불연속점에서 푸리에 급수의 수렴성
📂푸리에해석 불연속점에서 푸리에 급수의 수렴성 정리 구간 [ − L , L ) [-L,\ L) [ − L , L ) f ( t ) f(t) f ( t ) 조각마다 연속 이라고 하자. 불연속점을 t i ( i = 1 , ⋯ m ) t_{i}\ (i=1,\ \cdots m ) t i ( i = 1 , ⋯ m ) f ( a − ) f(a-) f ( a − ) f ( a + ) f(a+) f ( a + ) f ( t ) f(t) f ( t ) t i t_{i} t i
a 0 2 + ∑ n = 1 ∞ ( a n cos n π t i L + b n sin n π t i L ) = f ( t i + ) + f ( t i − ) 2
\dfrac{a_{0}}{2}+\sum \limits_{n=1}^{\infty}\left( a_{n} \cos \dfrac{n \pi t_{i} }{L} +b_{n}\sin\dfrac{n\pi t_{i}}{L} \right) = \dfrac{f(t_{i}+)+f(t_{i}-)}{2}
2 a 0 + n = 1 ∑ ∞ ( a n cos L nπ t i + b n sin L nπ t i ) = 2 f ( t i + ) + f ( t i − )
f f f f f f t t t f f f
증명 임의의 불연속 점을 t i = t t_{i}=t t i = t
푸리에 급수와 디리클레 커널의 관계
S N f ( t ) = 1 L ∫ − L L f ( x ) D N ( π ( x − t ) L ) d x
S^{f}_{N}(t)=\dfrac{1}{L}\int_{-L}^{L}f(x)D_{N}\left(\dfrac{\pi (x-t)}{L}\right)dx
S N f ( t ) = L 1 ∫ − L L f ( x ) D N ( L π ( x − t ) ) d x
위 관계식을 통해 아래의 식을 얻는다.
lim N → ∞ S N ( t ) = lim N → ∞ 1 L ∫ − L L f ( x ) D N ( π ( x − t ) L ) d x = lim N → ∞ 1 L ∫ − L − t L − t f ( λ + t ) D N ( π λ L ) d λ = lim N → ∞ 1 L ∫ − L L f ( λ + t ) D N ( π λ L ) d λ = lim N → ∞ 1 L ∫ − L 0 f ( λ + t ) D N ( π λ L ) d λ + lim N → ∞ 1 L ∫ 0 L f ( λ + t ) D N ( π λ L ) d λ
\begin{align}
& \lim_{N \rightarrow \infty} S_{N} (t) \nonumber \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(x)D_{N}\left( \dfrac{\pi (x-t)}{L} \right) dx \nonumber \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L-t}^{L-t} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L} \right) d\lambda \nonumber \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L} \right) d\lambda \nonumber \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{0} f(\lambda + t)D_{N}\left(\dfrac{\pi \lambda }{L}\right) d\lambda +\lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L} f(\lambda +t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda
\end{align}
= = = = N → ∞ lim S N ( t ) N → ∞ lim L 1 ∫ − L L f ( x ) D N ( L π ( x − t ) ) d x N → ∞ lim L 1 ∫ − L − t L − t f ( λ + t ) D N ( L πλ ) d λ N → ∞ lim L 1 ∫ − L L f ( λ + t ) D N ( L πλ ) d λ N → ∞ lim L 1 ∫ − L 0 f ( λ + t ) D N ( L πλ ) d λ + N → ∞ lim L 1 ∫ 0 L f ( λ + t ) D N ( L πλ ) d λ
두번째 등호는 x − t = λ x-t=\lambda x − t = λ ( 2 L ) (2L) ( 2 L )
디리클레 커널의 적분
1 L ∫ − L L D N ( π ( x − t ) L ) d x = 1
\dfrac{1}{L}\int_{-L}^{L}D_{N}\left( \dfrac{\pi (x-t)}{L} \right)dx = 1
L 1 ∫ − L L D N ( L π ( x − t ) ) d x = 1
디리클레 커널은 우함수이므로 위 식으로부터 다음의 식을 얻는다.
1 L ∫ 0 L D N ( π λ L ) d λ = 1 L ∫ − L 0 D N ( π λ L ) d λ = 1 2
\dfrac{1}{L} \int_{0}^{L} D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda=\dfrac{1}{L} \int_{-L}^{0} D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda=\dfrac{1}{2}
L 1 ∫ 0 L D N ( L πλ ) d λ = L 1 ∫ − L 0 D N ( L πλ ) d λ = 2 1
따라서 아래의 식이 성립한다.
