# Fejér's theorem

In mathematics, Fejér's theorem, named after Hungarian mathematician Lipót Fejér, states the following:
Fejér's Theorem — Let $f:\mathbb {R} \to \mathbb {C}$ be a continuous function with period $2\pi$, let $s_{n}(f)$ be the nth partial sum of the Fourier series of $f$, and let $\sigma _{n}(f)$ be the sequence of Cesàro means of the sequence $s_{n}(f)$, that is the sequence of arithmetic means of $s_{0}(f),...,s_{n}(f)$. Then the sequence $\sigma _{n}(f)$ converges uniformly to $f$ on $\mathbb {R}$ as n tends to infinity.

## Explanation of Fejér's Theorem's

Explicitly, we can write the Fourier series of f as $f(x)=\sum _{n=-\infty }^{\infty }c_{n}\,e^{inx}$where the nth partial sum of the Fourier series of f may be written as
    $s_{n}(f,x)=\sum _{k=-n}^{n}c_{k}e^{ikx},$
where the Fourier coefficients $c_{k}$ are
    $c_{k}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)e^{-ikt}dt.$
Then, we can define
    $\sigma _{n}(f,x)={\frac {1}{n}}\sum _{k=0}^{n-1}s_{k}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)F_{n}(t)dt$
with F$_n$ being the nth order Fejér kernel.
Then, Fejér's theorem asserts that
$\lim _{n\to \infty }\sigma _{n}(f,x)=f(x)$
with uniform convergence. With the convergence written out explicitly, the above statement becomes
$\forall \epsilon >0\,\exists \,n_{0}\in \mathbb {N} :n\geq n_{0}\implies |f(x)-\sigma _{n}(f,x)|<\epsilon ,\,\forall x\in \mathbb {R}$

