If the Fourier sequence $\{H_n\}$ of a function $f$ converges almost everywhere to a function $g$, then $f=g$ almost everywhere?

Question

Let $\{H_n\}$ be the Fourier sequence of an $L^2$-function $f$, with $H_n$ being the $n$-th Fourier series of $f$. If $f$ converges almost everywhere to a function $g$, then does it hold that $f = g$ almost everywhere?

Answer 1

It is true :

By the containment of $L^p$ spaces see David M. Bressoud's book (A Radical approach to Lebesgue's theory of Integration) on page 260 :

$ (\int_a^b |H_n|^p dx)^{\frac {1}{p}} \le (b-a)^{\frac {(q-p)}{pq}} (\int_a^b |H_n|^q dx)^{\frac {1}{q}}$ you can take p=1 and q=2

By Bessel's Inequality $\int_a^b |H_n|^2 dx \le \int_a^b |f|^2 dx $

This implies {$H_n $} and $f$ are in $L^1 \bigcap L^2$ over a bounded interval [a b] and {$H_n $} is uniformly integrable and since $H_n -f$ converges almost everywhere , one can apply Vitali Convergence theorem

It is also a known result in $L^2$ (Riesz-Fischer theorem) $\lim\limits_{n\mapsto \infty} \int_a^b (H_n -f)^2 dx = 0$

and by the inequalities above :$\lim\limits_{n\mapsto \infty} \int_a^b |(H_n -f)|dx = 0$

by Vitali convergence theorem $\lim\limits_{n\mapsto \infty} \int_a^b |(H_n -f)| dx =\int_a^b \lim\limits_{n\mapsto \infty} |(H_n -f)| dx= 0$

implying $\lim\limits_{n\mapsto \infty} (H_n -f) =0 $ almost everywhere

Therefore $ \lim\limits_{n\mapsto \infty} H_n = g = f$ almost everywhere