Fejér's theorem

In mathematics, Fejér's theorem, named after Hungarian mathematician Lipót Fejér, states the following:

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
 * $$s_n(f,x)=\sum_{k=-n}^nc_ke^{ikx},$$

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 Fn 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

$$\forall \epsilon > 0 \, \exist\, n_0 \in \mathbb{N}: n \geq n_0 \implies | f(x) - \sigma_n(f,x)| < \epsilon, \, \forall x \in  \mathbb{R}$$

Proof of Fejér's Theorem
We first prove the following lemma:

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

$$s_n(f,x)=\sum_{k=-n}^n c_ke^{ikx} = \sum_{k=-n}^n [\frac{1}{2\pi}\int_{-\pi}^\pi f(t)e^{-ikt}dt ] e^{ikx} = \frac{1}{2\pi} \int_{-\pi}^\pi f(t) \sum_{k=-n}^n e^{ik(x-t)} \, dt = \frac{1}{2\pi} \int_{-\pi}^\pi f(t) \, D_n(x-t) \, dt.$$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:

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:

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

$$\forall \epsilon > 0 \, \exist\, n_0 \in \mathbb{N}: n \geq n_0 \implies | f(x) - \sigma_n(f,x)| < \epsilon, \, \forall x \in  \mathbb{R}$$

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 $$\forall \epsilon > 0,\exist \delta > 0: |x-y| \leq \delta \implies |f(x)-f(y)| \leq \epsilon$$. 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.

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 corrollary relating which goes as follows:

A more general form of the theorem applies to functions which are not necessarily continuous. 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


 * $$\sigma_n(x_0) \to \frac{1}{2}\left(f(x_0+0)+f(x_0-0)\right).$$

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 σn is replaced with (C, &alpha;) mean of the Fourier series.