In mathematics, Fejér's theorem ,[ 1] [ 2] named after Hungarian mathematician Lipót Fejér , states the following:[ 3]
Explanation of Fejér's Theorem's[ edit ]
Explicitly, we can write the Fourier series of f as
f
(
x
)
=
∑
n
=
−
∞
∞
c
n
e
i
n
x
{\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
s
n
(
f
,
x
)
=
∑
k
=
−
n
n
c
k
e
i
k
x
,
{\displaystyle s_{n}(f,x)=\sum _{k=-n}^{n}c_{k}e^{ikx},}
where the Fourier coefficients
c
k
{\displaystyle c_{k}}
are
c
k
=
1
2
π
∫
−
π
π
f
(
t
)
e
−
i
k
t
d
t
.
{\displaystyle c_{k}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)e^{-ikt}dt.}
Then, we can define
σ
n
(
f
,
x
)
=
1
n
∑
k
=
0
n
−
1
s
k
(
f
,
x
)
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
F
n
(
t
)
d
t
{\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 n th order Fejér kernel .
Then, Fejér's theorem asserts that
lim
n
→
∞
σ
n
(
f
,
x
)
=
f
(
x
)
{\displaystyle \lim _{n\to \infty }\sigma _{n}(f,x)=f(x)}
with uniform convergence. With the convergence written out explicitly, the above statement becomes
∀
ϵ
>
0
∃
n
0
∈
N
:
n
≥
n
0
⟹
|
f
(
x
)
−
σ
n
(
f
,
x
)
|
<
ϵ
,
∀
x
∈
R
{\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} }
Proof of Fejér's Theorem[ edit ]
We first prove the following lemma:
Lemma 1 — The nth partial sum of the Fourier series
s
n
(
f
,
x
)
{\displaystyle s_{n}(f,x)}
may be written using the Dirichlet Kernel as:
s
n
(
f
,
x
)
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
D
n
(
t
)
d
t
{\displaystyle 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
)
{\displaystyle D_{n}(x)}
, the Dirichlet Kernel :
D
n
(
x
)
=
∑
k
=
−
n
n
e
i
k
x
.
{\displaystyle 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
)
{\displaystyle s_{n}(f,x)}
above
s
n
(
f
,
x
)
=
∑
k
=
−
n
n
c
k
e
i
k
x
=
∑
k
=
−
n
n
[
1
2
π
∫
−
π
π
f
(
t
)
e
−
i
k
t
d
t
]
e
i
k
x
=
1
2
π
∫
−
π
π
f
(
t
)
∑
k
=
−
n
n
e
i
k
(
x
−
t
)
d
t
=
1
2
π
∫
−
π
π
f
(
t
)
D
n
(
x
−
t
)
d
t
.
{\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
s
n
(
f
,
x
)
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
D
n
(
t
)
d
t
.
{\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:
Lemma 2 — The nth Cesaro sum
σ
n
(
f
,
x
)
{\displaystyle \sigma _{n}(f,x)}
may be written using the Fejér Kernel as:
σ
n
(
f
,
x
)
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
F
n
(
t
)
d
t
{\displaystyle \sigma _{n}(f,x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x-t)F_{n}(t)dt}
Proof : Recall the definition of the Fejér Kernel
F
n
(
x
)
{\displaystyle F_{n}(x)}
F
n
(
x
)
=
1
n
∑
k
=
0
n
−
1
D
k
(
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
σ
n
(
f
,
x
)
{\displaystyle \sigma _{n}(f,x)}
σ
n
(
f
,
x
)
=
1
n
∑
k
=
0
n
−
1
s
k
(
f
,
x
)
=
1
n
∑
k
=
0
n
−
1
1
2
π
∫
−
π
π
f
(
x
−
t
)
D
k
(
t
)
d
t
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
[
1
n
∑
k
=
0
n
−
1
D
k
(
t
)
]
d
t
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
F
n
(
t
)
d
t
{\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
∀
ϵ
>
0
∃
n
0
∈
N
:
n
≥
n
0
⟹
|
f
(
x
)
−
σ
n
(
f
,
x
)
|
<
ϵ
,
∀
x
∈
R
{\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
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
{\displaystyle |\sigma _{n}(f,x)-f(x)|}
. We begin by invoking Lemma 2:
σ
n
(
f
,
x
)
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
F
n
(
t
)
d
t
.
{\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
σ
n
(
f
,
x
)
−
f
(
x
)
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
F
n
(
t
)
d
t
−
f
(
x
)
=
1
2
π
∫
−
π
π
f
(
x
−
t
)
F
n
(
t
)
d
t
−
f
(
x
)
1
2
π
∫
−
π
π
F
n
(
t
)
d
t
=
1
2
π
∫
−
π
π
f
(
x
)
F
n
(
t
)
d
t
=
1
2
π
∫
−
π
π
[
f
(
x
−
t
)
−
f
(
x
)
]
F
n
(
t
)
d
t
.
