A Fourier series that does not converge in $L^1$

Question

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 $\int _{\mathbb{T}}|\sum _{p}^{q}\hat{f}(n) cos(nt)| dt$

Answer 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 $L^1$.