The Fourier transform and its inverse can be defined as $$\mathcal{F}(f(x))=F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{-ikx} \ dx \ \ \text{and} \ \ \mathcal{F}^{-1}(F(k))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k)e^{ikx} \ dk$$ respectively. Now, the Fourier transform of a constant, $a$, is $$\mathcal{F}(a)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} ae^{-ikx} \ dx=a\sqrt{2\pi}\delta(k),$$ where $\delta$ denotes the Dirac delta function. My question is, what is $\mathcal{F}^{-1}(a)$? Is it, by symmetry of the Fourier transform and its inverse, $$\mathcal{F}^{-1}(a)=a\sqrt{2\pi}\delta(-x)?$$

Simple proof:

Let $x=k_1$, then $$\mathcal{F}(f(k_1))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(k_1)e^{-ikk_1} \ dk_1.$$Now let $k=-x_1$, then $$\mathcal{F}(f(k_1))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(k_1)e^{ix_1k_1} \ dk_1.$$ Does this provide proof of my argument?

  • 1
    $\begingroup$ The Fourier transform of $1$ is $\sqrt{2\pi} \delta$ because for any $\varphi$ (nice enough : smooth and $L^1$..) the Fourier transform of $1 . \varphi$ is $\mathcal{F}(\varphi) \ast \delta$. In other words for a distribution its Fourier transform is the unique distribution that fits well in the convolution theorem(s). The Fourier inversion theorem stays true for distributions, that is from $\mathcal{F}(\delta) = \frac{1}{\sqrt{2\pi}}$ you know $\mathcal{F}^{-1}( \frac{1}{\sqrt{2\pi}}) = \delta$ $\endgroup$ – reuns May 11 at 15:18
  • $\begingroup$ @reuns but what about the argument of the delta function (this is what I am concerned about)? I want to know if you replace $k$ with $-x$ when taking the inverse Fourier transform of a function to the frequency space (hence $\delta(-x)$). $\endgroup$ – Stuart-James Burney May 12 at 6:15
  • $\begingroup$ The Dirac delta "function" is not a function, it is a measure. When you speak of "inverse," you need to be specific about the domain and range of the function [Fourier Transform] you are speaking of. $\endgroup$ – ncmathsadist May 13 at 12:19
  • $\begingroup$ When we take the fourier transform of a test function, standard Gaussian for example, $$\int_{-\infty}^{\infty}f(x)e^{itx}\mbox{d}x=e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}$$ and when $\sigma\rightarrow\infty$ we have $$|e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}|\rightarrow 1.$$ $\endgroup$ – rodger_kicks May 13 at 12:36

The defining formula $$\mathcal{F}f(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) \, e^{-ikx} \, dx$$ only works when $f \in L^1(\mathbb{R}).$ The Fourier transform can however be extended to several other cases. For example, for a tempered distribution $u$ it is defined to be the distribution $\mathcal{F}u$ satisfying $$\int_{-\infty}^{\infty} \mathcal{F}u(x) \, \varphi(x) \, dx = \int_{-\infty}^{\infty} u(x) \, \mathcal{F}\varphi(x) \, dx,$$ for all test functions $\varphi$ in Schwartz space $\mathcal{S}(\mathbb{R}) \subset L^1(\mathbb{R}).$

One example of a tempered distribution is $\delta.$ Its Fourier transform is thus given by $$ \int_{-\infty}^{\infty} \mathcal{F}\delta(x) \, \varphi(x) \, dx = \int_{-\infty}^{\infty} \delta(x) \, \mathcal{F}\varphi(x) \, dx = \mathcal{F}(0) = \frac{1}{\sqrt{2\pi}} \left. \int_{-\infty}^{\infty} \varphi(x) \, e^{-ikx} \, dx \right|_{k=0} = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \varphi(x) \, dx , $$ i.e. $\mathcal{F}\delta = \frac{1}{\sqrt{2\pi}}$ (constant function).

For $\varphi \in \mathcal{S}(\mathbb{R})$ we have $\mathcal{F}^2\varphi(x) = \varphi(-x),$ and it's easy to show from this that also for tempered distributions we have $\mathcal{F}^2 u(x) = u(-x).$ Therefore, for the constant function $1$ we have $$ \mathcal{F}1(x) = \mathcal{F}\{\sqrt{2\pi}\,\mathcal{F}\delta\}(x) = \sqrt{2\pi}\,\mathcal{F}^2\delta(x) = \sqrt{2\pi}\,\delta(-x) = \sqrt{2\pi}\,\delta(x). $$

  • $\begingroup$ So for the inverse Fourier transform of a function, say $e^{-a|k|}$, is $$\mathcal{F}^{-1}(e^{-a|k|})=\sqrt{\frac{2}{\pi}}\frac{a}{a^2+x^2}?$$ $\endgroup$ – Stuart-James Burney May 13 at 22:17
  • $\begingroup$ That is correct. We have $$\mathcal{F}(e^{-a|x|}) = \sqrt{\frac{2}{\pi}}\frac{a}{a^2+k^2}$$ and $$\mathcal{F}^{-1}(e^{-a|k|}) = \sqrt{\frac{2}{\pi}}\frac{a}{a^2+x^2}.$$ $\endgroup$ – md2perpe May 14 at 15:15

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.