## Proof of Fejér's Theorem

We first prove the following lemma:
Lemma 1 — The nth partial sum of the Fourier series $s_{n}(f,x)$ may be written using the Dirichlet Kernel as: $s_{n}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,D_{n}(t)\,dt$
Proof: Recall the definition of $D_{n}(x)$, the Dirichlet Kernel:$D_{n}(x)=\sum _{k=-n}^{n}e^{ikx}.$We substitute the integral form of the Fourier coefficients into the formula for $s_{n}(f,x)$ above
$s_{n}(f,x)=\sum _{k=-n}^{n}c_{k}e^{ikx}=\sum _{k=-n}^{n}[{\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)e^{-ikt}dt]e^{ikx}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)\sum _{k=-n}^{n}e^{ik(x-t)}\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)\,D_{n}(x-t)\,dt.$Using a change of variables we get
$s_{n}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,D_{n}(t)\,dt.$
This completes the proof of Lemma 1.
We next prove the following lemma:
Lemma 2 — The nth Cesaro sum $\sigma _{n}(f,x)$ may be written using the Fejér Kernel as: $\sigma _{n}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)F_{n}(t)dt$
Proof: Recall the definition of the Fejér Kernel $F_{n}(x)$
$F_{n}(x)={\frac {1}{n}}\sum _{k=0}^{n-1}D_{k}(x)$As in the case of Lemma 1, we substitute the integral form of the Fourier coefficients into the formula for $\sigma _{n}(f,x)$
$\sigma _{n}(f,x)={\frac {1}{n}}\sum _{k=0}^{n-1}s_{k}(f,x)={\frac {1}{n}}\sum _{k=0}^{n-1}{\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,D_{k}(t)\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,[{\frac {1}{n}}\sum _{k=0}^{n-1}D_{k}(t)]\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,F_{n}(t)\,dt$This completes the proof of Lemma 2.
We next prove the 3rd Lemma:
Lemma 3 — The Fejer Kernel has the following 3 properties:
* a) ${\frac {1}{2\pi }}\int _{-\pi }^{\pi }F_{n}(x)\,dx=1$
* b) $F_{n}(x)\geq 0$
* c) For all fixed $\delta >0$, $\lim _{n\to \infty }\int _{\delta \leq |x|\leq \pi }F_{n}(x)\,dx=0$
Proof: a) Given that $F_{n}$ is the mean of $D_{n}$, the integral of which is 1, by linearity, the integral of $F_{n}$ is also equal to 1.
b) As $D_{n}(x)$ is a geometric sum, we get an simple formula for $D_{n}(x)$ and then for $F_{n}(x)$,using De Moivre's formula :
$F_{n}(x)={\frac {1}{n}}\sum _{k=0}^{n-1}{\frac {\sin((2k+1)x/2)}{\sin(x/2)}}={\frac {1}{n}}{\frac {\sin ^{2}(nx/2)}{\sin ^{2}(x/2)}}\geq 0$
c) For all fixed $\delta >0$,
$\int _{\delta \leq |x|\leq \pi }F_{n}(x)\,dx={\frac {2}{n}}\int _{\delta \leq x\leq \pi }{\frac {\sin ^{2}(nx/2)}{\sin ^{2}(x/2)}}\,dx\leq {\frac {2}{n}}\int _{\delta \leq x\leq \pi }{\frac {1}{\sin ^{2}(x/2)}}\,dx$
This shows that the integral converges to zero, as $n$ goes to infinity.
This completes the proof of Lemma 3.
We are now ready to prove Fejér's Theorem. First, let us recall the statement we are trying to prove
$\forall \epsilon >0\,\exists \,n_{0}\in \mathbb {N} :n\geq n_{0}\implies |f(x)-\sigma _{n}(f,x)|<\epsilon ,\,\forall x\in \mathbb {R}$
We want to find an expression for $|\sigma _{n}(f,x)-f(x)|$. We begin by invoking Lemma 2:
$\sigma _{n}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,F_{n}(t)\,dt.$By Lemma 3a we know that
$\sigma _{n}(f,x)-f(x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,F_{n}(t)\,dt-f(x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,F_{n}(t)\,dt-f(x){\frac {1}{2\pi }}\int _{-\pi }^{\pi }F_{n}(t)\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)\,F_{n}(t)\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }[f(x-t)-f(x)]\,F_{n}(t)\,dt.$
Applying the triangle inequality yields
$|\sigma _{n}(f,x)-f(x)|=|{\frac {1}{2\pi }}\int _{-\pi }^{\pi }[f(x-t)-f(x)]\,F_{n}(t)\,dt|\leq {\frac {1}{2\pi }}\int _{-\pi }^{\pi }|[f(x-t)-f(x)]\,F_{n}(t)|\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }|f(x-t)-f(x)|\,|F_{n}(t)|\,dt,$and by Lemma 3b, we get
$|\sigma _{n}(f,x)-f(x)|={\frac {1}{2\pi }}\int _{-\pi }^{\pi }|f(x-t)-f(x)|\,F_{n}(t)\,dt.$We now split the integral into two parts, integrating over the two regions $|t|\leq \delta$ and $\delta \leq |t|\leq \pi$.
$|\sigma _{n}(f,x)-f(x)|=\left({\frac {1}{2\pi }}\int _{|t|\leq \delta }|f(x-t)-f(x)|\,F_{n}(t)\,dt\right)+\left({\frac {1}{2\pi }}\int _{\delta \leq |t|\leq \pi }|f(x-t)-f(x)|\,F_{n}(t)\,dt\right)$The motivation for doing so is that we want to prove that $\lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|=0$. We can do this by proving that each integral above, integral 1 and integral 2, goes to zero. This is precisely what we'll do in the next step.
We first note that the function f is continuous on [-π,π]. We invoke the theorem that every periodic function on [-π,π] that is continuous is also bounded and uniformily continuous. This means that $\forall \epsilon >0,\exists \delta >0:|x-y|\leq \delta \implies |f(x)-f(y)|\leq \epsilon$. Hence we can rewrite the integral 1 as follows
${\frac {1}{2\pi }}\int _{|t|\leq \delta }|f(x-t)-f(x)|\,F_{n}(t)\,dt\leq {\frac {1}{2\pi }}\int _{|t|\leq \delta }\epsilon \,F_{n}(t)\,dt={\frac {1}{2\pi }}\epsilon \int _{|t|\leq \delta }\,F_{n}(t)\,dt$Because $F_{n}(x)\geq 0,\forall x\in \mathbb {R}$ and $\delta \leq \pi$${\frac {1}{2\pi }}\epsilon \int _{|t|\leq \delta }\,F_{n}(t)\,dt\leq {\frac {1}{2\pi }}\epsilon \int _{-\pi }^{\pi }\,F_{n}(t)\,dt$By Lemma 3a we then get for all n
${\frac {1}{2\pi }}\epsilon \int _{-\pi }^{\pi }\,F_{n}(t)\,dt=\epsilon$This gives the desired bound for integral 1 which we can exploit in final step.
For integral 2, we note that since f is bounded, we can write this bound as $M=\sup _{-\pi \leq t\leq \pi }|f(t)|$
${\frac {1}{2\pi }}\int _{\delta \leq |t|\leq \pi }|f(x-t)-f(x)|\,F_{n}(t)\,dt\leq {\frac {1}{2\pi }}\int _{\delta \leq |t|\leq \pi }2M\,F_{n}(t)\,dt={\frac {M}{\pi }}\int _{\delta \leq |t|\leq \pi }F_{n}(t)\,dt$We are now ready to prove that $\lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|=0$. We begin by writing
$|\sigma _{n}(f,x)-f(x)|\leq \epsilon \,+{\frac {M}{\pi }}\int _{\delta \leq |t|\leq \pi }F_{n}(t)\,dt$Thus,$\lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|\leq \lim _{n\to \infty }\epsilon \,+\lim _{n\to \infty }{\frac {M}{\pi }}\int _{\delta \leq |t|\leq \pi }F_{n}(t)\,dt$By Lemma 3c we know that the integral goes to 0 as n goes to infinity, and because epsilon is arbitrary, we can set it equal to 0. Hence $\lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|=0$, which completes the proof.

