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

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
SUMMARY

The problem presented in Problem of the Week #123 involves demonstrating that an increasing sequence of continuous functions $\{f_n\}$ converging pointwise to a continuous function $f$ on the interval $[a,b]$ also converges uniformly on that interval. The solution utilizes the Heine-Borel theorem, establishing that the sets $A_n = \{x \in [a,b] : f(x) - f_n(x) < \varepsilon\}$ form an open cover of $[a,b]$. By leveraging the compactness of $[a,b]$, it is shown that there exists a maximum index $n_0$ such that $f_n$ converges uniformly to $f$ for all $n \geq n_0$.

PREREQUISITES
  • Understanding of pointwise and uniform convergence of functions
  • Familiarity with the Heine-Borel theorem
  • Knowledge of continuous functions on closed intervals
  • Basic concepts of compactness in real analysis
NEXT STEPS
  • Study the properties of uniform convergence in real analysis
  • Explore the implications of the Heine-Borel theorem in different contexts
  • Investigate examples of increasing sequences of functions and their convergence
  • Learn about the relationship between continuity and convergence in functional analysis
USEFUL FOR

Mathematicians, students of real analysis, and educators seeking to deepen their understanding of convergence concepts and their applications in functional analysis.

Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
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!
 
Physics news on Phys.org
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 $x\in [a,b]$,

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

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]
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K