In mathematics, Fejér's theorem ,[1] [2] 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}}
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 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)} 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)} 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)} 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)|}
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 }
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)} s n ( f , x ) {\displaystyle s_{n}(f,x)} [4] L 2 ( − π , π ) {\displaystyle L^{2}(-\pi ,\pi )} Carleson's theorem , was proven in 1966 by L. Carleson.[4]
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 (C, α) mean of the Fourier series (Zygmund 1968 , Theorem III.5.1).
References [ edit ] ^ Lipót Fejér, « Sur les fonctions intégrables et bornées » , C.R. Acad. Sci. Paris ^ 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 . 
From Wikipedia the free encyclopedia
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f66446500795c337fdc3be37053fc095896bc13b)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92a612156ef1f6b7718a6149d9b587d43177aa17)
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db0b6b0ed79b27abdeb68777756e91b562c06b59)
![{\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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53ea7f9f0bea60afbee4c4f3346c09ee780326f5)
