$ \frac{1}{2\pi } \int dk e^{ik x } = \delta(x)$ in the distributional sense

Question

It is often said or heard. The right hand side is well understood as a distribution or linear functional.

How to make sense of the left hand side as a distribution?

Answer 1

$$ \text{"}\;\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{ikx}\;dk = \delta(x) \quad\text{in the distributional sense}\;\text{"} $$ presumably means $$ \frac{1}{2\pi}\int_{-\infty}^{+\infty}\left(\int_{-\infty}^{+\infty} \varphi(x) e^{ikx}\;dx\right)\;dk = \varphi(0) $$ for all appropriate "test functions" $\varphi$. Can you prove this? For example, if $\varphi$ is $C^1$ with compact support?

Answer 2

The LHS is not a distribution; this equality is an abuse of notation. It is an abuse of notation because of a very obvious reason: for any $x\in\Bbb{R}$, $k\mapsto e^{ikx}$ is not in $L^1(\Bbb{R})$, so it makes zero sense to even put it under an integral sign.

Now, what is the heuristic/what is the intended meaning of such an equality? Well, I hope you know that the Fourier transform $\mathcal{F}$ can be viewed as a mapping in several different ways, but the most general is by first defining it on Schwarz space $\mathcal{S}(\Bbb{R})$ (or $L^1(\Bbb{R})$)by the usual formula $\mathcal{F}(f)(k)=\int_{\Bbb{R}}f(x)e^{-2\pi i kx}\,dx$, or whatever your favorite convention is for the $2\pi$'s. This gives a continuous linear mapping $\mathcal{S}(\Bbb{R})\to \mathcal{S}(\Bbb{R})$. By dualizing, we get a continuous linear map $\mathcal{S}'(\Bbb{R})\to \mathcal{S}'(\Bbb{R})$. The dual space is known is the space of tempered-distributions. In fact the Fourier transform $\mathcal{F}:\mathcal{S}'(\Bbb{R})\to \mathcal{S}'(\Bbb{R})$ is an isomorphism.

The space of tempered distributions is a large function space because it contains 'well-behaved' functions in the sense that it contains (an isomorphic copy of) $L^1(\Bbb{R})$ and hence also the Schwarz functions $\mathcal{S}(\Bbb{R})$; but it also contains much more functions, such as polynomials. In particular the constant function $1\in \mathcal{S}'(\Bbb{R})$. The statement of that equality is that the Fourier transform maps this constant function $1$ (viewed as a tempered distribution via integration $\langle 1,\phi\rangle:=\int_{\Bbb{R}}1\cdot \phi=\int_{\Bbb{R}}\phi$ for all $\phi\in \mathcal{S}(\Bbb{R})$) to the Dirac distribution $\delta$, i.e \begin{align} \mathcal{F}(1)=\delta. \end{align} This is the rigorous intent of that statement. And heuristically speaking, this would coincide with what we would get if we naively (but erroneously) pretended that the constant function $1$ belongs to $L^1(\Bbb{R})$ and said 'by the usual definition of Fourier transform, $\mathcal{F}(1)(x)=\int_{\Bbb{R}}1\cdot e^{ikx}\,dk=\int_{\Bbb{R}}e^{ikx}\,dk$'.

Another thing is that this 'equality' is a nice heurisitc because the Fourier inversion formula appears out of this: \begin{align} \int_{\Bbb{R}}\hat{f}(k)e^{2\pi ikx}\,dk&=\int_{\Bbb{R}}\int_{\Bbb{R}}f(y)e^{-2\pi i ky}\,dy\, e^{2\pi i kx}\,dk\\ &=\int_{\Bbb{R}}\int_{\Bbb{R}}f(y)e^{2\pi i k(x-y)}\,dk\,dy\tag{'Fubini'}\\ &=\int_{\Bbb{R}}f(y)\delta(x-y)\,dy\\ &=f(x). \end{align} All these equal signs should be treated heuristically (except the first, provided $\hat{f}\in L^1$). So, bottom line is that the notation is heuristic and handwavy and shouldn't be taken literally. At the same time, it does produce the correct results which is why you often see it.