Partial Converse to the Uniform Limit Theorem (3.26.10)

James Munkres Topology, chapter 3.26, exercise 10:

(a) Let $f_n : X→ℝ$ be a sequence of continuous functions, such that $f_n(x)→f(x)$ for each $x∈X$. If $f$ is continuous, and if the sequence $(f_n)$ is monotone increasing, and if $X$ is compact, show that the convergence is uniform.

(b) Give examples to show that the converse fails to hold if either of the requirements of $X$ being compact or $f_n$ being monotone increasing is deleted.

Proof: (a) For each $n$, let $K_n=\{x∈X~|~|f_n(x)-f(x)|≥ε\}$. It is routine to show $K_n$ is closed, by showing its complement is open: Suppose $|f_n(x)-f(x)|<ε$. Then $f(x)∈(f_n(x)-ε,f_n(x)+ε)=V$, so let $U$ be a neighborhood of $f(x)$ contained in $V$, and we see $f^{-1}(U)⊆X$ is open such that $f(U)⊆V$ so that $U$ is a neighborhood of $x$ also contained in the complement of $K_n$. Now, since $(f_n)$ is monotone increasing, and hence $f(x)≥f_n(x)$ by convergence, we see $K_{n+1}⊆K_n$ for all $n$. If we assume each $K_n$ is nonempty, then finite intersections $∩K_i=K_{\text{max }i}$ are also nonempty, so since $X$ is compact, there exists $x∈X$ such that $x∈K_n$ for all $n$, i.e. $|f_n(x)-f(x)|≥ε$ for all $n$. However, this is impossible since $f_n(x)→f(x)$ for all $x$. Hence $K_N=ø$ for some $N$, hence $K_n=ø$ for all $n≥N$, that is, $|f_n(x)-f(x)|<ε$ for all $x$, for sufficiently large $n$.

(b) Suppose $X$ need not be compact; then the sequence of functions $(f_n)$ defined by $$f_n(x)= \{ \begin{array}{lr} 0 & : x ≤ n\\ n-x & : x > n \end{array}$$ is seen to be monotone increasing with $f_n(x)→0$ for all $x$, yet does not converge uniformly to the zero function.

Suppose $f_n$ need not be monotone increasing; then let $[0,1]=∪[a_n,b_n]$ be any countably infinite partition of $[0,1]$ into disjoint closed intervals where $a_n≠b_n$ for all $n$. Then when $c_n=\dfrac{a_n+b_n}{2}$, define a sequence of functions $(f_n)$ by $$f_n(x)= \{ \begin{array}{lr} 0 & : x ∉ [a_n,b_n]\\ \dfrac{x-a_n}{c_n-a_n} & : x ∈ [a_n,c_n]\\ \dfrac{x-b_n}{c_n-b_n} & : x ∈ [c_n,b_n] \end{array}$$ It is clear by the pasting lemma that each $f_n$ is a continuous function $[0,1]→[0,1]$. As well, it is clear $f_n(x)≠0$ for at most one value of $n$, so that $f_n(x)→0$ for each $x$. However, since $f_n(c_n)=1$ for each $n$, we do not have uniform convergence to the zero function.$~\square$