A question bout Fourier coefficients

Question

Let $\{a_n\}_n\in\mathbb{Z}$ be a sequence of complex numbers and let $$p_n(e^{it})=\sum_{k=-n}^{n}(1-\frac{|k|}{n+1})a_ke^{ikt},\;\;\;\;(e^{it}\in\mathbb{T}).$$ Let $1\leq p\leq\infty$. Suppose that the sequence $(p_n)_{n\geq 0}$ is convergent in the $L^p(\mathbb{T})$ to a function $f\in L^p(\mathbb{T})$. Then how can I show that $\{a_n\}_n\in\mathbb{Z}$ are actually the Fourier coefficients of $f$?

Answer 1

Convergence in $L^{p}$ implies that $\int f(e^{it})e^{ikt} \, dt =\lim \int p_n(e^{it})e^{ikt} \, dt$, You can calculate this directly from definition of $p_n$. Justification for $p>1$:

$|\int f(e^{it})e^{ikt} \, dt - \int p_n(e^{it})e^{ikt} \, dt|$ $$\leq \int|f(e^{it})-p_n(e^{it})|\, dt \leq \sqrt ({2\pi})^{1/q} (\int|f(e^{it})-p_n(e^{it})|^{p}\, dt)^{1/p} \to 0$$ by Holder's inequality, where $q$ is defined by $\frac 1 p +\frac 1 q=1$. Hence $\int f(e^{it})e^{ikt} \, dt =\lim \int p_n(e^{it})e^{ikt} \, dt$. This argument is not needed when $p=1$.