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:
enter image description here
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?

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    $\begingroup$ @Willie Wong: I protest against the replacement of the tag "delta-function" by "distribution-theory". I understand your motivation but kindly request to undo it. $\endgroup$ Commented Dec 9, 2013 at 11:32
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    $\begingroup$ @HandeBruijn why? There's nothing in this that is not explained by standard theory of distributions. You take the limit in two different topologies and you get two different limits, nothing so strange about it. (In fact similar things happen even if you work with things other than the Dirac $\delta$, so I don't really see this question as about something special with the delta function per se.) $\endgroup$ Commented Dec 9, 2013 at 11:36

4 Answers 4


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.

  • $\begingroup$ That last sentence of yours does the job. Yes, maybe there is always a tool in that immense mathematical toolbox which is useful for some purpose. The only problem is to dig it up. I've been educated with the idea that an integral is just an integral. But now it seems that there are several sorts of integrals. If measure theory gives us the integral that's "relevant", then why not provide only that sort of integral, I'd say ? Instead of all sorts of "integration in some sense" and "kind of convergence that is used". But I suppose that's asked too much and I will accept your answer. $\endgroup$ Commented Dec 8, 2013 at 14:26
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    $\begingroup$ @HandeBruijn: you're not the first one to express the sentiment that people should be taught Lebesgue integration instead of Riemann, which is arguably no less intuitive. On the other hand, it requires some better understanding of elementary set theory, so it might be confusing for non-mathematicians. $\endgroup$
    – tomasz
    Commented Dec 8, 2013 at 16:53

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.

  • $\begingroup$ I'm reluctant to accept your answer because of this remark: saying that the two limits are equal makes no sense. Mathematically perhaps, but physicists may have other objectives. One thing is that they have to compare, numerically, real numbers coming out of experiments with real numbers coming out of a theory. At least the former have limited precision and are endowed with errors. In order for mathematical theories to be useful, they have to be stable against these errors. If we assume an incredibly small error at $x=0$ ("shifted an infinitesimal distance"), then your argument collapses. $\endgroup$ Commented Dec 8, 2013 at 13:58
  • $\begingroup$ A second reason why I'm reluctant to accept your answer is this: The physicists can "get away with it" because they formally treat $\delta$ as a function, although it is not. Let me say that there is no science more mathematical than physics. Physicists have invented the delta-function, while mathematicians have been reluctant for quite some time to accept it. And when they did it, they "rebuilt" it into something, well, uhm .. different. The mere fact that "delta-function" has not been in the standard tag list until yesterday (I've used the privilege to add it) says a lot. $\endgroup$ Commented Dec 8, 2013 at 14:06

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.


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.

  • $\begingroup$ Indeed, some iterated limits do commute, such as $\lim_{x\to 0}\lim_{y\to 0} xy/(x^2+y^2)$. But the main problem is that, in many cases, they are not stable against errors and therefore cannot be evaluated reliably numerically. This makes them rather worthless from a practical point of view. For the example at hand, substitute polar coordinates $(x,y)=(r\cos(\phi),r\sin(\phi))$ to get the limit of $\sin(\phi)/2$ for $r\to 0$ which is singular and hence very sensitive to errors at the origin. So, from a practical point of view, it's better to say that the limits do not exist there. $\endgroup$ Commented Dec 10, 2013 at 11:21
  • $\begingroup$ Why oh why is it forbidden to edit comments? The above formula in polar coordinates must be: $\;\sin(2\phi)/2\;$. $\endgroup$ Commented Dec 10, 2013 at 17:44

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