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.
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.