What is $\lim\limits_{t \rightarrow \infty}e^{-{it}}$

Question

Let $f(t)=e^{-it}$, $t \geq 0$. We know that f doesn't converge when t goes to infinity. But does it converge as a distribution?

Answer 1

Let me collect the info that you can also find scattered in the comments. There are several versions of what we could be talking about here:

1.) If we really mean "does $f(t)=e^{-it}$ for fixed $t$ define a distribution"? The answer for this is yes. Indeed, any function $g$ which is integrable on all compacta defines a distribution via $$ T_g(\varphi) := \int_\mathbb{R} g(x) \varphi(x) $$ where the integral exists for all test functions $\varphi$ (as they have compact support). In particular the function $f_t(x) = e^{-it}$ (constant function) defines a distribution.

2.) We could now also ask whether the family of distributions defined by $T_{f(t)}$ converges in the space of distributions. That is clearly not the case as we have $$ T_{f(t)}(\varphi) = e^{-it} \int_\mathbb{R} \varphi(x) dx, $$ hence we cannot take the limit $t\rightarrow \infty$.

3.) Finally we can consider the same question, but with $F_t(x)=e^{-itx}$. In this case we have $$ T_{F_t}(\varphi) = \hat{\varphi}(t). $$ By the Riemann-Lebesgue lemma we have for any $g\in L^1(\mathbb{R})$ $$ \lim_{\vert \xi \vert \rightarrow \pm \infty} \hat{g}(\xi) = 0. $$ In particular, we have this convergence for test functions.

Answer 2

When one talks about convergence in the space of distributions one usually has a parameter that is not the ordinary space variable. Examples:

1. $e^{ikx} \to 0$ as $k\to\pm\infty$ since $\langle e^{ikx}, \varphi(x) \rangle \to 0$ for every test function $\varphi$ as $k\to\pm\infty.$
2. $\frac{1}{\epsilon}e^{-x^2/\epsilon^2} \to \sqrt{\pi}\,\delta(x)$ as $\epsilon\to 0.$

So what would we mean with something like $\lim_{x\to\pm\infty} f(x)$ in the space of distributions? I would define it as $$\lim_{R\to\pm\infty} \langle f(x), \varphi(x-R) \rangle.$$ For this particular example, $f(x)=e^{-ix}$ this limit does not exist. The value will just "spin around" on some circle centered at origin in the complex plane.

But for the similar function $g(x)=e^{-ix^2}$ (or with a plus sign) the limit exists and equals $0$. Therefore I would say that $\lim_{x\to\pm\infty} g(x) = 0$ in a distributional sense.

An even more interesting example is $h(x) = e^{ie^{x^2}}.$ This function is bounded and continuous and therefore is a tempered distribution, meaning that it has a Fourier transform. Therefore its derivative $h'(x) = 2x\,ie^{x^2}\,e^{ie^{x^2}}$ is also a tempered distribution, despite being far from bounded. But in the distributional sense defined above, the factor $e^{ie^{x^2}}$ tends to $0$ so fast as $x\to\pm\infty$ (by its oscillations) that also $e^{x^2}e^{ie^{x^2}}$ tends to $0$ in the distributional sense.

(I must admit that my last example made me a bit confused, so I hope that what I wrote is correct. My confusion lies in me thinking that if a tempered distribution like $h'$ is multiplied with a bounded function like $e^{-ie^{x^2}}$ then the result should also be a tempered distribution, which is not the case here. EDIT: I realized that a tempered distribution can not be multiplied with bounded functions in general. Also the derivatives of all orders need to be bounded. Here the factor $e^{-ie^{x^2}}$ fails.)