Is uniform convergence a necessary condition for interchangeability of limits

Question

I am aware of the standard result (namely, please see Terence Tao's Analysis II, Page 54, Proposition 3.3.3 - Interchange of limits & uniform limits) that uniform convergence implies the interchangeability of limits, however I am wondering if the converse holds.

i.e. If we can interchange the limits of a function, does it imply the uniform convergence of this function please?

I don’t think it holds but I am struggling to construct a counterexample.

Many thanks in advance!

Answer 1

If I understand you correctly you are looking for something like this sequence:

$x_m=\frac{1}{m}$

And applying this to the function which doesn't uniformly converge but pointwise converges:

$ \forall x\in [0,1] : f_n(x)=x^n $

the limits are interchangeable and look something like this:

$ \lim_{n\rightarrow \infty}\lim_{m\rightarrow \infty}f_n(x_m)=\lim_{n\rightarrow \infty}f_n(0)=0$

$ \lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}f_n(x_m)=\lim_{m\rightarrow \infty}f(x_m)=0$

When $f$ is the pointwise limit function which $f(x)=0$ for $x\neq 1$ otherwise $f(1)=1$.

We actually can say more, we know the limits are interchangeable do to continuity, and continuity isn't preserved under pointwise convergence.

Answer 2

Aside from the excellent answer above, I have constructed a similar counterexample as follows:

Consider $f_n (x) = x^n, x \in [0, 1)$, which pointwise converges to 0 but not uniformly.

But notice that we can interchange the limit and integration:

$$ \lim_{n \to \infty} \int_0^1 x^n \,dx = \lim_{n \to \infty} \frac{x^{n + 1}}{n + 1} \big\vert_{0}^1 = \lim_{n \to \infty} \frac{1}{n + 1} = 0 $$

And

$$ \int_0^1 \lim_{n \to \infty} x^n \,dx = \int_0^1 0 \,dx = 0 $$

Answer 3

I assume you mean the following statement: If $U\subseteq \mathbb R^n$ and a sequence of functions $f_n:U\to\mathbb R^m$ converges uniformly to $f$, and a sequence $x_m\in U$ converges to $x\in U$, then

$$\lim_{m\to\infty}\lim_{n\to\infty}f_n(x_m)=\lim_{n\to\infty}\lim_{m\to\infty}f_n(x_m).$$

The converse is not true, since this theorem can be strengthened: locally uniform convergence of $f_n$ (meaning that for all $x\in U$ there is an open ball around $x$ such that $f_n$ restricted to that ball converges uniformly) is already sufficient to interchange limits, so full uniform convergence can't be necessary.