# Relation between two notions intermediate between “pointwise convergence” and “uniform convergence”

(I asked this on MSE a week ago, but did not get any answers there, so I'm trying here.)

Let $$X$$ be a topological space. I will define four ways in which a sequence $$(f_n)$$ of continuous functions $$X \to \mathbb{R}$$ might converge to a continuous function $$f\colon X \to \mathbb{R}$$, four notions intermediate between pointwise convergence (labeled (1) below) and uniform convergence (labeled (4) below). (Note: I am listing all four below in order to justify why I care about (2) and (3), but only (2) and (3) are involved in the actual question.) Specifically:

1. $$f_n$$ tends to $$f$$ pointwise, namely: for all $$x\in X$$ we have $$f_n(x) \to f(x)$$ (with no assumption of uniformity of any kind), viꝫ., for all $$\varepsilon>0$$ and all $$x\in X$$ there is $$n_0$$ such that when $$n\geq n_0$$ we have $$|f_n(x)-f(x)|<\varepsilon$$.

2. We now demand that (for $$\varepsilon>0$$ fixed), $$n_0$$ take the same value on each element of an open cover $$(U_i)_{i\in I}$$ of $$X$$. More precisely: for all $$\varepsilon>0$$ there is a covering $$(U_i)_{i\in I}$$ of $$X$$ by open sets such that for all $$i\in I$$ there is $$n_0$$ such that when $$x \in U_i$$ and $$n\geq n_0$$ we have $$|f_n(x)-f(x)|<\varepsilon$$.

3. We now demand that the same open cover $$(U_i)_{i\in I}$$ work for every $$\varepsilon>0$$; this is the same as demanding that $$f_n\to f$$ uniformly on each $$U_i$$, viꝫ. there is a covering $$(U_i)_{i\in I}$$ of $$X$$ by open sets such that for all $$\varepsilon>0$$ and all $$i\in I$$ there is $$n_0$$ such that when $$x \in U_i$$ and $$n\geq n_0$$ we have $$|f_n(x)-f(x)|<\varepsilon$$. (I suppose we could say that $$f_n \to f$$ “locally uniformly”, but I'm not sure whether this is standard terminology.)

4. We now demand that the covering consist of just $$X$$, i.e., that $$f_n \to f$$ uniformly, viꝫ. there for all $$\varepsilon>0$$ there is $$n_0$$ such that when $$x \in X$$ and $$n\geq n_0$$ we have $$|f_n(x)-f(x)|<\varepsilon$$.

To summarize:

1. $$\forall \varepsilon>0. \forall x\in X. \exists n_0\in \mathbb{N}. \forall n≥n_0. |f_n(x)-f(x)|<\varepsilon$$

2. $$\forall \varepsilon>0. \exists (U_i) \text{ open covering}. \forall i\in I. \exists n_0\in \mathbb{N}. \forall x\in U_i. \forall n≥n_0. |f_n(x)-f(x)|<\varepsilon$$

3. $$\exists (U_i) \text{ open covering}. \forall \varepsilon>0. \forall i\in I. \exists n_0\in \mathbb{N}. \forall x\in U_i. \forall n≥n_0. |f_n(x)-f(x)|<\varepsilon$$

4. $$\forall \varepsilon>0. \exists n_0\in \mathbb{N}. \forall x\in X. \forall n≥n_0. |f_n(x)-f(x)|<\varepsilon$$

It's clear that $$(4)\Rightarrow(3)\Rightarrow(2)\Rightarrow(1)$$.

• A counterexample showing that (3) does not imply (4) is given by $$X = \mathbb{R}$$ and $$f_n(x) = \exp(-(x-n)^2)$$, which converges in sense (3) but not uniformly (4) toward $$f = 0$$.

• A counterexample showing that (1) does not imply (2) is given by $$X = [0,1]$$ and $$f_n(x) = \max(0, \min(nx, 2-nx))$$ (graphs here) which converge pointwise (1) toward $$f = 0$$ but there is no neighborhood $$V$$ of $$0$$ such that $$\exists n_0\in \mathbb{N}. \forall x\in V. \forall n\geq n_0. |f_n(x)|<\frac{1}{2}$$ so (2) does not hold.

Question: Does (2) have a standard name? Are (2) and (3) perhaps in fact equivalent? If yes, what is a proof? If no, what is a counterexample?

• I was curious enough to go off wandering, and, well, I guess now I know that 'viz.' for 'videlicet' is not just some bizarre misspelling of 'z' for 'd' but a corruption of vi…ꝫ, with the ꝫ short for 'et'. Apr 4, 2022 at 14:42
• Yup! I seem to be very lonely, so far, in writing “viꝫ” for “videlicet”, but I hope the trend of actually using this nice Unicode character will catch! Apr 4, 2022 at 15:01

Consider the Baire space $$\mathbf N^\mathbf N$$ and let $$f_n$$ be the following function: If your sequence $$x$$ starts with $$k$$ zeroes followed by the entry $$n$$ then $$f_n(x) = 1/(k+1)$$, otherwise $$f_n(x) = 0$$.
The $$f_n$$ are continuous: away from the sequence that is constantly 0, they are even locally constant. At $$0^\infty$$ you can easily check continuity using cylinder sets.
The $$f_n$$ satisfy (2): If $$\epsilon > 1/(k+1)$$, we can take as our covering the $$k$$-cylinder sets: for $$v \in \mathbf N^k$$, let $$U_v$$ be the set of sequences that start with $$v$$. If $$v = 0^k$$, then $$\sup f_n(U_v) = 1/(k+1) < \epsilon$$. For $$v \not=0^k$$, we can wait until n is larger than the first nonzero entry of $$v$$, and get $$f_n(U_v) = 0$$ for all $$n \geq n_0$$ for some $$n_0$$. So the $$U_v$$ form the required covering.
The $$f_n$$ do not satisfy (3): Some set of the open covering must contain $$0^\infty$$, and so it must contain a cylinder set $$U_v$$ with $$v = 0^k$$ for some $$k$$. But $$\sup f_n(U_v) = 1/(k+1)$$ for all $$n$$, so you don't have local uniform convergence.
In my construction I heavily used that $$\mathbf N^\mathbf N$$ is not locally compact, I am curious whether (2) and (3) are equivalent for locally compact spaces.