How to show the Fourier Inverse of the Fourier transform is the identity transformation?

Question

Let $\mathcal{F}\left[f(t)\right](x)$ be the Fourier Transform of $f$, defined regularly as

$$\mathcal{F}\left[f(t)\right](x)=\int_{-\infty}^{\infty}f(t)e^{-itx}\,dt$$

And let $\mathcal{F}^{-1}\left[g(x)\right](t)$ be the Inverse Fourier Transform of $g$, defined regularly as $$ \mathcal{F}^{-1}\left[g(x)\right](t)= \int_{-\infty}^{\infty}g(x)e^{ixt}\,dx$$

Show that $\mathcal{F}^{-1}\left[\mathcal{F}\left[f(t)\right](x)\right] (t) = f(t)$. In shorter notation, show that

$$\mathcal{F}^{-1} \mathcal{F} = I$$

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I can't seem to find nor work out myself a proper proof of this identity. Could you possibly provide me with one, or even a good hint to one?

Answer 1

There is no $e^{ixt}e^{-ixt}$ but $e^{ixt}e^{-ixw}$ in Fourier transform.
The key is $ \int_{-\infty}^{\infty} e^{-jwt} \, dw = 2\pi \delta(t) $.
I'm going to show:
$ f(s) = \frac{1}{2\pi} \int_{-\infty}^{\infty} [ \int_{-\infty}^{\infty} f(t)e^{-jwt} \, dt ] e^{jws} \, dw \\ =\frac{1}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(t)e^{-jw(t-s)} \, dt dw \\ =\frac{1}{2\pi} \int_{-\infty}^{\infty}f(t) 2\pi \delta(t-s) \, dt \\ =f(s) $

Answer 2

Hint: Try composing the two formally and noting that $e^{ixt}e^{-ixt}=1$

Answer 3

Let the Fourier transform be $F(w) = \int_{-\infty}^\infty f(t)e^{-jwt}dt$ (eqn 1)

We want to prove that $f(t) = \frac{1}{2\pi}\int_{-\infty}^\infty F(w)e^{jwt}dw$ (eqn 2)

Check out the proof of the "inversion formula for characteristic function", the proof of the following equation is similar to the proof of the inversion formula.

$\int_{t1}^{t2}f(t)dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{jwt2}-e^{jwt1}}{jw}F(w)dw$ (eqn 3, assume t1<t2)

The way to prove (3) is to define

$J_T= \frac{1}{2\pi}\int_{-T}^{T}\frac{e^{jwt2}-e^{jwt1}}{jw}F(w)dw = \frac{1}{2\pi}\int_{-T}^{T}\frac{e^{jwt2}-e^{jwt1}}{jw}\int_{-\infty}^{\infty}f(t)e^{-jwt}dtdw$

swap the order of the integral, and find that

$J_T=\frac{1}{\pi}\int_{-\infty}^{\infty}[\int_{0}^{T}(\frac{sin(w(t-t1))}{w}-\frac{sin(w(t-t2))}{w})dw]f(t)dt$

$\lim_{T\to\infty}J_T=\int_{t1}^{t2}f(t)dt$, so (3) is proved

Let $t2 = t1 + \Delta t$, divide both side of equation (3) by $\Delta t$ , and get $f(t) = \frac{1}{2\pi}\int_{-\infty}^\infty F(w)e^{jwt}dw$