If $f\in L^1(\mathbb{T})$ is such that $\forall n<0, \hat{f}(n)=0$, does the Fourier series of $f$ converges in $L^1(\mathbb{T})$ norm to $f$

fourier seriesharmonic-analysis

Let $\mathbb{T}$ be the 1-torus. Using the uniform boundedness principle, from the fact that $L^1(\mathbb{T})$ is a homogeneous Banach space and from
$$\|D_N*\|_{1\rightarrow1}\ge\|D_N*F_n\|_{1}\rightarrow\|D_N\|_1, n\rightarrow\infty,$$
(where $D_N$ is the Dirichlet kernel and $F_n$ is the Fejer kernel), it follows that there exists $f\in L^1(\mathbb{T})$ such that the (symmetric) Fourier series of $f$ doesn't converge to $f$ in the $L^1(\mathbb{T})$ norm (see e.g. Katznelson – An introduction to harmonic analysis). What if we add the condition that all the Fourier coefficients of negative indexes are null? Since the space $$B:=\{f\in L^1(\mathbb{T})\ |\ \forall n<0, \hat{f}(n)=0 \}$$ is still a homogeneous Banach space but $F_n \notin B$, the previous argument doesn't apply, so I'm wondering if it is true or not that there exists a $f\in B$ such that the Fourier series of $f$ doesn't converge to $f$ in $B$-norm (i.e. in $L^1(\mathbb{T})$ norm). Any suggestions?

Best Answer

The article "Kehe Zhu - Duality of Bloch spaces and norm convergence of Taylor series" answers negatively to this question.

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