# Example of a function whose Fourier Series falls short to merge at One factor

Can one consider an instance of a continual $2\pi$ routine function whose Fourier collection falls short to merge on $\mathbb{R}$.

I referred this in the wikipedia web page yet no make use : It could be intriguing to keep in mind that Jean - Pierre Kahane and also Yitzhak Katznelson confirmed that for any kind of offered set E of action absolutely no, there exists a continual function ƒ such that the Fourier collection of ƒ falls short to merge on any kind of factor of E.

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2019-05-09 11:26:43
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Answers: 2

All, please see this instance.

Allow $G_{n}$ represent the group of this $2n$ numbers, $$\frac{1}{2n-1},\frac{1}{2n-3},...,\frac{1}{3},1,-1,-\frac{1}{3},\cdots,-\frac{1}{2n-1}$$

We take a purely raising series of favorable integers ${\lambda_n}$ and also take into consideration the teams $G_{\lambda_1},G_{\lambda_2},\cdots,$. We increase each variety of the team $G_{\lambda_n}$ by $n^{-2}$ and also get the series $$\frac{1}{1^{2}(2\lambda_{1}-1)}, \cdots,-\frac{1}{1^{2}(2\lambda_{1}-1)}, \frac{1}{2^{2}(2\lambda_{2}-1)},...,-\frac{1}{2^2(2\lambda_{2}-1)},....,$$

claim $\alpha_{1},\alpha_{2},\cdots$. Our purpose is to show that $$\sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$. is the fourier collection of a continual function. We organize the terms in the list below means $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2\lambda_{1}+1}^{2\lambda_{1}+2\lambda_{2}} \alpha_{n}\cos{nx} + \sum\limits_{n=2\lambda_{1}+2\lambda_{2}+1}^{2\lambda_{1}+2\lambda_{2}+2\lambda_{3}} \alpha_{n} \cos{nx}\cdots$$

The last collection can be created as $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$

where $$\phi(n,r,x)= \frac{\cos{(r+1)x}}{2n-1} + \frac{\cos{(r+2)x}}{2n-3} + \cdots + \frac{\cos{(r+n)x}}{1} - \frac{\cos{(r+n+1)x}}{1} - \cdots - \frac{\cos{(r+2n)x}}{2n-1}$$

Now one can show that there is a constant $M$ (independent of $n,r$ and also $x$) such that $|\phi(n,r,x)|\leq M$. From this it adheres to that the organized collection $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$. merges definitely on $\mathbb{R}$, claim to $f(x)$, and also $f$ is continual on $\mathbb{R}$. It is additionally very easy to examine that $$f(x) \sim \sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$

We will ultimately show that ${\lambda_n}$ can be picked to make sure that the above collection deviates at absolutely no, that is $S_{n} = \alpha_{1} + \alpha_{2} + \cdots + \alpha_{n}$ deviates to infinity.

Given that $$S_{2\lambda_{1}+2\lambda_{2}+ \cdots + 2 \lambda_{n-1} + \lambda_{n}} = \frac{1}{n^2} \Bigl( \frac{1}{2\lambda_{n}-1} + \cdots + \frac{1}{3} + 1 \Bigr)$$ acts as $\frac{\ln{\lambda_{n}}}{{2n^{2}}}$ as $n \to \infty$, it suffices to take $\lambda_{n}=n^{n^2}$. After that the fourier collection does not merge to $f$ at $x=2k\pi, \ k\in \mathbb{Z}$.

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2019-05-11 23:28:29
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No. Any kind of continual function remains in $L^2$. The Carleson-Hunt theorem states that an $L^2$ function's Fourier collection merges virtually anywhere to the function.

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2019-05-10 08:34:22
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