## Iterated Limits Schizophrenia

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?