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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

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

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?

Stella Biderman
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Arturo don Juan
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  • First of all, you should specify what types of functions you have in mind: Schwartz space functions, $L^1$-functions, $L^2$-functions, and so on. It affects the meaning of saying two functions are "equal." Second of all, have you looked in any books on Fourier analysis? Such an equality for some class of functions would be discussed in almost any of them. – KCd Jan 13 '15 at 06:09
  • I'm only getting this off of a section on a book I have on introductory complex analysis. The function $f(t)$ must be continuous, but that is the only constraint that I can see in this introductory section, and I have in mind Schwartz space functions. – Arturo don Juan Jan 13 '15 at 06:16
  • So no, I have not looked in any books on Fourier Analysis. I thought the fourier-analysis tag was appropriate nevertheless. – Arturo don Juan Jan 13 '15 at 06:17
  • I agree the tag is appropriate, but it feels like you have not put in effort yourself to figure this out with your own studying yet. Take the time to look in a book or online notes about Fourier analysis, which quite likely will have a proof of this very result. Just look up Fourier inversion formula in the index. – KCd Jan 13 '15 at 06:20

3 Answers3

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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) $

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

Stella Biderman
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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 $f(t) = \frac{1}{2\pi}\int_{-\infty}^\infty F(w)e^{jwt}dw$

Ziao
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