# fourier transform is a bijective transformation

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?

• 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. – Bungo Nov 21 '16 at 18:56
• Can you help me please? – user384789 Nov 21 '16 at 19:03
• 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. – Bungo Nov 21 '16 at 19:06
• 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. – Bungo Nov 21 '16 at 19:10