Fejér's theorem

Explicitly, we can write the Fourier series of f as

$${\displaystyle f(x)=\sum _{n=-\infty }^{\infty }c_{n}\,e^{inx}}$$

where the nth partial sum of the Fourier series of f may be written as

${\displaystyle s_{n}(f,x)=\sum _{k=-n}^{n}c_{k}e^{ikx},}$

where the Fourier coefficients ${\displaystyle c_{k}}$ are

${\displaystyle c_{k}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)e^{-ikt}dt.}$

Then, we can define

${\displaystyle \sigma _{n}(f,x)={\frac {1}{n}}\sum _{k=0}^{n-1}s_{k}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)F_{n}(t)dt}$

with F_n being the nth order Fejér kernel.

Then, Fejér's theorem asserts that

$${\displaystyle \lim _{n\to \infty }\sigma _{n}(f,x)=f(x)}$$

with uniform convergence. With the convergence written out explicitly, the above statement becomes

$${\displaystyle \forall \epsilon >0\,\exists \,n_{0}\in \mathbb {N} :n\geq n_{0}\implies |f(x)-\sigma _{n}(f,x)|<\epsilon ,\,\forall x\in \mathbb {R} }$$

We first prove the following lemma:

Proof: Recall the definition of ${\displaystyle D_{n}(x)}$, the Dirichlet Kernel:

$${\displaystyle D_{n}(x)=\sum _{k=-n}^{n}e^{ikx}.}$$

We substitute the integral form of the Fourier coefficients into the formula for ${\displaystyle s_{n}(f,x)}$ above

$${\displaystyle s_{n}(f,x)=\sum _{k=-n}^{n}c_{k}e^{ikx}=\sum _{k=-n}^{n}[{\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)e^{-ikt}dt]e^{ikx}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)\sum _{k=-n}^{n}e^{ik(x-t)}\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)\,D_{n}(x-t)\,dt.}$$

Using a change of variables we get

$${\displaystyle s_{n}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,D_{n}(t)\,dt.}$$

This completes the proof of Lemma 1.

We next prove the following lemma:

Proof: Recall the definition of the Fejér Kernel ${\displaystyle F_{n}(x)}$

$${\displaystyle F_{n}(x)={\frac {1}{n}}\sum _{k=0}^{n-1}D_{k}(x)}$$

As in the case of Lemma 1, we substitute the integral form of the Fourier coefficients into the formula for ${\displaystyle \sigma _{n}(f,x)}$

$${\displaystyle \sigma _{n}(f,x)={\frac {1}{n}}\sum _{k=0}^{n-1}s_{k}(f,x)={\frac {1}{n}}\sum _{k=0}^{n-1}{\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,D_{k}(t)\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,[{\frac {1}{n}}\sum _{k=0}^{n-1}D_{k}(t)]\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,F_{n}(t)\,dt}$$

This completes the proof of Lemma 2.

We next prove the 3rd Lemma:

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

$${\displaystyle \forall \epsilon >0\,\exists \,n_{0}\in \mathbb {N} :n\geq n_{0}\implies |f(x)-\sigma _{n}(f,x)|<\epsilon ,\,\forall x\in \mathbb {R} }$$

We want to find an expression for ${\displaystyle |\sigma _{n}(f,x)-f(x)|}$. We begin by invoking Lemma 2:

$${\displaystyle \sigma _{n}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,F_{n}(t)\,dt.}$$

By Lemma 3a we know that

$${\displaystyle \sigma _{n}(f,x)-f(x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,F_{n}(t)\,dt-f(x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)\,F_{n}(t)\,dt-f(x){\frac {1}{2\pi }}\int _{-\pi }^{\pi }F_{n}(t)\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)\,F_{n}(t)\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }[f(x-t)-f(x)]\,F_{n}(t)\,dt.}$$

Applying the triangle inequality yields

$${\displaystyle |\sigma _{n}(f,x)-f(x)|=|{\frac {1}{2\pi }}\int _{-\pi }^{\pi }[f(x-t)-f(x)]\,F_{n}(t)\,dt|\leq {\frac {1}{2\pi }}\int _{-\pi }^{\pi }|[f(x-t)-f(x)]\,F_{n}(t)|\,dt={\frac {1}{2\pi }}\int _{-\pi }^{\pi }|f(x-t)-f(x)|\,|F_{n}(t)|\,dt,}$$

and by Lemma 3b, we get

$${\displaystyle |\sigma _{n}(f,x)-f(x)|={\frac {1}{2\pi }}\int _{-\pi }^{\pi }|f(x-t)-f(x)|\,F_{n}(t)\,dt.}$$

