Fejér's theorem

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

Fejér's Theorem — Let [math]\displaystyle{ f: \mathbb{R} \to \mathbb{C} }[/math] be a continuous function with period [math]\displaystyle{ 2 \pi }[/math], let [math]\displaystyle{ s_n(f) }[/math] be the nth partial sum of the Fourier series of [math]\displaystyle{ f }[/math], and let [math]\displaystyle{ \sigma_n(f) }[/math] be the sequence of Cesàro means of the sequence [math]\displaystyle{ s_n(f) }[/math], that is the sequence of arithmetic means of [math]\displaystyle{ s_0(f), ..., s_n(f) }[/math]. Then the sequence [math]\displaystyle{ \sigma_n(f) }[/math] converges uniformly to [math]\displaystyle{ f }[/math] on [math]\displaystyle{ \mathbb{R} }[/math] as n tends to infinity.

Explanation of Fejér's Theorem's

Explicitly, we can write the Fourier series of f as [math]\displaystyle{ f(x)= \sum_{n=- \infty}^{\infty} c_n \, e^{inx} }[/math]where the nth partial sum of the Fourier series of f may be written as

[math]\displaystyle{ s_n(f,x)=\sum_{k=-n}^nc_ke^{ikx}, }[/math]

where the Fourier coefficients [math]\displaystyle{ c_k }[/math] are

[math]\displaystyle{ c_k=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)e^{-ikt}dt. }[/math]

Then, we can define

[math]\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 }[/math]

with F_n being the nth order Fejér kernel.

Then, Fejér's theorem asserts that

[math]\displaystyle{ \lim_{n\to \infty} \sigma_n (f, x) = f(x) }[/math]

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

[math]\displaystyle{ \forall \epsilon \gt 0 \, \exist\, n_0 \in \mathbb{N}: n \geq n_0 \implies | f(x) - \sigma_n(f,x)| \lt \epsilon, \, \forall x \in \mathbb{R} }[/math]

Proof of Fejér's Theorem

We first prove the following lemma:

Lemma 1 — The nth partial sum of the Fourier series [math]\displaystyle{ s_n(f,x) }[/math] may be written using the Dirichlet Kernel as: [math]\displaystyle{ s_n(f,x) = \frac{1}{2\pi} \int_{-\pi}^\pi f(x-t) \, D_n(t) \, dt }[/math]

Proof: Recall the definition of [math]\displaystyle{ D_n(x) }[/math], the Dirichlet Kernel:[math]\displaystyle{ D_n(x) = \sum_{k=-n}^n e^{ikx}. }[/math]We substitute the integral form of the Fourier coefficients into the formula for [math]\displaystyle{ s_n(f,x) }[/math] above

[math]\displaystyle{ 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. }[/math]Using a change of variables we get

[math]\displaystyle{ s_n(f,x) = \frac{1}{2\pi} \int_{-\pi}^\pi f(x-t) \, D_n(t) \, dt. }[/math]

This completes the proof of Lemma 1.

We next prove the following lemma:

Lemma 2 — The nth Cesaro sum [math]\displaystyle{ \sigma_n(f,x) }[/math] may be written using the Fejér Kernel as: [math]\displaystyle{ \sigma_n(f,x)=\frac{1}{2\pi}\int_{-\pi}^\pi f(x-t)F_n(t)dt }[/math]

Proof: Recall the definition of the Fejér Kernel [math]\displaystyle{ F_n(x) }[/math]

[math]\displaystyle{ F_n(x) = \frac{1}{n} \sum_{k=0}^{n-1}D_k(x) }[/math]As in the case of Lemma 1, we substitute the integral form of the Fourier coefficients into the formula for [math]\displaystyle{ \sigma_n(f,x) }[/math]

[math]\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 }[/math]This completes the proof of Lemma 2.

