Fourier Transform and Convolution


To prove that,

$$\mathcal{F}(f\otimes g) = \mathcal{F}(f)\mathcal{F}(g) \leftrightarrow\mathcal{F}(f\cdot g) = \mathcal{F}(f)\otimes\mathcal{F}(g)$$

Given that,

$$f\otimes g = \int^\infty_{-\infty}f(\tau)g(t-\tau)d\tau \\ \mathcal{F}(f(t)) = \int^\infty_{-\infty}f(t)e^{-i\omega t}dt$$

So,

\begin{align*} &\mathcal{F}(f\otimes g)=\int^\infty_{-\infty}dt\int^\infty_{-\infty}d\tau f(\tau)g(t-\tau)e^{-i\omega t} \\ &=\int^\infty_{-\infty}f(\tau)e^{-i\omega\tau}d\tau\int^\infty_{-\infty}g(t-\tau)e^{-i\omega(t-\tau)}d(t-\tau) = \mathcal{F}(f)\mathcal{F}(g) \end{align*}

Next, \(\hat{f}(\omega)=\mathcal{F}(f(t))\)

\begin{align*} \hat{f}\otimes\hat{g} &= \int^\infty_{-\infty}\hat{f}(\lambda)\hat{g}(\omega-\lambda)d\lambda \\ \mathcal{F}^{-1}(\hat{f}\otimes\hat{g}) &= \int^\infty_{-\infty}d\lambda\int^\infty_{-\infty}d\omega \hat{f}(\lambda)\hat{g}(\omega-\lambda)e^{i\omega t} \\ &= \int^\infty_{-\infty}\hat{f}(\lambda)e^{i\lambda t}d\lambda\int^\infty_{-\infty}\hat{g}(\omega-\lambda)e^{i(\omega-\lambda)t}d(\omega-\lambda) = f\cdot g\\ \mathcal{F}(f\cdot g) &= \hat{f}\otimes\hat{g}\\ \end{align*}

References