0
$\begingroup$

Fourier transformation picture

How to show that Fourier transform (please find picture 1) is a bijective transformation?

For injective, I did the following. Is it right? $${F}_1(x)=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f_1(x)e^{-i\omega x}dx}$$ $${F}_2(x)=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f_2(x)e^{-i\omega x}dx}$$ Suppose that ${F}_1(x)={F}_2(x)$, then $$\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f_1(x)e^{-i\omega x}dx}=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f_2(x)e^{-i\omega x}dx}$$ $$=>\int^{\infty}_{-\infty}{f_1(x)e^{-i\omega x}dx}-\int^{\infty}_{-\infty}{f_2(x)e^{-i\omega x}dx}=0$$ $$=>\int^{\infty}_{-\infty}({f_1(x)-f_2(x))e^{-i\omega x}dx}=0$$ $$=>f_1(x)=f_2(x)$$

How can we show that it is surjective?

$\endgroup$
  • $\begingroup$ On $L^2$, the integral definition of the Fourier transform may not converge, so there's rather more to the proof than what you have shown. $\endgroup$ – Bungo Nov 21 '16 at 18:56
  • $\begingroup$ Can you help me please? $\endgroup$ – user384789 Nov 21 '16 at 19:03
  • $\begingroup$ Well, proving that the Fourier transform is bijective on $L^2$ is actually a fairly long and technical proof if I recall correctly. What facts do you have available? Do you at least know the definition of the Fourier transform on $L^2$? This is typically defined by a density argument using the fact that $L^1 \cap L^2$ is dense in $L^2$, and the fact that we can calculate the Fourier transform of a function in $L^1 \cap L^2$ using the usual integral definition. $\endgroup$ – Bungo Nov 21 '16 at 19:06
  • $\begingroup$ Injectivity is a consequence of Plancherel's theorem: since $\|f_1 - f_2\|_2 = \|\hat{f_1} - \hat{f_2}\|_2$, the Fourier transforms of two $L^2$ functions are equal a.e. iff the original functions are equal a.e. $\endgroup$ – Bungo Nov 21 '16 at 19:10
2
$\begingroup$

Assuming that the functions involved are nice enough (which appears to be done implicitly in the attached text anyway), this follows from the Fourier inversion theorem, which provides an explicit inverse for the Fourier transformation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy