Limit of Fourier transform of indicator function

Question

I'm an engineering student and currently looking into gentle introduction of Fourier analysis with a book "Lectures on the Fourier Transform and Its Application" by Brad Osgood. I now know that the definition of the statement \begin{align} \lim_{n\rightarrow \infty}(\mathcal{F}\boldsymbol{1}_{n[-1,1]})(s) = \lim_{n\rightarrow \infty}\int_{-n}^n e^{-2\pi i s x} dx = \delta(s) \end{align} is \begin{align} \lim_{n\rightarrow\infty}\int_{-\infty}^\infty \left(\int_{-n}^n e^{-2\pi i s x} dx\right) \varphi(s) ds = \varphi(0) \end{align} for every test functions $\varphi$. Tweaking the lower- and upper-limits of the integral, we can think of \begin{align} (\mathcal{F}\boldsymbol{1}_{nI})(s) = \int_{nI} e^{-2\pi i s x} dx \end{align} where $I=[a,b]$ is an interval. It is clear that the limit in distribution makes sense if and only if $b=-a$. "If" part is well-known and "only if" part follows from "if" part: \begin{align} \int_{-\infty}^\infty \left(\int_{na}^{nb} e^{-2\pi i s x} dx\right) \varphi(s) ds = e^{\frac{n(a+b)}{2}}\underbrace{\int_{-\infty}^\infty \bigg(\int_{-\frac{n(b-a)}{2}}^{\frac{n(b-a)}{2}} e^{-2\pi i s x} dx\bigg) \varphi(s) ds}_{\rightarrow \varphi(0)}. \end{align} Given $b\neq-a$, the multiplicative term $e^{\frac{n(a+b)}{2}}$ does not converge, hence the entire limit does not converge unless $\varphi(0)=0$. Choosing any test function with $\varphi(0)\neq0$ outputs the desired result. I now find my blatant fault! Thanks.

We can further think of the following multidimensional extension: \begin{align} \lim_{n\rightarrow \infty}(\mathcal{F}\boldsymbol{1}_{nI})(s) = \lim_{n\rightarrow \infty} \int_{nI} e^{-2\pi i (s\cdot x)} dx \end{align} for $I\subset\mathbb{R}^n$. The question is then when the above limit makes sense? One trivial sufficient condition is that $I$ is a multidimensional interval centered at origin. Any comments would be much appreciated.

Edit: more elaboration on the argument regarding the well-definedness of the limit in one-dimension.

Answer 1

The distribution-theoretic definition/interpretation of $\int e^{-2\pi is\cdot x} \, dx$ is as the Fourier transform of the distribution induced by the function $x\mapsto 1.$ The Fourier transform of a distribution $u$ is defined as $\langle \mathcal{F}u, \varphi \rangle := \langle u, \mathcal{F}\varphi \rangle$ for every test function $\varphi\in\mathcal{S}$ (the Schwartz space).

This implies that for any sequence $u_k$ of tempered distributions converging to a tempered distribution $u,$ their Fourier transforms $\mathcal{F}$ converge to the Fourier transform of $u$:

$$ \langle \mathcal{F}u_k, \varphi \rangle = \langle u_k, \mathcal{F}\varphi \rangle \to \langle u, \mathcal{F}\varphi \rangle = \langle \mathcal{F}u, \varphi \rangle. $$

Therefore, if $A_k$ is a sequence of measurable sets growing to all of $\mathbb{R}^n$ then $$\mathcal{F}\mathbf{1}_{A_k} \to \mathcal{F}\mathbf{1}_{\mathbb{R}^n},$$ where $\mathbf{1}_A$ is the indicator function of the set $A,$ i.e. $\mathbf{1}_A(x)=1$ if $x\in A,$ and $\mathbf{1}_A(x)=0$ otherwise.

So you could take $I$ to be any open neighborhood of origin, like the unit ball centered at origin $I=B_1(0).$

Answer 2

By Fubini's theorem \begin{align} \int_{-\infty}^\infty \left(\int_{-n}^n e^{-2\pi i s x} dx\right) \varphi(s) ds &= \int^n_{-n}\Big(\int_\mathbb{R}e^{-2\pi i s x}\varphi(s)\,ds\Big)\,dx\\ &=\int^n_{-n}\widehat{\varphi}(x)\,dx\xrightarrow{n\rightarrow\infty}\int_{\mathbb{R}}\widehat{\varphi}(x)\,dx \end{align}

A same result is obtained if instead of $[-n,n]$ you use any interval $[a,b]$ and let $a\rightarrow-\infty$ and $b\rightarrow\infty$.

Since $\varphi\in\mathcal{C}_{00}(\mathbb{R})$, the Fourier inversion theorem implies that $\widehat{\varphi}\in L_1(\mathbb{R})\cap\mathcal{C}_0(\mathbb{R})$ and that $$\varphi(t)=\int_\mathbb{R} e^{i2\pi xt}\widehat{\varphi}(t)\,dt$$ In particular, $\varphi(0)=\int_{\mathbb{R}}\widehat{\varphi}(t)\,dt$