Exponential Function Set and Trigonometric Function Set are Orthogonal Bases
📂Fourier Analysis Exponential Function Set and Trigonometric Function Set are Orthogonal Bases Theorem Both sets { e i n x } n = − ∞ ∞ \left\{ e^{inx} \right\}_{n=-\infty}^\infty { e in x } n = − ∞ ∞ { cos n x } n = 0 ∞ ∪ { sin n x } n = 1 ∞ \left\{ \cos nx\ \right\}_{n=0}^\infty \cup \left\{ \sin nx \right\}_{n=1}^\infty { cos n x } n = 0 ∞ ∪ { sin n x } n = 1 ∞ orthonormal bases of L 2 ( − π , π ) L^{2}(-\pi,\ \pi) L 2 ( − π , π ) { cos n x } n = 0 ∞ \left\{ \cos nx \right\}_{n=0}^{\infty} { cos n x } n = 0 ∞ { sin n x } n = 1 ∞ \left\{ \sin nx \right\}_{n=1}^{\infty} { sin n x } n = 1 ∞ L 2 ( 0 , π ) L^{2}(0,\ \pi) L 2 ( 0 , π )
Explanation This fact explains the reason why it is valid to express a given function as a series of trigonometric functions in Fourier series .
Proof Let ϕ n ( x ) = e i n x \phi_{n}(x)=e^{inx} ϕ n ( x ) = e in x f ∈ L 2 ( − π , π ) f \in L^{2}(-\pi,\ \pi) f ∈ L 2 ( − π , π ) ϵ \epsilon ϵ
Lemma
For any f ∈ L 2 ( a , b ) f \in L^{2}(a,\ b) f ∈ L 2 ( a , b ) { f n } \left\{ f_{n} \right\} { f n } ∣ f n − f ∣ → 0 | f_{n} - f | \rightarrow 0 ∣ f n − f ∣ → 0 [ a , b ] [a,\ b] [ a , b ]
Then, according to the lemma, there exists f ~ \tilde{f} f ~
∣ f − f ~ ∣ < ϵ 3
\begin{equation}
| f-\tilde{f} | < \frac{\epsilon}{3}
\label{eq1}
\end{equation}
∣ f − f ~ ∣ < 3 ϵ
Let’s say c n = 1 2 π ⟨ f , ϕ n ⟩ c_{n}=\frac{1}{2\pi}\langle f,\ \phi_{n} \rangle c n = 2 π 1 ⟨ f , ϕ n ⟩ c ~ n = 1 2 π ⟨ f ~ , ϕ n ⟩ \tilde{c}_{n}=\frac{1}{2\pi}\langle \tilde{f},\ \phi_{n} \rangle c ~ n = 2 π 1 ⟨ f ~ , ϕ n ⟩ Fourier coefficients of f f f f ~ \tilde{f} f ~ f ~ \tilde{f} f ~ ∑ c ~ n ϕ n \sum \tilde{c}_{n}\phi_{n} ∑ c ~ n ϕ n converges uniformly to f ~ \tilde{f} f ~ since it converges uniformly, it converges in norm . Hence, the following equation holds.
∥ f ~ − ∑ − N N c ~ n ϕ n ∥ < ϵ 3
\begin{equation}
\left\| \tilde{f} -\sum \limits_{-N}^{N} \tilde{c}_{n}\phi_{n} \right\| < \dfrac{\epsilon}{3}
\label{eq2}
\end{equation}
f ~ − − N ∑ N c ~ n ϕ n < 3 ϵ
And the following equation holds.
