Fejér's theorem

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 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 , \exist, n_0 \in \mathbb{N}: n \geq n_0 \implies | f(x) - \sigma_n(f,x)|

Proof of Fejér's Theorem

We first prove the following lemma:

: 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_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:

• 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 a 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(n x / 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(n x / 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 , \exist, n_0 \in \mathbb{N}: n \geq n_0 \implies | f(x) - \sigma_n(f,x)|

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 will do in the next step.

We first note that the function f is continuous on [-\pi,\pi]. We invoke the theorem that every periodic function on [-\pi,\pi] 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 proved in 1966 by L. Carleson. We can however prove a corollary 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 L^1(-\pi,\pi). If the left and right limits f(x_0\pm 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 \sigma_n is replaced with (C, α) mean of the Fourier series .

References

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References

1. C.R. Acad. Sci. Paris]]'', 10 décembre 1900, 984-987, .
2. Leopold Fejér, [https://babel.hathitrust.org/cgi/pt?id=inu.30000106096518&view=1up&seq=61 Untersuchungen über Fouriersche Reihen], ''[[Mathematische Annalen]]'', [https://gdz.sub.uni-goettingen.de/id/PPN235181684_0058 vol. 58], 1904, 51-69.
3. (1988-07-21). "Introduction". Cambridge University Press.
4. Rogosinski, W. W.. (December 1965). "An elementary companion to a theorem of J. Mercer". [[Journal d'Analyse Mathématique]].