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I'm not at all pure math minded so I don't know which direction to take this problem. It reads:

Let $F$ denote the Fourier transform, i.e. $F[f(x)]=\hat f (\omega)$. Prove the following statements:

a) Shift property on the $x$ axis: $F[f(x-a)]=e^{-ia\omega}F[f(x)]$

b: Shift property on the $\omega$ axis: $F[e^{ia\omega}f(x)]=\hat f(\omega-a)$

c) Dilation: $F[f(ax)](\omega)=\frac{1}{|a|}F[f](\frac{\omega}{a})$

We have defined the Fourier transform as $F[f]=\hat f(a)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{-iax}dx$.

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(a) $$F[f(x-a)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x-a)e^{-i\omega x}dx$$

Define $y=x-a$, $dy=dx$ The limits doesn't change.

$$F[f(x-a)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(y)e^{-i\omega(y+a)}dy=$$

$$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(y)e^{-i\omega a}e^{-i\omega y}dy=$$

$$=\frac{e^{-i\omega a}}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(y)e^{-i\omega y}dy=e^{-i\omega a}F[f(x)]$$

(b) Is not $e^{ia\omega}$ but $e^{iax}$ $$F[e^{iax}f(x)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{iax}f(x)e^{-i\omega x}dx=$$

$$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{-i(\omega-a) x}dx=\hat f(\omega-a)$$

(c)

$a>0$ $$F[f(ax)](\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(ax)e^{-i\omega x}dx=$$

Define $y=ax$, with $dx=dy/a$

$$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(1/a)f(y)e^{-i\omega(y/a)}dy=$$

$$=\frac{1}{a}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(y)e^{-i(\omega/a)x}dy=\frac{1}{\vert a\vert}\hat f\left(\frac{\omega}{a}\right)$$

$a=-\vert a\vert\neq 0$

$$F[f(ax)](\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(ax)e^{-i\omega x}dx=$$

Define $y=ax$, with $dx=dy/a$. The limits change as the variable changes sign($x\to ax=-\vert a\vert x$)

$$=\frac{1}{\sqrt{2\pi}}\int_{\infty}^{-\infty}(1/a)f(y)e^{i\omega(y/a)}dy=$$

$$=-\frac{1}{a}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(y)e^{-i(\omega/a)y}dy=$$

$$=\frac{1}{\vert a\vert}\hat f\left(\frac{\omega}{a}\right)$$

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