I'm reading Katznelson's book "Harmonic Analysis" and there is an exercise that I can't solve :



> 
> Show that if the sequence $\left\{ N\_j \right\} $ tends to infinity
>  fast enough, then the Fourier series of the function $$ f(t)=\sum\_{j>0}2^{-j }K\_{N\_j}$$ does not converge in $L^1(\mathbb{T})$.
> 
> 
> 


$K\_{N\_j}$ denotes the Féjer Kernel.


In a first time I tried to compute the Fourier coefficients of $f$. If $N\_l<n\leq N\_{l+1}$ then $\hat{f}(n)=\sum \_{j> l}2^{-j}(1-\frac{|n|}{N\_j+1})=\hat{f}(-n)$. 
So I have tried to show that $(S\_n(f))$ is not a Cauchy sequence with the $L^1$ norm but I failed to find a good enough lower bound of $\int \_{\mathbb{T}}|\sum \_{p}^{q}\hat{f}(n) cos(nt)| dt$



A Fourier series that does not converge in $L^1$
Let us give sufficient conditions on the growth rate of $N\_l$ for the Fourier series to diverge. 


We consider $p = N\_l + 1$ and $q = \frac1M N\_{l+1}$. For $p < |n| \leq q$ we have that $$ \hat{f}(n) = \sum\_{j > l} 2^{-j}(1 - \frac{|n|}{N\_j + 1}) \geq (1 - \frac{1}{M}) \sum\_{j > l} 2^{-j} = (1 - \frac{1}M) \cdot 2^{-l}$$
we also have
$$ \hat{f}(n) - (1 - \frac1M) \cdot 2^{-l} \leq \frac1M \cdot 2^{-l}.$$


So 


$$ \left| \sum\_{p}^q \hat{f}(n) \cos(nt)\right| \geq (1 - \frac1M)2^{-l} \left|\sum\_{p}^{q} \cos(nt)\right| - \frac{q-p}{M} 2^{-l} = (1 - \frac1M)2^{-l} |D\_q(t) - D\_p(t)| - \frac{q-p}{M} 2^{-l}.$$


Now you can use the divergence of the Dirichlet kernel (this follows from the fact that the $L^1$ norm of the Dirichlet kernel grows like $\log n$). 


1. Given $N\_l$, we choose $q$ such that $\| D\_{q} - D\_{N\_{l}+1}\|\_{L^1} \geq 2^{l}$.
2. Next we choose $M$ such that $M > 2 (q - N\_{l} - 1)$.
3. We choose $N\_{l+1} > M q$.
4. repeat with $l \mapsto l+1$.


Then we have 


$$ \|S\_q(f) - S\_p(f)\|\_{L^1} \geq \frac12 ( 1 - 2^{-l}) $$


for the $p,q$ associated to each $l$, showing that the Fourier series cannot converge in $L^1$. 


