# Determining Conditions for the Exchange of Limit and Integral

I encountered this problem: Consider the following function： $$f(x,\Delta k)=\lim_{\Delta k \to \infty} \frac{\sin ^{2}\frac{x\Delta k}{2}}{\Delta k\left ( \frac{x}{2} \right )^{2} }\equiv 2\pi\delta(x)$$ It can be verified that $$f(x)$$ satisfies the following condition: $$\delta(x)=\begin{cases} 0 &(x\ne0) \\ \infty &(x=0) \end{cases}$$ Please prove: $$\int_{-\infty}^{\infty}f(x)\delta(x)\mathrm{d}x=f(0)$$ I've noticed that if I use the limit form of the given function and interchange the integral and limit, I end up with a divergent integral instead of the correct result. This seems to be because integration and taking limits cannot be interchanged. I would like to know if there's a simple method to determine when these two operations can be interchanged. I know that if the limit converges uniformly, then interchange of limit and integral is permissible, but this cannot be applied directly to this problem. I also wonder if this implies that if I want to represent $$\delta(x)$$ using the limit expression, do I need to ensure certain conditions on my $$f(x)$$ (such as ensuring the integral is finite)? If it doesn't, do I need to switch to a more effective limit expression?

• Use the distribution theorem to solve this problem. Can you calrify more about the function $f(x)$, is it a test function? Feb 19 at 15:25
• @AliMezher Yes, In the final formula, $f(x)$ is a test function. No restrictions are imposed on $f(x)$ in the problem statement. Feb 20 at 3:36

## 1 Answer

Point 1. We typically expect the limit and integral to be interchangeable if there is no "redistribution of mass", either because

1. all the mass distributions are confined in a region of finite mass (dominated convergence), or
2. the mass grows monotonically towards a specific direction (monotone convergence).

OP's situation is a typical example in which mass redistributes, hence, interchange of limit and integral should not be expected.

To get around this issue in OP's case, one may reparametrize the approximate delta functions so that the mass no longer redistributes under this reprametrization.

Point 2. Since the Dirac delta describes an allocation of a unit point mass, it cannot be properly described only using the language of functions. Consequently, one has to work with a more generalized setting to define and discuss the properties of the Dirac delta. The usual setting is to consider $$\delta(x)$$ as a distribution (a.k.a. generalized function), an object that can encode both macroscopic (function-like) and microscopic (point-mass-like) allocation of signed mass.

Formally, a distribution is a continuous linear functional over a suitable space of test functions. This means that we describe an allocation of mass via "how it feels like under various probing/testing methods". So, if $$f(x)$$ is a distribution, then its property can be studied by how the map

$$\varphi \quad \mapsto \quad \int \varphi(x)f(x) \, \mathrm{d}x$$

behaves. Specifically, if this map behaves similar to the evaluation map $$\varphi \mapsto \varphi(0)$$, then we may possibly say that $$f(x)$$ is close to the Dirac delta.

Enough digresson, so let met return to OP's question. Let $$\varphi \in C_b(\mathbb{R})$$ be a test function which is continuous and bounded on $$\mathbb{R}$$. Then by writing $$g(x) = f(x, 1) = \bigl( \frac{\sin(x/2)}{x/2} \bigr)^2$$ and noting that $$f(x, \Delta k) = g(x \Delta k) \Delta k$$,

\begin{align*} \int_{\mathbb{R}} \varphi(x) f(x, \Delta k) \, \mathrm{d}x &= \int_{\mathbb{R}} \varphi(x) g(x \Delta k) \Delta k \, \mathrm{d}x \\ &= \int_{\mathbb{R}} \varphi(t/\Delta k) g(t) \, \mathrm{d}t \tag{t = x \Delta k}. \end{align*}

Since $$\varphi$$ is bounded and $$g$$ is integrable on $$\mathbb{R}$$, the integrand of the last integral is dominated by an integrable function. So by the dominated convergence theorem, the limit of the last integral as $$\Delta k \to \infty$$ is

\begin{align*} \lim_{\Delta k \to \infty} \int_{\mathbb{R}} \varphi(t/\Delta k) g(t) \, \mathrm{d}t &= \int_{\mathbb{R}} \lim_{\Delta k \to \infty} \varphi(t/\Delta k) g(t) \, \mathrm{d}t \\ &= \int_{\mathbb{R}} \varphi(0) g(t) \, \mathrm{d}t \\ &= 2\pi \varphi(0), \end{align*}

where we utilized the fact that $$\int_{\mathbb{R}} g(t) \, \mathrm{d}t = 2\pi$$. Therefore

$$f(x, \Delta k) \quad \to \quad 2\pi \delta(x)$$

in distribution sense as $$\Delta k \to \infty$$. (In fact, this shows that the above convergence occurs in weak sense.)

• Thank you for your response! It seems that some knowledge of real analysis is required here, which unfortunately I haven't studied. I did some searching and found that both dominated convergence and monotone convergence seem to require that the function itself (as well as its limit) be integrable. Feb 20 at 4:23
• So, if I choose a test function that is divergent and non-integrable, such as $\exp(x)$ , can I find a suitable form of the limit function (or distribution? I'm not sure if my terminology is correct), such that it produces the effect of a delta function (in the sense of exchanging the order of integration and limit)? Feb 20 at 4:23