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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?

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    $\begingroup$ 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, 2019 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 $\delta(-x)$). $\endgroup$
    – Steven
    May 12, 2019 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, 2019 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 $$|e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}|\rightarrow 1.$$ $\endgroup$
    – rami_salazar
    May 13, 2019 at 12:36

1 Answer 1

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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 $$ \mathcal{F}1(x) = \mathcal{F}\{\sqrt{2\pi}\,\mathcal{F}\delta\}(x) = \sqrt{2\pi}\,\mathcal{F}^2\delta(x) = \sqrt{2\pi}\,\delta(-x) = \sqrt{2\pi}\,\delta(x). $$

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  • $\begingroup$ So for the inverse Fourier transform of a function, say $e^{-a|k|}$, is $$\mathcal{F}^{-1}(e^{-a|k|})=\sqrt{\frac{2}{\pi}}\frac{a}{a^2+x^2}?$$ $\endgroup$
    – Steven
    May 13, 2019 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, 2019 at 15:15

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