1 2 f ( t + ) = f ( t + ) 1 L ∫ 0 L D N ( π λ L ) d λ = 1 L ∫ 0 L f ( t + ) D N ( π λ L ) d λ 1 2 f ( t − ) = f ( t − ) 1 L ∫ − L 0 D N ( π λ L ) d λ = 1 L ∫ − L 0 f ( t − ) D N ( π λ L ) d λ
\begin{equation}
\begin{aligned}
\dfrac{1}{2}f(t+) &= f(t+)\dfrac{1}{L} \int_{0}^{L} D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda=\dfrac{1}{L} \int_{0}^{L} f(t+)D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda
\\ \dfrac{1}{2}f(t-) &= f(t-)\dfrac{1}{L} \int_{-L}^{0} D_{N} \left( \dfrac{\pi \lambda}{L} \right) d\lambda=\dfrac{1}{L} \int_{-L}^{0} f(t-)D_{N} \left( \dfrac{\pi \lambda }{L} \right) d\lambda
\end{aligned}
\end{equation}
2 1 f ( t + ) 2 1 f ( t − ) = f ( t + ) L 1 ∫ 0 L D N ( L πλ ) d λ = L 1 ∫ 0 L f ( t + ) D N ( L πλ ) d λ = f ( t − ) L 1 ∫ − L 0 D N ( L πλ ) d λ = L 1 ∫ − L 0 f ( t − ) D N ( L πλ ) d λ
( 1 ) (1) ( 1 ) ( 2 ) (2) ( 2 )
lim N → ∞ 1 L ∫ 0 L f ( λ + t ) D N ( π λ L ) d λ − 1 2 f ( t + ) = lim N → ∞ 1 L ∫ 0 L ( f ( λ + t ) − f ( t + ) ) D N ( π λ L ) d λ = lim N → ∞ 1 L ∫ 0 L ( f ( λ + t ) − f ( t + ) ) ) sin ( N + 1 2 ) π λ L 2 sin π λ 2 L d λ = lim N → ∞ 1 L ∫ 0 L g + ( λ ) sin [ ( N + 1 2 ) π λ L ] d λ
\begin{align*}
& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda - \dfrac{1}{2}f(t+) \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L}\Big( f(\lambda + t) -f(t+) \Big) D_{N}\left(\dfrac{\pi \lambda }{L}\right) d\lambda \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L}\Big( f(\lambda + t) -f(t+)) \Big)\dfrac{\sin\left(N+\dfrac{1}{2}\right) \dfrac{\pi \lambda}{L}}{2\sin \dfrac{\pi \lambda}{2L}} d\lambda \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L}g_+(\lambda) \sin\left[ \left(N+\dfrac{1}{2}\right) \dfrac{\pi \lambda}{L}\right] d\lambda
\end{align*}
= = = N → ∞ lim L 1 ∫ 0 L f ( λ + t ) D N ( L πλ ) d λ − 2 1 f ( t + ) N → ∞ lim L 1 ∫ 0 L ( f ( λ + t ) − f ( t + ) ) D N ( L πλ ) d λ N → ∞ lim L 1 ∫ 0 L ( f ( λ + t ) − f ( t + )) ) 2 sin 2 L πλ sin ( N + 2 1 ) L πλ d λ N → ∞ lim L 1 ∫ 0 L g + ( λ ) sin [ ( N + 2 1 ) L πλ ] d λ
이때, g + ( λ ) = f ( λ + t ) − f ( t + ) λ λ 2 sin π λ 2 L g_+(\lambda) = \dfrac{ f(\lambda + t) -f(t+) }{\lambda}\dfrac{\lambda}{2\sin \dfrac{\pi \lambda}{2L}} g + ( λ ) = λ f ( λ + t ) − f ( t + ) 2 sin 2 L πλ λ
푸리에 급수와 디리클레 커널의 관계
S N f ( t ) = 1 L ∫ − L L f ( x ) D n ( π ( x − t ) L ) d x
S_{N}^{f} (t)=\dfrac{1}{L}\int_{-L}^{L}f(x)D_{n}\left(\dfrac{\pi (x-t)}{L}\right)dx
S N f ( t ) = L 1 ∫ − L L f ( x ) D n ( L π ( x − t ) ) d x
같은 방법으로 계산하면 다음과 같다.
