Inverse Fourier Transform of Fourier Transform

Question

Given the Fourier Tranform defined as $$\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi):=\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}e^{-2\pi i \xi t}dt=\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}$$ They say inverse Fourier Transform is given by $$\color{blue}{\mathcal{F}^{-1}\{}\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}\color{blue}{\}}(t)=\int_{-\infty}^{+\infty}\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}e^{2\pi i \xi t}d\xi=\color{green}{f(}t\color{green}{)}$$ I'm expecting that $$\color{blue}{\mathcal{F}^{-1}\{}\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi)\color{blue}{\}}(t)=\color{green}{f(}t\color{green}{)}$$ But actually $$\color{blue}{\mathcal{F}^{-1}\{}\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi)\color{blue}{\}}(t)=\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}e^{-2\pi i \xi t}dt)e^{2\pi i \xi t}d\xi=$$$$=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}dtd\xi=\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}d\xi)\color{green}{f(}t\color{green}{)}dt=\int_{-\infty}^{+\infty}0\cdot \color{green}{f(}t\color{green}{)}dt=0 \text{ ??}$$

Answer 1

The first integral should be \begin{align} I(u) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-i2\pi\xi t}e^{i2\pi\xi u}dtd\xi\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-i2\pi\xi t+i2\pi\xi u}dtd\xi\\ &=\int_{-\infty}^{\infty}f(t)\left(\int_{-\infty}^{\infty}e^{i2\pi\xi (u-t)}d\xi\right)dt\\ \end{align} We have $$\delta(u-t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(u-t)x}dx = \int_{-\infty}^{\infty}e^{i2\pi(u-t)\xi}d\xi $$ Thereby it becomes \begin{align} I(u) &= \int_{-\infty}^{\infty}f(t)\delta(u-t)dt\\ &= f(u) \end{align} The function as mapping is returned, doesn't matter how its argument is denoted

Answer 2

Let $f \in L^1(\mathbb{R})$ so that $\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} \, dx$ is defined. Then $$ |\hat{f}(\xi)| = \left|\int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} dx\right| \leq \int_{-\infty}^{\infty} \left|f(x) \, e^{-ix\xi}\right| \, dx = \int_{-\infty}^{\infty} \left|f(x) \right| \, dx < \infty $$ so $\hat{f}\in L^\infty(\mathbb{R})$ but we cannot be sure that $\hat{f} \in L^1(\mathbb{R}).$ Therefore, let us define $\hat{f_\epsilon}(\xi) := e^{-\epsilon \xi^2/2}\hat{f}(\xi)$ so that $\hat{f_\epsilon} \in L^1(\mathbb{R}),$ and at the end take the limit $\epsilon\to 0.$ We then have \begin{align} \int_{-\infty}^{\infty} \hat{f_\epsilon}(\xi) \, e^{iy\xi} \, d\xi &= \int_{-\infty}^{\infty} e^{-\epsilon \xi^2/2}\hat{f}(\xi) \, e^{iy\xi} \, d\xi \\ &= \int_{-\infty}^{\infty} e^{-\epsilon \xi^2/2} \left( \int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} \, dx \right) \, e^{iy\xi} \, d\xi \\ \\ &= \int_{-\infty}^{\infty} f(x) \left( \int_{-\infty}^{\infty} e^{-\epsilon \xi^2/2} \, e^{-ix\xi} \, e^{iy\xi} \, d\xi \right) \, dx \\ &= \int_{-\infty}^{\infty} f(x) \sqrt{\frac{2\pi}{\epsilon}} e^{-(x-y)^2/(2\epsilon)} \, dx \\ \\ &\overset{\{ x=y+z\sqrt{2\epsilon} \}}{=} \int_{-\infty}^{\infty} f(y+z\sqrt{2\epsilon}) \sqrt{\frac{2\pi}{\epsilon}} e^{-z^2} \, \sqrt{2\epsilon} \, dz \\ &= 2\sqrt{\pi} \int_{-\infty}^{\infty} f(y+z\sqrt{2\epsilon}) \, e^{-z^2} \, dz \\ &\to 2\sqrt{\pi} \int_{-\infty}^{\infty} f(y) \, e^{-z^2} \, dz = 2\sqrt{\pi} \int_{-\infty}^{\infty} \, e^{-z^2} \, dz \, f(y) = 2\pi \, f(y). \end{align}