Fourier transfrom: Understanding example with solution

Question

The fourier transform $\hat{f}(\omega)$ of the complex, quadratic and integrable function $f(t)$ is:

$\hat{f}(\omega)=\frac{1}{\sqrt{2\pi} }\int_{-\infty }^{\infty } \! f(t)e^{-i \omega t} \, dt $.

Give a proof of the linearity of the fourier transform. For $f(t)=a_{1}f_{1}(t)+a_{2}f_{2}(t)$ is valid $\hat{f}(\omega)=a_{1}\hat{f}_{1}(\omega)+a_{2}\hat{f}_{2}(\omega)$.

Solution:

$\hat{f}(\omega)=\frac{1}{\sqrt{2\pi} }\int_{-\infty }^{\infty } \! f(t)e^{-i \omega t} \, dt $.

$\hat{f}(\omega)=\frac{1}{\sqrt{2\pi} }\int_{-\infty }^{\infty } \! (f(t)a_{1}f_{1}(t)+a_{2}f_{2}(t))e^{-i \omega t} \, dt $.

$ \hat{f}(\omega)=\frac{1}{\sqrt{2\pi} }\left[a_{1}\int_{-\infty }^{\infty } \! f_{1}(t)e^{-i \omega t} \, dt +a_{2}\int_{-\infty }^{\infty } \! f_{2}(t)e^{-i \omega t} \, dt\right] $

$\hat{f}(\omega)=a_{1}\hat{f}_{1}(\omega)+a_{2}\hat{f}_{2}(\omega)$

Problem: I dont understand the last step. Where is $\frac{1}{\sqrt{2\pi}} $ and how they got $\hat{f_{1}}$ and $\hat{f_{2}}$ in the last step?

Answer 1

$\frac 1{\sqrt{2\pi}}$ is the normalization constant here in order for the transformation to remain Unitary.
Don't worry about that because a constant does not affect linearity.

By this definition: $\hat{f}(\omega)=\frac{1}{\sqrt{2\pi} }\int_{-\infty }^{\infty } \! f(t)e^{-i \omega t} \, dt$

and in the last step you have derived an expression with $\frac 1{\sqrt{2\pi}}\left[a_{1}\int_{-\infty }^{\infty } \! f_{1}(t)e^{-i \omega t} \, dt +a_{2}\int_{-\infty }^{\infty } \! f_{2}(t)e^{-i \omega t} \, dt\right]$

Directly from the definition, the first term is $\hat{f}(\omega)=a_{1}\hat{f}_{1}(\omega)+a_{2}\hat{f}_{2}(\omega)$. The square root of $2\pi$ "goes away" into the definition of $ \hat{f}_1$ and $\hat{f}_2$