∥ ∑ − N N c ~ n ϕ n − ∑ − N N c n ϕ n ∥ 2 = ∥ ∑ − N N ( c ~ n − c n ) ϕ n ∥ 2 = ∑ − N N ∣ c ~ n − c n ∣ 2 ∣ ϕ n ∣ 2 = ∑ − N N 1 2 π ∣ ⟨ f ~ − f , ϕ n ⟩ ∣ 2 ∣ ϕ n ∣ 2 = ∑ − N N 1 2 π ∣ ⟨ f ~ − f , ϕ n ⟩ ∣ 2 ≤ ∑ − ∞ ∞ 1 2 π ∣ ⟨ f ~ − f , ϕ n ⟩ ∣ 2 ≤ ∣ f ~ − f ∣ 2 < ( ϵ 3 ) 2
\begin{align}
\left\| \sum\limits_{-N}^{N}\tilde{c}_{n}\phi_{n} - \sum\limits_{-N}^{N}c_{n}\phi_{n}\right\| ^{2} &= \left\| \sum\limits_{-N}^{N} (\tilde{c}_{n}-c_{n}) \phi_{n} \right\| ^{2} \nonumber \\
&= \sum \limits_{-N}^{N} | \tilde{c}_{n} -c_{n} |^{2} | \phi_{n} |^{2} \nonumber \\
&= \sum \limits_{-N}^{N} \dfrac{1}{2\pi} | \langle \tilde{f}-f,\ \phi_{n} \rangle |^{2} | \phi_{n}|^{2} \nonumber \\
&= \sum \limits_{-N}^{N} \dfrac{1}{2\pi} | \langle \tilde{f}-f,\ \phi_{n} \rangle |^{2} \nonumber \\
&\le \sum \limits_{-\infty}^{\infty} \dfrac{1}{2\pi} | \langle \tilde{f}-f,\ \phi_{n} \rangle |^{2} \nonumber \\
&\le | \tilde{f} - f | ^{2} \nonumber \\
&< \left( \dfrac{\epsilon}{3} \right)^{2} \label{eq3}
\end{align}
− N ∑ N c ~ n ϕ n − − N ∑ N c n ϕ n 2 = − N ∑ N ( c ~ n − c n ) ϕ n 2 = − N ∑ N ∣ c ~ n − c n ∣ 2 ∣ ϕ n ∣ 2 = − N ∑ N 2 π 1 ∣ ⟨ f ~ − f , ϕ n ⟩ ∣ 2 ∣ ϕ n ∣ 2 = − N ∑ N 2 π 1 ∣ ⟨ f ~ − f , ϕ n ⟩ ∣ 2 ≤ − ∞ ∑ ∞ 2 π 1 ∣ ⟨ f ~ − f , ϕ n ⟩ ∣ 2 ≤ ∣ f ~ − f ∣ 2 < ( 3 ϵ ) 2
The fourth equality holds because of ∣ ϕ n ∣ = 1 | \phi_{n} |=1 ∣ ϕ n ∣ = 1 Bessel’s inequality . The last line holds due to the assumption ( eq1 ) \eqref{eq1} ( eq1 ) ( 1 ) (1) ( 1 ) ( 2 ) (2) ( 2 ) ( 3 ) (3) ( 3 ) ∣ f − ∑ c n ϕ n ∣ → 0 | f -\sum c_{n}\phi_{n} | \rightarrow 0 ∣ f − ∑ c n ϕ n ∣ → 0
f − ∑ − N N c n ϕ n = ( f − f ~ ) + ( f ~ − ∑ − N N c ~ n ϕ n ) + ( ∑ − N N c ~ n ϕ n − ∑ − N N c n ϕ n )
f-\sum \limits_{-N}^{N}c_{n}\phi_{n} = (f-\tilde{f}) + \left( \tilde{f} -\sum _{-N}^{N} \tilde{c}_{n}\phi_{n} \right) + \left( \sum _{-N}^{N} \tilde{c}_{n}\phi_{n} - \sum _{-N}^{N}c_{n}\phi_{n}\right)
f − − N ∑ N c n ϕ n = ( f − f ~ ) + ( f ~ − − N ∑ N c ~ n ϕ n ) + ( − N ∑ N c ~ n ϕ n − − N ∑ N c n ϕ n )
Since the above equation holds, by the triangle inequality,
∥ f − ∑ − N N c n ϕ n ∥ ≤ ∥ f − f ~ ∥ + ∥ f ~ − ∑ − N N c ~ n ϕ n ∥ + ∥ ∑ − N N c ~ n ϕ n − ∑ − N N c n ϕ n ∥ = ϵ 3 + ϵ 3 + ϵ 3 = ϵ
\begin{align*}
\left\| f-\sum\limits_{-N}^{N}c_{n}\phi_{n} \right\| &\le \| f-\tilde{f} \| + \left\| \tilde{f} -\sum _{-N}^{N} \tilde{c}_{n}\phi_{n} \right\| + \left\| \sum _{-N}^{N} \tilde{c}_{n}\phi_{n} - \sum _{-N}^{N}c_{n}\phi_{n} \right\|
\\ &= \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3}
\\ &= \epsilon
\end{align*}
f − − N ∑ N c n ϕ n ≤ ∥ f − f ~ ∥ + f ~ − − N ∑ N c ~ n ϕ n + − N ∑ N c ~ n ϕ n − − N ∑ N c n ϕ n = 3 ϵ + 3 ϵ + 3 ϵ = ϵ
Therefore,
∥ f − ∑ − ∞ ∞ c n ϕ n ∥ → 0
\left\| f- \sum \limits_{-\infty}^{\infty} c_{n}\phi_{n} \right\| \rightarrow 0
f − − ∞ ∑ ∞ c n ϕ n → 0
And according to condition (b) , { ϕ n } \left\{ \phi_{n} \right\} { ϕ n }
■
The rest of the cases are essentially the same or can be proven by almost the same process, so they are omitted.
