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Fix $0<p<1$. Let $D$ be the complex unit disk and let $\mathbb{T}$ be the torus. Let ${\frak{H}}^p(\mathbb{T})$ denotes the $L^p(\mathbb{T})$ closure of the holomorphic trigonometric polynomials, let $H^p(D)$ be the Hardy space on the disk and let $S'(\mathbb{T})$ denotes the space of distributions on the torus.

It is known (see for example Duren - Theory of $H^p$ Spaces, theorem 3.3) that the map that sends $f\in H^p(D)$ to its boundary values is a isometric isomorphism between $H^p(D)$ and ${\frak{H}}^p(\mathbb{T})$.

On the other hand, thanks to a theorem of Hardy and Littlewood (see for example Duren - Theory of $H^p$ spaces, theorem 6.2) it is known that for each $f\in H^p(D)$, $f_r:t\mapsto f(re^{it})$ converges distributionally for $r\rightarrow1^-$ to a distribution $F$ on $\mathbb{T}$, and that the map $$H^p(D)\rightarrow S'(\mathbb{T}), f\mapsto F$$ is linear and continuous.

So, by composing the inverse of the first map with the second map, we get a continuous linear map $${\frak{H}}^p(\mathbb{T}) \rightarrow S'(\mathbb{T}) $$ such that its restriction to ${\frak{H}}^p(\mathbb{T})\cap L^1(\mathbb{T})$ is the canonical embedding in $S'(\mathbb{T})$ (the one obtained by integral pairing) and this map is obviously the only continuous one that has this property thanks to the density of ${\frak{H}}^p(\mathbb{T})\cap L^1(\mathbb{T})$ in ${\frak{H}}^p(\mathbb{T})$.

Composing this map with Fourier transform, we are able to define Fourier coefficients of functions in ${\frak{H}}^p(\mathbb{T})$ (notice that this is the only sensible way to define them, thanks to the uniqueness property stated above).

So the question: is it true that for each $\varphi\in{\frak{H}}^p(\mathbb{T})$, denoting by $e_n$ the $n$th character and by $\hat\varphi(n)$ the $n$th Fourier coefficient of $\varphi$, that: $$||\sum_{n=0}^{N}\hat\varphi(n)e_n-\varphi||_p \rightarrow0, N\rightarrow+\infty?$$

I can't find any result about this question in any book I consulted... can anyone give me an answer and any reference?

Bob
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The answer is no, as suggested by fedja in this related question. As a counterexample, just get $$\varphi(t)=\cot(t/2).$$ In fact, define: $$f(z)=\frac{1+z}{1-z}=1+2\sum_{n=1}^\infty z^n.$$ We have that $\forall p\in(0,1), f\in H^p(D)$ and the boundary values of $f$ are represented by $\varphi$, so $\varphi\in{\frak{H}}^p(\mathbb{T})$ and $\hat{\varphi}(0)=1$ and $\forall n>0, \hat{\varphi}(n)=2$. Now: $$\forall N\in\mathbb{N}, \int_{-\pi}^\pi {|\sum_{n=0}^N\hat\varphi}(n)e^{int}-\varphi(t)|^p\frac{dt}{2\pi}=\frac{1}{2^{1-p}\pi}\int_{-\pi}^\pi|1-e^{it}|^{-p}dt.$$ Then: $$||\sum_{n=0}^{N}\hat\varphi(n)e_n-\varphi||_p^p\rightarrow\frac{1}{2^{1-p}\pi}\int_{-\pi}^\pi|1-e^{it}|^{-p}dt>0, N\rightarrow+\infty,$$ so the Fourier series of $\varphi$ doesn't converge to $\varphi$ in the $L^p(\mathbb{T})$ norm, neither gets closer to $\varphi$ in $L^p(\mathbb{T})$ norm as $N$ increases.

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Bob
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