lim N → ∞ 1 L ∫ − L 0 f ( λ + t ) D N ( π λ L ) d λ − 1 2 f ( t − ) = lim N → ∞ 1 L ∫ − L 0 g − ( λ ) sin [ ( N + 1 2 ) π λ L ] d λ
\begin{align*}
& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{0} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda - \dfrac{1}{2}f(t-) \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{0}g_-(\lambda) \sin\left[ \left(N+\dfrac{1}{2}\right) \dfrac{\pi \lambda}{L}\right] d\lambda
\end{align*}
= N → ∞ lim L 1 ∫ − L 0 f ( λ + t ) D N ( L πλ ) d λ − 2 1 f ( t − ) N → ∞ lim L 1 ∫ − L 0 g − ( λ ) sin [ ( N + 2 1 ) L πλ ] d λ
이때, g − ( λ ) = f ( λ + t ) − f ( t − ) λ λ 2 sin π λ 2 L g_-(\lambda) = \dfrac{ f(\lambda + t) -f(t-) }{\lambda}\dfrac{\lambda}{2\sin \dfrac{\pi \lambda}{2L}} g − ( λ ) = λ f ( λ + t ) − f ( t − ) 2 sin 2 L πλ λ g ± ( λ ) g_\pm(\lambda) g ± ( λ )
lim λ → 0 λ 2 sin π λ 2 L = lim λ → 0 π λ 2 L sin π λ 2 L L π = L π
\lim \limits_{\lambda \rightarrow 0} \dfrac{\lambda}{2\sin \dfrac{\pi \lambda}{2L}} =\lim \limits_{\lambda \rightarrow 0} \dfrac{\dfrac{\pi \lambda}{2L}}{\sin \dfrac{\pi \lambda}{2L}} \dfrac{L}{\pi}=\dfrac{L}{\pi}
λ → 0 lim 2 sin 2 L πλ λ = λ → 0 lim sin 2 L πλ 2 L πλ π L = π L
이므로 임의의 λ ∈ [ − L , L ) \lambda \in [-L,\ L) λ ∈ [ − L , L ) M 1 > 0 M_{1}>0 M 1 > 0
∣ λ 2 sin π λ 2 L ∣ ≤ M 1 < ∞
\left| \dfrac{\lambda}{2\sin \dfrac{\pi \lambda}{2L}}\right| \le M_{1} <\infty
2 sin 2 L πλ λ ≤ M 1 < ∞
0 0 0 0 0 0 f f f t t t λ ∈ ( 0 , L ) \lambda \in (0,\ L) λ ∈ ( 0 , L ) M 2 M_2 M 2
∣ f ( λ + t ) − f ( t + ) λ ∣ ≤ M 2 < ∞
\left| \dfrac{f(\lambda+t) -f(t+)}{\lambda} \right| \le M_2 <\infty
λ f ( λ + t ) − f ( t + ) ≤ M 2 < ∞
마찬가지로 구간 내에서 조각마다 연속이다.위의 두 사실에 의해 g + ( λ ) g_+(\lambda) g + ( λ ) [ 0 , L ) [0,\ L) [ 0 , L )
lim N → ∞ 1 L ∫ 0 L f ( λ + t ) D N ( π λ L ) d λ − 1 2 f ( t + ) = lim N → ∞ 1 L ∫ 0 L g + ( λ ) sin [ ( N + 1 2 ) π λ L ] d λ = 0
\begin{align*}
& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda - \dfrac{1}{2}f(t+) \\
=& \lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L}g_+(\lambda) \sin\left[ \left(N+\dfrac{1}{2}\right) \dfrac{\pi \lambda}{L} \right]d\lambda \\
=&\ 0
\end{align*}
= = N → ∞ lim L 1 ∫ 0 L f ( λ + t ) D N ( L πλ ) d λ − 2 1 f ( t + ) N → ∞ lim L 1 ∫ 0 L g + ( λ ) sin [ ( N + 2 1 ) L πλ ] d λ 0
g + ( λ ) g_+(\lambda) g + ( λ )
리만-르벡 보조정리
함수 f ( t ) f(t) f ( t ) [ − L , L ) [-L,\ L) [ − L , L )
lim n → ∞ a n = lim n → ∞ 1 L ∫ − L L f ( t ) cos n π L t d t = 0
\lim \limits_{n \rightarrow \infty} a_{n} = \lim \limits_{n \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(t) \cos \dfrac{n \pi}{L}t dt=0
n → ∞ lim a n = n → ∞ lim L 1 ∫ − L L f ( t ) cos L nπ t d t = 0
lim n → ∞ b n = lim n → ∞ 1 L ∫ − L L f ( t ) sin n π L t d t = 0
\lim \limits_{n \rightarrow \infty} b_{n} = \lim \limits_{n \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{L} f(t) \sin \dfrac{n \pi}{L}t dt=0
n → ∞ lim b n = n → ∞ lim L 1 ∫ − L L f ( t ) sin L nπ t d t = 0
따라서
lim N → ∞ 1 L ∫ 0 L f ( λ + t ) D N ( π λ L ) d λ = 1 2 f ( t + )
\lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{0}^{L} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda=\dfrac{1}{2}f(t+)
N → ∞ lim L 1 ∫ 0 L f ( λ + t ) D N ( L πλ ) d λ = 2 1 f ( t + )
같은 방법으로 아래의 식을 얻을 수 있다.
lim N → ∞ 1 L ∫ − L 0 f ( λ + t ) D N ( π λ L ) d λ = 1 2 f ( t − )
\lim \limits_{N \rightarrow \infty} \dfrac{1}{L} \int_{-L}^{0} f(\lambda + t)D_{N}\left( \dfrac{\pi \lambda }{L}\right) d\lambda=\dfrac{1}{2}f(t-)
N → ∞ lim L 1 ∫ − L 0 f ( λ + t ) D N ( L πλ ) d λ = 2 1 f ( t − )
두 식을 합치면
lim N → ∞ S N ( t ) = 1 2 ( f ( t + ) + f ( t − ) )
\lim \limits_{N \rightarrow \infty} S_{N}(t) = \dfrac{1}{2}\big(f(t+)+f(t-)\big)
N → ∞ lim S N ( t ) = 2 1 ( f ( t + ) + f ( t − ) )
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