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I want to compute $K_N * K_N$, where $K_N$ is the Nth Fejér Kernel $$K_N(t) = \frac{1}{N} \sum_{n=0}^{N-1} \sum_{k=-n}^n e^{ikt}.$$

So I tried to determine $\widehat{K_N * K_N} = \hat{K}_N\hat{K}_N,$ and since $K_N =\frac{1}{N} \sum_{n=0}^{N-1} D_n(t)$, where $D_n(t)$ is the Dirichlet kernel, I thought $\hat{K}_N(n)$ could be written as $\frac{1}{N}(\hat{D_0}(n) + \dots + \hat{D}_{N-1}(n))$, where $\hat{D}_n(n) = 1$ for $|n| \leq N$, and zero otherwise.

But this gives me $\hat{K}_N = 1$ and that is apparently wrong since the solution says that $\hat{K}_N(n)\hat{K}_N(n) = (1 - \frac{|n|}{N})^2$ if $|n| \leq N$.

(Note that I am working with convolutions of $2\pi$-periodic functions.)

So I am wondering 1) Why my way of solving this gives the wrong answer, 2) how to actually solve it to get the right answer.

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The Fejér kernel can be written as $$ F_n(x) = \sum_{|j|\leq n}\left(1-\frac{|j|}{n}\right) e^{ijx} $$ hence if $f$ is a $2\pi$-periodic function in $L^2(-\pi,\pi)$ we have: $$ (f*F_n)(x) = \sum_{|j|\leq n}\left(1-\frac{|j|}{n}\right)\widehat{f}_{\!j}\, e^{ijx} $$ by the convolution property you mention, $\widehat{f*g}=\widehat{f}\cdot\widehat{g}$. By taking $f=F_n$ in the above identity $$ (F_n*F_n)(x) = \sum_{|j|\leq n}\left(1-\frac{|j|}{n}\right)^2 e^{ijx} $$ readily follows.

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    $\begingroup$ The way you write the Fejér kernel in the first row is new to me, and I can't seem to find out a way to show that it, in fact, is possible to write it that way. Would you mind explaining it to me? $\endgroup$ Commented May 28, 2017 at 20:11
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    $\begingroup$ @Henry.M: that is pretty straightfoward: for a fixed $k$, how many times does the term $e^{ikx}$ appear in the sum $$\sum_{n=0}^{N-1}\sum_{k=-n}^{n}e^{ikx}$$ ? $\endgroup$ Commented May 28, 2017 at 20:14
  • $\begingroup$ I want to say N-k times... $\endgroup$ Commented May 28, 2017 at 20:52
  • $\begingroup$ @Henry.M: then you can easily check why my first representation is correct. $\endgroup$ Commented May 28, 2017 at 21:04

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