We now split the integral into two parts, integrating over the two regions ${\displaystyle |t|\leq \delta }$ and ${\displaystyle \delta \leq |t|\leq \pi }$.

$${\displaystyle |\sigma _{n}(f,x)-f(x)|=\left({\frac {1}{2\pi }}\int _{|t|\leq \delta }|f(x-t)-f(x)|\,F_{n}(t)\,dt\right)+\left({\frac {1}{2\pi }}\int _{\delta \leq |t|\leq \pi }|f(x-t)-f(x)|\,F_{n}(t)\,dt\right)}$$

The motivation for doing so is that we want to prove that ${\displaystyle \lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|=0}$. 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 ${\displaystyle \forall \epsilon >0,\exists \delta >0:|x-y|\leq \delta \implies |f(x)-f(y)|\leq \epsilon }$. Hence we can rewrite the integral 1 as follows

$${\displaystyle {\frac {1}{2\pi }}\int _{|t|\leq \delta }|f(x-t)-f(x)|\,F_{n}(t)\,dt\leq {\frac {1}{2\pi }}\int _{|t|\leq \delta }\epsilon \,F_{n}(t)\,dt={\frac {1}{2\pi }}\epsilon \int _{|t|\leq \delta }\,F_{n}(t)\,dt}$$

Because ${\displaystyle F_{n}(x)\geq 0,\forall x\in \mathbb {R} }$ and ${\displaystyle \delta \leq \pi }$

$${\displaystyle {\frac {1}{2\pi }}\epsilon \int _{|t|\leq \delta }\,F_{n}(t)\,dt\leq {\frac {1}{2\pi }}\epsilon \int _{-\pi }^{\pi }\,F_{n}(t)\,dt}$$

By Lemma 3a we then get for all n

$${\displaystyle {\frac {1}{2\pi }}\epsilon \int _{-\pi }^{\pi }\,F_{n}(t)\,dt=\epsilon }$$

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 ${\displaystyle M=\sup _{-\pi \leq t\leq \pi }|f(t)|}$

$${\displaystyle {\frac {1}{2\pi }}\int _{\delta \leq |t|\leq \pi }|f(x-t)-f(x)|\,F_{n}(t)\,dt\leq {\frac {1}{2\pi }}\int _{\delta \leq |t|\leq \pi }2M\,F_{n}(t)\,dt={\frac {M}{\pi }}\int _{\delta \leq |t|\leq \pi }F_{n}(t)\,dt}$$

We are now ready to prove that ${\displaystyle \lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|=0}$. We begin by writing

$${\displaystyle |\sigma _{n}(f,x)-f(x)|\leq \epsilon \,+{\frac {M}{\pi }}\int _{\delta \leq |t|\leq \pi }F_{n}(t)\,dt}$$

Thus,

$${\displaystyle \lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|\leq \lim _{n\to \infty }\epsilon \,+\lim _{n\to \infty }{\frac {M}{\pi }}\int _{\delta \leq |t|\leq \pi }F_{n}(t)\,dt}$$

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 ${\displaystyle \lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|=0}$, which completes the proof.

In fact, Fejér's theorem can be modified to hold for pointwise convergence.[3]

Sadly however, the theorem does not work in a general sense when we replace the sequence ${\displaystyle \sigma _{n}(f,x)}$ with ${\displaystyle s_{n}(f,x)}$. This is because there exist functions whose Fourier series fails to converge at some point.[4] However, the set of points at which a function in ${\displaystyle L^{2}(-\pi ,\pi )}$ diverges has to be measure zero. This fact, called Lusins conjecture or Carleson's theorem, was proven in 1966 by L. Carleson.[4] We can however prove a corrollary relating which goes as follows:

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 L¹(-π,π). If the left and right limits f(x₀±0) of f(x) exist at x₀, or if both limits are infinite of the same sign, then

${\displaystyle \sigma _{n}(x_{0})\to {\frac {1}{2}}\left(f(x_{0}+0)+f(x_{0}-0)\right).}$

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).

• Zygmund, Antoni (1968), Trigonometric Series (2nd ed.), Cambridge University Press (published 1988), ISBN 978-0-521-35885-9.