Prob 10 Sec 26 in Munkres' TOPOLOGY, 2nd ed: How to give examples of this result failing?

Question

Let $X$ be a compact topological space. Let $f_n \colon X \to \mathbb{R}$ be a sequence of continuous functions such that $f_n(x) \leq f_{n+1}(x)$ for all $x \in X$ and for all $n \in \mathbb{N}$. Let $f \colon X \to \mathbb{R}$ be a continuous function such that, for each $x \in X$, the sequence $f_n(x) \to f(x)$ as $n \to \infty$. Then the sequence $f_n$ converges uniformly to $f$ on $X$.

This is problem 10 (a), Sec. 26, in the book Topplogy by James R. Munkres, 2nd edition. I've managed to give a proof of this result.

However, what if we require that $f_n(x) \geq f_{n+1}(x)$ for all $x \in X$ and for all $n \in \mathbb{N}$? My assertion is the convergence is still uniform.

Here's my proposed proof:

Let $\epsilon > 0$ be given. Let $x \in X$. Since $f_n(x) \to f(x)$ as $n \to \infty$, there exists $N(x) \in \mathbb{N}$ such that $\vert f_n(x) - f(x) \vert < \epsilon$ if $n \geq N(x)$.

Since $f_n(x)$ is a monotically decreasing sequence, we must have $f_n(x) \geq f(x)$ for all $n$. So $$\vert f_n(x) - f(x) \vert = f_n(x) - f(x) < \epsilon \ \ \ \mbox{ if } \ n \geq N(x). $$ In particular, $$f_{N(x)} (x) - f(x) < \epsilon.$$ Since $f_{N(x)} - f \colon X \to \mathbb{R}$ is continuous, the set $$U_x \colon= \left( f_{N(x)} - f \right)^{-1} \left( \ (-\infty, \epsilon) \ \right)$$ is open in $X$ and the point $x$ belongs to $U_x$.

In this way, we have an open covering $\{ U_x \colon x \in X \}$ of $X$. Since $X$ is compact, there is a finite subcollection $U_{x_1}, \ldots, U_{x_k}$ that also covers $x$.

Let $N \colon= \max \left( N(x_1), \ldots, N(x_k) \right)$.

Let $y \in X$. Then $y \in U_{x_i}$ for some $i = 1, \ldots, k$. So, $$f_{N(x_i)} (y) - f(y) < \epsilon. $$

So, if $n \geq N$, then we have $$ \begin{align*} \vert f_n(y) - f(y) \vert &= f_n(y) - f(y) \\ &\leq f_N(y) - f(y) \\ &\leq f_{N(x_i)} (y) - f(y) \\ &< \epsilon. \end{align*} $$

Thus, for every given $\epsilon > 0$, there exists a natural number $N$ such that, if $n \geq N$, then we have $$\vert f_n(y) - f(y) \vert < \epsilon$$ for all $y \in X$.

Hence $f_n$ converges uniformly to $f$ on $X$.

Is the above proof correct?

But the result breaks down in the following example.

Let $X \colon= [0,1]$, and, for each $n \in \mathbb{N}$, let $f_n \colon X \to \mathbb{R}$ be defined by $$f_n(x) \colon= x^n \ \ \ \mbox{ for all } \ x \in X.$$ Now, for all $n \in \mathbb{N}$, $$f_n(0) = 0 = f_{n+1}(0),$$ and $$f_n(1) = 1 = f_{n+1}(1);$$ for $x \in (0,1)$, we have $$f_n(x) = x^n > x^{n+1} = f_{n+1}(x).$$ The sequence $f_n$ converges pointwise to the function $f \colon X \to \mathbb{R}$ defined by $$ f(x) \colon= \begin{align*} 0 & \ \mbox{ if } \ 0 \leq x < 1; \\ 1 & \ \mbox{ if } \ x = 1. \end{align*} $$ but this convergence is not uniform. But $f$ is not continuous. So the hypotheses of our result are not met fully.

(i) What if $X$ is not compact?

(ii) What if the neither of the monotonicity conditions is satisfied?

Please give an elementary enough example in each case where the result breaks down.

Answer 1

Your example of $f_n(x) = x^n$ also shows that compactness is necessary, if you restrict the domain to the noncompact set $[0,1)$. Then $f_n(x)$ converges pointwise to the continuous function $f(x) = 0$.

To show that something like monotonicity is necessary, consider $X = [0,1]$, and set $$ f_n(x) = \begin{cases} 0 & x \le 1- \frac{2}{n} \\ nx-n+2 & 1-\frac{2}{n} < x \le 1 - \frac{1}{n}\\ n-nx & 1-\frac{1}{n} < x \le 1\\ \end{cases} $$ Here the nontrivial part of the graph of $f_n$ is an increasingly narrow triangle of height 1. For all $x\in [0,1]$, $f_n(x) \to 0$, but since $f_n\left(1-\frac 1 n\right) = 1$, the convergence is not uniform.

Answer 2

1) Counterexample when $X$ is not compact:

Take $X=\mathbb{R}$ with the usual topology, and let $f_n(x) = \min \{ x,n\}$.

Then $f_n \leq f_{n+1}$ for all $n$, and $(f_n)$ converges to the function $f(x)=x$ pointwise, but certainly not uniformly (since $\sup_{x\in \mathbb{R}} |f_n(x)-x| = \infty$ for all $n$).

2) Counterexample when monotonicity is not satisfied:

Let $X = [0,1]$, and define $f_n(x) = \frac{nx}{1+n^2x^2}$.

Then $f_n \to 0$ pointwise on $[0,1]$, so the limit is continuous, but convergence is not uniform since $f_n(\frac{1}{n}) = \frac{1}{2}$, which does not converge to $0$.

Answer 3

• The proof seems ok. A simple proof would be setting $g_n:=-f_n$ and using the result for monotonically increasing sequences.

• The example of the powers you gave with $X=[0,1]$ is not really a counter-example as the limit function $f$ is not continuous.

• If you use the same power functions with the non-compact domain $X:=[0,1)$ you'll find a sequence of continuous functions which converges to a continuous limit function non-uniformly.

• Define $f_n:[0,1] \to \mathbb{R}$ as the piecewise linear function passing through the points $(0,0),(1/n,1),(2/n,0),(1,0)$. The sequence $(f_n)_{n=2}^\infty$ is a sequence of continuous functions that converge uniformly to the continuous zero function. However, the convergence is not uniform.