{\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
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
=
|
1
2
π
∫
−
π
π
[
f
(
x
−
t
)
−
f
(
x
)
]
F
n
(
t
)
d
t
|
≤
1
2
π
∫
−
π
π
|
[
f
(
x
−
t
)
−
f
(
x
)
]
F
n
(
t
)
|
d
t
=
1
2
π
∫
−
π
π
|
f
(
x
−
t
)
−
f
(
x
)
|
|
F
n
(
t
)
|
d
t
,
{\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
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
=
1
2
π
∫
−
π
π
|
f
(
x
−
t
)
−
f
(
x
)
|
F
n
(
t
)
d
t
.
{\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
|
t
|
≤
δ
{\displaystyle |t|\leq \delta }
and
δ
≤
|
t
|
≤
π
{\displaystyle \delta \leq |t|\leq \pi }
.
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
=
(
1
2
π
∫
|
t
|
≤
δ
|
f
(
x
−
t
)
−
f
(
x
)
|
F
n
(
t
)
d
t
)
+
(
1
2
π
∫
δ
≤
|
t
|
≤
π
|
f
(
x
−
t
)
−
f
(
x
)
|
F
n
(
t
)
d
t
)
{\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
lim
n
→
∞
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
=
0
{\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
∀
ϵ
>
0
,
∃
δ
>
0
:
|
x
−
y
|
≤
δ
⟹
|
f
(
x
)
−
f
(
y
)
|
≤
ϵ
{\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
1
2
π
∫
|
t
|
≤
δ
|
f
(
x
−
t
)
−
f
(
x
)
|
F
n
(
t
)
d
t
≤
1
2
π
∫
|
t
|
≤
δ
ϵ
F
n
(
t
)
d
t
=
1
2
π
ϵ
∫
|
t
|
≤
δ
F
n
(
t
)
d
t
{\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
F
n
(
x
)
≥
0
,
∀
x
∈
R
{\displaystyle F_{n}(x)\geq 0,\forall x\in \mathbb {R} }
and
δ
≤
π
{\displaystyle \delta \leq \pi }
1
2
π
ϵ
∫
|
t
|
≤
δ
F
n
(
t
)
d
t
≤
1
2
π
ϵ
∫
−
π
π
F
n
(
t
)
d
t
{\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
1
2
π
ϵ
∫
−
π
π
F
n
(
t
)
d
t
=
ϵ
{\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
M
=
sup
−
π
≤
t
≤
π
|
f
(
t
)
|
{\displaystyle M=\sup _{-\pi \leq t\leq \pi }|f(t)|}
1
2
π
∫
δ
≤
|
t
|
≤
π
|
f
(
x
−
t
)
−
f
(
x
)
|
F
n
(
t
)
d
t
≤
1
2
π
∫
δ
≤
|
t
|
≤
π
2
M
F
n
(
t
)
d
t
=
M
π
∫
δ
≤
|
t
|
≤
π
F
n
(
t
)
d
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
lim
n
→
∞
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
=
0
{\displaystyle \lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|=0}
. We begin by writing
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
≤
ϵ
+
M
π
∫
δ
≤
|
t
|
≤
π
F
n
(
t
)
d
t
{\displaystyle |\sigma _{n}(f,x)-f(x)|\leq \epsilon \,+{\frac {M}{\pi }}\int _{\delta \leq |t|\leq \pi }F_{n}(t)\,dt}
Thus,
lim
n
→
∞
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
≤
lim
n
→
∞
ϵ
+
lim
n
→
∞
M
π
∫
δ
≤
|
t
|
≤
π
F
n
(
t
)
d
t
{\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
lim
n
→
∞
|
σ
n
(
f
,
x
)
−
f
(
x
)
|
=
0
{\displaystyle \lim _{n\to \infty }|\sigma _{n}(f,x)-f(x)|=0}
, which completes the proof.
Modifications and Generalisations of Fejér's Theorem[ edit ]
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
σ
n
(
f
,
x
)
{\displaystyle \sigma _{n}(f,x)}
with
s
n
(
f
,
x
)
{\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
L
2
(
−
π
,
π
)
{\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 corollary 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 L 1 (-π,π). If the left and right limits f (x 0 ±0) of f (x ) exist at x 0 , or if both limits are infinite of the same sign, then
σ
n
(
x
0
)
→
1
2
(
f
(
x
0
+
0
)
+
f
(
x
0
−
0
)
)
.
{\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 σn is replaced with (C, α) mean of the Fourier series (Zygmund 1968 , Theorem III.5.1).
^ Lipót Fejér, « Sur les fonctions intégrables et bornées » , C.R. Acad. Sci. Paris , 10 décembre 1900, 984-987, .
^ Leopold Fejér, Untersuchungen über Fouriersche Reihen , Mathematische Annalen , vol. 58 , 1904, 51-69.
^ a b "Introduction" , An Introduction to Hilbert Space , Cambridge University Press, pp. 1–3, 1988-07-21, retrieved 2022-11-14
^ a b Rogosinski, W. W.; Rogosinski, H. P. (December 1965). "An elementary companion to a theorem of J. Mercer". Journal d'Analyse Mathématique . 14 (1): 311–322. doi :10.1007/bf02806398 . ISSN 0021-7670 .