# Converse of Fejer's Theorem

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.

• A hint may be that the Fejer kernel can be viewed as a triangle, where Dirichlet kernel would be a box. A triangle is a box convolved with itself and convolution of functions become nicer than the factors. – mathreadler Mar 20 '19 at 14:20

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.

• Thank you so much. This is so obvious now. – Namch96 Mar 20 '19 at 14:34

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.