## Modifications and Generalisations of Fejér's Theorem

In fact, Fejér's theorem can be modified to hold for pointwise convergence.
Modified Fejér's Theorem — Let $f\in L^{2}(-\pi ,\pi )$ be continuous at $x\in (-\pi ,\pi )$, then $\sigma _{n}(f,x)$ converges pointwise as n goes to infinity.
Sadly however, the theorem does not work in a general sense when we replace the sequence $\sigma _{n}(f,x)$ with $s_{n}(f,x)$. This is because there exist functions whose Fourier series fails to converge at some point. However, the set of points at which a function in $L^{2}(-\pi ,\pi )$ diverges has to be measure zero. This fact, called Lusins conjecture or Carleson's theorem, was proven in 1966 by L. Carleson. We can however prove a corollary relating which goes as follows:
Corollary — Let $s_{n}\in \mathbb {C} ,\,\forall n\in \,\mathbb {Z} _{+}$. If $s_{n}$ converges to s as n goes to infinity, then $\sigma _{n}$ converges to s as n goes to infinity.
A more general form of the theorem applies to functions which are not necessarily continuous (Zygmund 1968, Theorem III.3.4). Suppose that f is in L$^1$(-π,π). If the left and right limits f(x$_0$±0) of f(x) exist at x$_0$, or if both limits are infinite of the same sign, then
    $\sigma _{n}(x_{0})\to {\frac {1}{2}}\left(f(x_{0}+0)+f(x_{0}-0)\right).$
Existence or divergence to infinity of the Cesàro mean is also implied. By a theorem of Marcel Riesz, Fejér's theorem holds precisely as stated if the (C, 1) mean σ$_n$ is replaced with (C, α) mean of the Fourier series (Zygmund 1968, Theorem III.5.1).