Fejér's theorem

Template:Short description 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 $$f(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{e}^{inx}$$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

$c_k=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)e^{-ikt}dt.$

Then, we can define

${\sigma }_n(f,x)=\frac{1}{n}\sum _{k=0}^{n-1}{s}_k(f,x)=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x-t)F_n(t)dt$

with F_n being the nth order Fejér kernel.

Then, Fejér's theorem asserts that

$$\underset{n\to \mathrm{\infty }}{\lim }{\sigma }_n(f,x)=f(x)$$

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

$$\mathrm{\forall }ϵ>0\mathrm{\exists }{n}_0\in \mathbb{N}:n\ge n_0⟹|f(x)-{\sigma }_n(f,x)|<ϵ,\mathrm{\forall }x\in ℝ$$

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

$$s_n(f,x)=\sum _{k=-n}^n{c}_k{e}^{ikx}=\sum _{k=-n}^n[\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)e^{-ikt}dt]e^{ikx}=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)\sum _{k=-n}^n{e}^{ik(x-t)}dt=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)D_n(x-t)dt.$$Using a change of variables we get

$$s_n(f,x)=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x-t)D_n(t)dt.$$

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 ${\sigma }_n(f,x)$

$${\sigma }_n(f,x)=\frac{1}{n}\sum _{k=0}^{n-1}{s}_k(f,x)=\frac{1}{n}\sum _{k=0}^{n-1}\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x-t)D_k(t)dt=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x-t)[\frac{1}{n}\sum _{k=0}^{n-1}{D}_k(t)]dt=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x-t)F_n(t)dt$$This completes the proof of Lemma 2.

We next prove the 3rd Lemma:

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 :

$$F_n(x)=\frac{1}{n}\sum _{k=0}^{n-1}\frac{\sin ((2k+1)x/2)}{\sin (x/2)}=\frac{1}{n}\frac{{\sin }^2(nx/2)}{{\sin }^2(x/2)}\ge 0$$

c) For all fixed $\delta >0$,

$$\underset{\delta \le |x|\le \pi }{\int }{F}_n(x)dx=\frac{2}{n}\underset{\delta \le x\le \pi }{\int }\frac{{\sin }^2(nx/2)}{{\sin }^2(x/2)}dx\le \frac{2}{n}\underset{\delta \le x\le \pi }{\int }\frac{1}{{\sin }^2(x/2)}dx$$

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

$$\mathrm{\forall }ϵ>0\mathrm{\exists }{n}_0\in \mathbb{N}:n\ge n_0⟹|f(x)-{\sigma }_n(f,x)|<ϵ,\mathrm{\forall }x\in ℝ$$

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

$${\sigma }_n(f,x)=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x-t)F_n(t)dt.$$By Lemma 3a we know that

$${\sigma }_n(f,x)-f(x)=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x-t)F_n(t)dt-f(x)=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x-t)F_n(t)dt-f(x)\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}{F}_n(t)dt=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)F_n(t)dt=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}[f(x-t)-f(x)]F_n(t)dt.$$

Applying the triangle inequality yields

$$|{\sigma }_n(f,x)-f(x)|=|\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}[f(x-t)-f(x)]F_n(t)dt|\le \frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}|[f(x-t)-f(x)]F_n(t)|dt=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}|f(x-t)-f(x)||F_n(t)|dt,$$and by Lemma 3b, we get

$$|{\sigma }_n(f,x)-f(x)|=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}|f(x-t)-f(x)|F_n(t)dt.$$We now split the integral into two parts, integrating over the two regions $|t|\le \delta$ and $\delta \le |t|\le \pi$.

$$|{\sigma }_n(f,x)-f(x)|=(\frac{1}{2\pi }\underset{|t|\le \delta }{\int }|f(x-t)-f(x)|F_n(t)dt)+(\frac{1}{2\pi }\underset{\delta \le |t|\le \pi }{\int }|f(x-t)-f(x)|F_n(t)dt)$$The motivation for doing so is that we want to prove that $\underset{n\to \mathrm{\infty }}{\lim }|{\sigma }_n(f,x)-f(x)|=0$. 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 $\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0:|x-y|\le \delta ⟹|f(x)-f(y)|\le ϵ$. Hence we can rewrite the integral 1 as follows

$$\frac{1}{2\pi }\underset{|t|\le \delta }{\int }|f(x-t)-f(x)|F_n(t)dt\le \frac{1}{2\pi }\underset{|t|\le \delta }{\int }ϵF_n(t)dt=\frac{1}{2\pi }ϵ\underset{|t|\le \delta }{\int }{F}_n(t)dt$$Because $F_n(x)\ge 0,\mathrm{\forall }x\in ℝ$ and $\delta \le \pi$$$\frac{1}{2\pi }ϵ\underset{|t|\le \delta }{\int }{F}_n(t)dt\le \frac{1}{2\pi }ϵ\underset{-\pi }{\overset{\pi }{\int }}{F}_n(t)dt$$By Lemma 3a we then get for all n

$$\frac{1}{2\pi }ϵ\underset{-\pi }{\overset{\pi }{\int }}{F}_n(t)dt=ϵ$$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 $M=\underset{-\pi \le t\le \pi }{\sup }|f(t)|$

$$\frac{1}{2\pi }\underset{\delta \le |t|\le \pi }{\int }|f(x-t)-f(x)|F_n(t)dt\le \frac{1}{2\pi }\underset{\delta \le |t|\le \pi }{\int }2M{F}_n(t)dt=\frac{M}{\pi }\underset{\delta \le |t|\le \pi }{\int }{F}_n(t)dt$$We are now ready to prove that $\underset{n\to \mathrm{\infty }}{\lim }|{\sigma }_n(f,x)-f(x)|=0$. We begin by writing

$$|{\sigma }_n(f,x)-f(x)|\le ϵ+\frac{M}{\pi }\underset{\delta \le |t|\le \pi }{\int }{F}_n(t)dt$$Thus,$$\underset{n\to \mathrm{\infty }}{\lim }|{\sigma }_n(f,x)-f(x)|\le \underset{n\to \mathrm{\infty }}{\lim }ϵ+\underset{n\to \mathrm{\infty }}{\lim }\frac{M}{\pi }\underset{\delta \le |t|\le \pi }{\int }{F}_n(t)dt$$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 $\underset{n\to \mathrm{\infty }}{\lim }|{\sigma }_n(f,x)-f(x)|=0$, 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 corollary relating which goes as follows:

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

${\sigma }_n(x_0)\to \frac{1}{2}(f(x_0+0)+f(x_0-0)).$

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 Template:Harv.

References

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