We next prove the 3rd Lemma:

Lemma 3 — The Fejer Kernel has the following 3 properties:

• a) [math]\displaystyle{ \frac{1}{2\pi} \int_{-\pi}^\pi F_n (x) \, dx =1 }[/math]
• b) [math]\displaystyle{ F_n(x) \geq 0 }[/math]
• c) For all fixed [math]\displaystyle{ \delta \gt 0 }[/math], [math]\displaystyle{ \lim_{n \to \infty} \int_{\delta \leq |x| \leq \pi} F_n (x) \, dx = 0 }[/math]

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

[math]\displaystyle{ \forall \epsilon \gt 0 \, \exist\, n_0 \in \mathbb{N}: n \geq n_0 \implies | f(x) - \sigma_n(f,x)| \lt \epsilon, \, \forall x \in \mathbb{R} }[/math]

We want to find an expression for [math]\displaystyle{ |\sigma_n(f,x) - f(x) | }[/math]. We begin by invoking Lemma 2:

[math]\displaystyle{ \sigma_n(f,x)= \frac{1}{2\pi} \int_{-\pi}^\pi f(x-t) \, F_n(t) \, dt. }[/math]By Lemma 3a we know that

[math]\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. }[/math]

Applying the triangle inequality yields

[math]\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, }[/math]and by Lemma 3b, we get

[math]\displaystyle{ |\sigma_n(f,x) - f(x) |= \frac{1}{2\pi} \int_{-\pi}^\pi |f(x-t)-f(x)| \, F_n(t) \, dt. }[/math]We now split the integral into two parts, integrating over the two regions [math]\displaystyle{ |t| \leq \delta }[/math] and [math]\displaystyle{ \delta \leq |t| \leq \pi }[/math].

[math]\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) }[/math]The motivation for doing so is that we want to prove that [math]\displaystyle{ \lim_{n \to \infty} |\sigma_n(f,x) - f(x) |=0 }[/math]. 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 [math]\displaystyle{ \forall \epsilon \gt 0,\exist \delta \gt 0: |x-y| \leq \delta \implies |f(x)-f(y)| \leq \epsilon }[/math]. Hence we can rewrite the integral 1 as follows

[math]\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 }[/math]Because [math]\displaystyle{ F_n(x) \geq 0, \forall x\in \mathbb{R} }[/math] and [math]\displaystyle{ \delta \leq \pi }[/math][math]\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 }[/math]By Lemma 3a we then get for all n

[math]\displaystyle{ \frac{1}{2\pi}\epsilon \int_{-\pi}^\pi \, F_n(t) \, dt = \epsilon }[/math]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 [math]\displaystyle{ M=\sup_{-\pi \leq t \leq \pi} |f(t)| }[/math]

[math]\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 }[/math]We are now ready to prove that [math]\displaystyle{ \lim_{n \to \infty} |\sigma_n(f,x) - f(x) |=0 }[/math]. We begin by writing

[math]\displaystyle{ |\sigma_n(f,x) - f(x) | \leq \epsilon \, + \frac{M}{\pi} \int_{\delta \leq|t|\leq \pi}F_n(t) \, dt }[/math]Thus,[math]\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 }[/math]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 [math]\displaystyle{ \lim_{n \to \infty} |\sigma_n(f,x) - f(x) |=0 }[/math], 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 [math]\displaystyle{ f \in L^2(- \pi, \pi) }[/math] be continuous at [math]\displaystyle{ x \in (-\pi,\pi) }[/math], then [math]\displaystyle{ \sigma_n(f,x) }[/math] converges pointwise as n goes to infinity.

Sadly however, the theorem does not work in a general sense when we replace the sequence [math]\displaystyle{ \sigma_n (f,x) }[/math] with [math]\displaystyle{ s_n (f,x) }[/math]. 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 [math]\displaystyle{ L^2(-\pi, \pi) }[/math] 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:

Corollary — Let [math]\displaystyle{ s_n \in \mathbb{C}, \, \forall n \in \, \mathbb{Z}_+ }[/math]. If [math]\displaystyle{ s_n }[/math] converges to s as n goes to infinity, then [math]\displaystyle{ \sigma_n }[/math] 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). 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

[math]\displaystyle{ \sigma_n(x_0) \to \frac{1}{2}\left(f(x_0+0)+f(x_0-0)\right). }[/math]

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).

References

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