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