Convergence of Fourier Series in $L^1(\mathbb{T})$

Question

Suppose $f \in L^1(\mathbb{T})$ and the sequence of partial sums of its Fourier series converges (in $L^1(\mathbb{T})$) to $g$. How can I prove $f=g$?

Answer 1

The usual strategy with these sort of problems (knowing that something converges to some limit implies that it's the "correct" limit) is to go a weaker type of convergence where you know that you always get the correct limit, then use that to compare the given limit and the correct limit.

We say that a sequence $h_n$ converges to $h$ in the Cesaro sense if the average $$\sigma_n = \frac{h_0 + \ldots + h_{n-1}}{n}$$ converges to $h$.

It is a known result that partial sums converge in $L^1$ in the Cesaro sense to $f$. Also as normal convergence implies Cesaro convergence with the same limit, they also converge in $L^1$ in the Cesaro sense to $g$. Hence $f=g$ by uniqueness of limits in $L^1$.

Answer 2

The partial sum of the Fourier series is $$ S_{n}(f)(e^{i\theta}) = \sum_{k=-n}^{n} \left[\frac{1}{2\pi}\int_{0}^{2\pi}f(e^{i\theta})e^{-ik\theta'}\,d\theta'\right] e^{ik\theta},\;\;\; n \ge 0. $$ For all $n \ge |m|$, $$ \frac{1}{2\pi}\int_{0}^{2\pi} S_{n}(f)(e^{i\theta'})e^{-im\theta'}\,d\theta' = \frac{1}{2\pi}\int_{0}^{2\pi}f(e^{i\theta'})e^{-im\theta'}\,d\theta'. $$ Therefore, if $S_{n}(f)$ converges in $L^{1}$ to $g$, the Fourier coefficients of $g$ must be the same as those of $f$. This leads to a function $h=f-g \in L^{1}$ with Fourier coefficients that are all $0$.

Define a new function $$ k(\theta) = \int_{0}^{\theta}h(e^{i\theta'})\,d\theta'. $$ Then $k$ is absolutely continuous and periodic on $[0,2\pi]$, with $k'(\theta)=f(e^{i\theta})-g(e^{i\theta})$ a.e.. Therefore, for $n \ne 0$, $$ \left.\int_{0}^{2\pi}k(\theta')e^{-in\theta'}\,d\theta' = k(\theta')\frac{e^{-in\theta'}}{-in}\right|_{0}^{2\pi}+\frac{1}{in}\int_{0}^{2\pi}(f(e^{i\theta'})-g(e^{i\theta'}))e^{-in\theta}\,d\theta' =0. $$ Because $k \in L^{2}$ and $\{ e^{in\theta}\}$ is a complete orthogonal basis of $L^{2}$, then $k$ is equal a.e. to a constant multiple of $1$. Because $k$ is continuous with $k(0)=0$, then $k\equiv 0$ follows. Because $0=k'=f-g$ a.e., then $f=g$ a.e.. So the two functions $f$ and $g$ are equal in $L^{1}$.