Iterated Limits Schizophrenia

Question

Consider the functions $g_n(x)$, with $n\in\mathbb{N}$, $n \ge 1$ and $x\in\mathbb{R}$, defined as follows: $$ g_n(x) = \begin{cases} 2n^2x & \text{if }0 \le x < 1/(2n) \\ 2n - 2 n^2 x & \text{if } 1/(2n) \le x < 1/n \\ 0 & \text{everywhere else} \end{cases} $$ Standard mathematics argument:
These functions are triangular, and they all disappear outside of $[0,1]$, so I can compute $\int_0^1 g_n(x) \, dx = 1$ for every $n$. For every $x$, $\lim_{n \rightarrow \infty} g_n(x) = 0$.
So the limit and the integration can't be interchanged.

Here is an animated picture of the functions:

The function $g_n(x)$ becomes a sharp peak at $x=0$ for $n \rightarrow \infty$ and, geometrically, it certainly does not disappear or becomes zero. Instead, it seems that we get, in the end, what physicists know as a delta function. Informally: $$ \delta(x) = \begin{cases} 0 & \text{for } x \ne 0 \\ \infty & \text{for } x = 0 \end{cases} \qquad ; \qquad \int_{-\infty}^{+\infty} \delta(x) \, dx = 1 $$ Whatever definition might be the "rigorous" one, a delta function, roughly speaking, is just a very large peak near $x = 0$ with area normed to $1$. Furthermore, it is typical that the following function, triangular as well, is indeed supposed to converge to the delta function - instead of becoming zero - for $n \rightarrow \infty$ and nobody has any doubt about it. $$ D_n(x) = \begin{cases} n^2x + n & \text{if } -1/n \le x \le 0 \\ n - n^2x & \text{if } 0 \le x \le +1/n \\ 0 & \text{everywhere else} \end{cases} $$ The only thing that distinguishes $g_n(x)$ from $D_n(x)$ is that the maximum of the former is shifted an infinitesimal distance $\lim_{n \rightarrow \infty} 1/(2n)$ with respect to the maximum of the latter at $x=0$.
So it's easy to see that these functions become one and the same for $n \rightarrow \infty$: $$ \lim_{n \rightarrow \infty} g_n(x) = \lim_{n \rightarrow \infty} D_n(x) = \delta(x) $$ Therefore, in the end, we have two arguments that, unfortunately, also lead to different outcomes for the iterated limit.

• According to standard mathematics, the iterated limits do not commute: $$\int_0^1 \left[ \lim_{n \rightarrow \infty} g_n(x) \right] \, dx = 0 \qquad \text{and} \qquad \lim_{n \rightarrow \infty} \left[ \int_0^1 g_n(x) \, dx \right] = 1 $$
• According to this physicist, the iterated limits do commute: $$\int_0^1 \left[ \lim_{n \rightarrow \infty} g_n(x) \right] \, dx = 1 \qquad \text{and} \qquad \lim_{n \rightarrow \infty} \left[ \int_0^1 g_n(x) \, dx \right] = 1 $$

And the question is, of course the following. Is it possible to escape from this apparent paradox? How then?

Answer 1

That depends on what you mean by $\lim_{n\to \infty} g_n(x)$. The pointwise limit is zero, so the integral is zero. And if you put the parentheses so: $\lim_{n\to \infty} (g_n(x))$ then we have to conclude that the limit is zero and $$\int_0^1 \lim_{n\to \infty} (g_n(x))\,dx=\int_0^1 0\,dx=0$$

However, you can also read this as $\lim_{n\to \infty} (g_n\, dx)$ where $dx$ denotes the usual Lebesgue measure. This is a limit of measures and it does, in fact, converge to the point measure at $0$, so that $$\int_{[0,1]} \lim_{n\to \infty} (g_n\, dx)=\int_{[0,1]}d\delta_0=1$$ (I've changed the notation because $\int_0^1$ can be confusing in this case, as $\int_{(0,1)}d\delta_0=0$).

So the "paradox" actually comes from different kinds of limits, and the arising confusion comes from not explaining the kind of convergence that is used in each case.

Answer 2

The mathematics does not say the limits never commute in your case, just that they do not necessarily do so. It just so happens that the limit of your $g_n$'s is a distribution, the $\delta$-distribution, which correspons to the dirac measure, but it need not necessarily have been this nice. At least this is my understanding of the matter.

Answer 3

The equality $\lim_{n\to\infty}g_n(x)=\delta(x)$ is incorrect. In fact, $\lim_{n\to\infty}g_n(x)=0$ for all $x$ (including $x=0$). For $D_n$, $$ \lim_{n\to\infty}D_n(x)=\begin{cases}0,&\ x\ne0,\\ \infty,&\ x=0\end{cases} $$ So, saying that the two limits are equal makes no sense.

The physicists can "get away with it" because they formally treat $\delta$ as a function, although it is not. This is well-understood by mathematicians. One uses the functions $g_n$, $D_n$ to define linear functionals over the space of continuous functions. Given a continuous function $f$, one defines $$ \varphi_n:f\mapsto\int_0^1f(t)g_n(t)dt. $$ One can check that $\lim_{n\to\infty}\varphi_n(f)=f(0)$. So the functionals $\varphi_n$ converge pointwise to the functional $\delta:f\mapsto f(0)$. Physicists love an intuitive vision of things, so they claim that this new functional should also be integration against a function, that they call $\delta(x)$, the Dirac function. So, for them, $$\tag{1} \delta(f)=\int_0^1f(t)\delta(t)dt. $$ This last equality makes no sense mathematically. What happens is that one wants to think of $\delta(t)dt$ as a measure, and integrate $f$ against it. And this can be done: $$\tag{2} \delta(f)=\int_{[0,1]}f(t)d\delta(t), $$ where now one considers the measure $\delta$ given by $$ \delta(S)=\begin{cases}1,&\ 0\in S,\\ 0,&\ 0\not\in S\end{cases} $$ What happens then is that physicists use $(1)$, while they should be using $(2)$. But they use $(1)$ in a way that it just becomes notation for $(2)$ which is the right one mathematically.

Answer 4

There is a difference between $\lim\limits_{n\to\infty}g_n(x)$ and $\lim\limits_{n\to\infty}g_n$. Since $x$ is specified, $\lim\limits_{n\to\infty}g_n(x)$ means the pointwise limit of $g_n$; that is, the limit of $g_n$ under the topology of pointwise convergence, aka the product topology. With $\lim\limits_{n\to\infty}g_n$, the topology under which the limit is taken is not clear, but if taken in the sense of distributions, that is under the weak-* topology on $C_c^\infty$, the limit would be the Dirac delta distribution.

If rather than $$ \int_0^1\left[\lim_{n\to\infty}g_n(x)\right]\,\mathrm{d}x = 1\tag{1} $$ which is false, the physicist said, or meant, $$ \int_0^1\left[\lim_{n\to\infty}g_n\right](x)\,\mathrm{d}x = 1\tag{2} $$ then the limit could be taken in the sense of distributions, under which $(2)$ would be correct.