Approximating $L^1(\mathbb{T})$ functions with continuous functions

Question

$\newcommand{\T}{\mathbb{T}}\newcommand{\norm}[1]{\left\lVert#1\right\rVert}\newcommand{\eps}{\epsilon} $ Consider the space $L^1(\T)$ where $\T = \mathbb{R}\diagup\mathbb{Z}$ is the unit circle.

Given a function $f \in L^1(\T)$ and $\eps > 0$, does there exist a continuous $g \in L^1(\T)$ such that $\norm{f - g}_1 < \eps$?

The continuity of $g$ means the continuity of the periodic extension $\tilde{g}$ of $g$ to $\mathbb{R}$, defined as the (unique) function that agrees with $g$ on $[0, 1)$, and $\tilde{g}(x) = \tilde{g}(x + 1)$ for all $x \in \mathbb{R}$.

Answer 1

Yes. Let $F_n$ be Fejer's kernel. Then $$F_n\ast f(x)=\sum_{|k|\le n}\Bigl(1-\frac{|k|}{n}\Bigr)\hat f_k\,e^{i2k\pi x}$$ converges to $f$ almost everywhere and in $L^1$.