Converse of Fejer's Theorem

Question

I am taking a course in Fourier Analysis this year. One of the theorems my lecturer wrote down was the following:

Let $f \in \mathcal{L}^{1}([-\pi,\pi])$ be $2\pi$-periodic. Then, $f \in \mathcal{C}^{0}([-\pi,\pi])$ if and only if $\sigma_{N}f \rightarrow f$ uniformly as $N \rightarrow \infty$, where $\sigma_{N}f(x) = \tfrac{1}{2\pi}f * F_{N}(x)$ where $F_{N}$ is the Fejer kernel (https://en.wikipedia.org/wiki/Fej%C3%A9r_kernel) and $*$ represents convolution.

Now the forward direction is known as Fejer's Theorem and can be found on https://en.wikipedia.org/wiki/Fej%C3%A9r%27s_theorem.

However, for the backward direction, he wrote the proof is trivial. I am having trouble actually writing down a proof for this. And I cannot find a similar result online or in books, which makes me think it may be false (but I cannot come up with a counterexample too). Any help for this "trivial" direction would be appreciated.

P.S A reformulation of $\sigma_{N}f -f$ is $\sigma_{N}f(x) - f(x) = \tfrac{1}{2\pi} \int_{-\pi}^{\pi} (f(x-y)-f(x))F_{N}(y) dy$. And there is also a closed form of the Fejer kernel in the first link that I provided.

Answer 1

The lecturer meant to say that it is trivial because a uniformly convergent sequence of continuous functions converges to a continuous function.

Since $\sigma_Nf$ is a continuous function for every $N$, its uniform convergence to $f$ implies that $f$ is continuous, too.

Answer 2

This is getting here a little late (after the acceptance of a solution), but I thought I would expand on @uniquesolution's answer. To me, it seems that it is the continuity of $\sigma_N f$ that is the issue. For that I will make use of the following result:

Proposition: Let $g \in L^1([-\pi,\pi])$ and $h \in C^0([-\pi,\pi])$, then $f \ast g \in C^0([-\pi,\pi])$.

The theorem statement is as follows:

Theorem: Let $f \in L^1([-\pi,\pi])$, then $$\sigma_Nf \to f \text{ uniformly as } N \to \infty \quad \iff \quad f \in C^0([-\pi,\pi]).$$

Proof of "$\implies$":

Let $f \in L^1([-\pi,\pi])$, then $\sigma_N f \equiv \frac{1}{2\pi}F_N \ast f$. We know that since $F_N$ is a finite sum of continuous functions, $F_N$ continuous, i.e. $F_N \in C^0([-\pi,\pi])$ for each $N \in \mathbb{N}$. Then by the above proposition: since $f \in L^1([-\pi,\pi])$, we have that $F_N \ast f \in C^0([-\pi,\pi])$, and therefore $\sigma_Nf \in C^0([-\pi,\pi])$. Then the remaining part of the proof is identical to @uniquesolution's answer.