}\,\mathrm{d}x\\ &=f(0)+\sum_{n\in\mathbb{Z}}\int_0^1\int_t^1 g(t)\,e^{-2\pi inx}\,\mathrm{d}x\,\mathrm{d}t\\ &=f(0)+\sum_{n\in\mathbb{Z}}\int_0^1 g(t)\frac{e^{-2\pi int}-1}{2\pi in}\,\mathrm{d}t\\ &=f(0)-\sum_{n\in\mathbb{Z}}\int_0^1 g(t)\frac{\sin(2\pi nt)}{2\pi n}\,\mathrm{d}t\\ &=f(0)+\int_{1/2}^1g(t)\,\mathrm{d}t-\int_0^{1/2}g(t)\,\mathrm{d}t \end{align} $$ where $\displaystyle\sum_{n\in\mathbb{Z}}$ is interpreted as $\displaystyle\lim_{N\to\infty}\sum_{n=-N}^N$

Therefore, $$ 0\le\sum_{n\in\mathbb{Z}}\hat{f}(n)\le f(0)+\|g\|_{L^1} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.