Problem of the Week #123 - October 6th, 2014

[Chris L T521]

Here's this week's problem!

-----

Problem: Let $\{f_n\}$ be an increasing sequence of continuous functions on $[a,b]$ which converges pointwise on $[a,b]$ to the continuous function $f$ on $[a,b]$. Show that the convergence is uniform on $[a,b]$.

-----

Hint: [sp]Let $\epsilon>0$. For each natural number $n$, define $E_n=\{x\in[a,b]\mid f(x)-f_n(x)<\epsilon\}$. Show that $\{E_n\}$ is an open cover of $[a,b]$ and use the Heine-Borel theorem.[/sp]

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge. You can find his solution below.

[sp]Let $\varepsilon > 0$. Since $f_n$ is increasing and $f_n(x) \to f(x)$ for all $x\in [a,b]$,

\(\displaystyle [a,b] \subset \cup_{N\ge 1} \cap_{n \ge N} A_n,\)

where $A_n := \{x\in [a,b]: f(x) - f_n(x) < \varepsilon\}$. Again, since $\{f_n\}$ is increasing, $\{A_n\}$ is increasing. Thus for each $N$, $\cap_{n\ge N} A_n = A_N$. Therefore, $[a,b]$ is covered by the sets $A_n$. Since $f$ and each $f_n$ are continuous on $[a,b]$, the $A_n$ are open sets. Hence the collection $\{A_n\,:\, n\in \Bbb N\}$ is an open covering of $[a,b]$. Since $[a,b]$ is compact, there exist finitely many indices $n_1, n_2,\ldots, n_k$ such that $[a,b] \subset A_{n_1} \cup A_{n_2} \cup \cdots \cup A_{n_k}$. Let $n_0 = \max\{n_1,n_2,\ldots, n_k\}$. Then $[a,b] \subset A_{n_0}$, which means that $f(x) - f_{n_0}(x) < \varepsilon$ for all $x \in [a,b]$. Since $f_n$ is increasing, $f(x) - f_n(x) \le f(x) - f_{n_0}(x) < \varepsilon$ for all $x\in [a,b]$ and $n\ge n_0$. Since $\varepsilon$ was arbitrary, $f_n$ converges uniformly to $f$ on $[a,b]$.[/sp]