Periodische Funktion/Komplexe Exponentialfunktion/Orthonormalsystem/L^2-Raum/Fakt/Beweis

Beweis

Es ist

{}{\begin{aligned}\left\langle f_{m},f_{n}\right\rangle &=\int _{0}^{T}f_{m}{\overline {f_{n}}}dx\\&=\int _{0}^{T}{\frac {1}{\sqrt {T}}}e^{\omega {\mathrm {i} }mx}{\frac {1}{\sqrt {T}}}{\overline {e^{\omega {\mathrm {i} }nx}}}dx\\&={\frac {1}{T}}\int _{0}^{T}e^{\omega {\mathrm {i} }mx}e^{-\omega {\mathrm {i} }nx}dx\\&={\frac {1}{T}}\int _{0}^{T}e^{\omega {\mathrm {i} }(m-n)x}dx.\end{aligned}}

Bei ${}m=n$ ist dies

{}{\frac {1}{T}}\int _{0}^{T}e^{0}dx={\frac {1}{T}}\int _{0}^{T}1dx=1\,.

Bei ${}m\neq n$ ist dies

{}{\begin{aligned}{\frac {1}{T}}\int _{0}^{T}e^{\omega {\mathrm {i} }(m-n)x}dx&={\frac {1}{T\omega {\mathrm {i} }(m-n)}}{\left(e^{\omega {\mathrm {i} }(m-n)x}\right)}{|}_{0}^{T}\\&={\frac {1}{2\pi {\mathrm {i} }(m-n)}}{\left(e^{T\omega {\mathrm {i} }(m-n)}-e^{0}\right)}\\&={\frac {1}{2\pi {\mathrm {i} }(m-n)}}{\left(e^{2\pi {\mathrm {i} }(m-n)}-e^{0}\right)}\\&={\frac {1}{2\pi {\mathrm {i} }(m-n)}}{\left(1-1\right)}\\&=0.\end{aligned}}