Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $ \{ F_n \}_{n \in \mathbb{N}}$ a sequence of increasing functions of $\mathbb{R}$ in $\mathbb{R}$ that converges pointwise to a continuous and bounded function $F: I \longrightarrow \mathbb{R}$. Show that $\{ F_n \}_{n \in \mathbb{N}}$ converges uniformly to $F$

Any help is welcome!

Thanks!

share|improve this question
3  
What is the domain of definition $I$ of the function $F$? –  Jean-Sébastien Sep 29 '12 at 20:55
    
Is the sequence increasing, or are each $F_{n}$ increasing? –  Thomas E. Sep 29 '12 at 21:00
    
@ThomasE. each $F_n$ is increasing –  P. M. O. Sep 29 '12 at 21:24
add comment

2 Answers

up vote 2 down vote accepted

If $I$ is not compact, the claim is false: Let $I=(0,1)$ and $F_n(x)=x^n$. Then $F_n$ is a strictly increasing function. and $(F_n)$ converges pointwise to the continuous and bounded zero function. However, the convergence is not uniform as $\sup |F_n(x)-F(x)|=1$ for all $n$.


Now assume that $I=[a,b]$ is compact. (Thus the condition that $F$ be bounded is superfluous: It is a consequence of its continuity). Assume $\epsilon>0$ is given. For $x\in I$, the set $U_x:=F^{-1}\left((F(x)-\frac\epsilon9,F(x)+\frac\epsilon9)\right)$ is a relative open subset of $I$. Hence we can find $r_x>0$ such that the relative open set $V_x:=B(x,r_x)\cap I$ is $\subseteq U_x$. We have $I=\bigcup_{x\in I} V_x$. By compactness this there exists a finite subcover, i.e. we find $x_0<x_1<\cdots <x_m$ such that $I=\bigcup_{x\in I} V_x$. Wlog. $x_0=a$, $x_m=b$. Among all such sequences $(x_k)$ we select one with minimal $m$.

Assume there is a $k$ such that $V_{x_k}\cap V_{x_{k+1}}=\emptyset$. Then there is a point $x$ between $V_{x_k}$ and $V_{x_{k+1}}$ (we may take for example $x=x_k+r_k$) that is covered by a $V_{x_i}$ with either $i<k$ or $i>k+1$. In the first case, we see that $V_{x_k}\subset V_{x_i}$ and hence $x_k$ can be dropped; in the second case, we can similarly drop $x_{k+1}$. In both cases we obtain a shorter sequence, contrary to the assumption that $m$ is minimal (note that we do not drop $x_0$ or $x_m$).

From $V_{x_k}\cap V_{x_{k+1}}\ne\emptyset$ we conclude that $|F(x_k)-F(x_{k+1})|<\frac{2\epsilon}9$ (because $|F(x_k)-F(x)|<\frac\epsilon9$ and $|F(x_{k+1})-F(x)|<\frac\epsilon9$ for some $x\in V_{x_k}\cap V_{x_{k+1}}$). In fact, $x_k\le x\le x_{k+1}$ implies that $|F(x)-F(x_k)|<\frac\epsilon9$ or $|F(x)-F(x_{k+1})|<\frac\epsilon9$ (in other words: $|F(x)-F(x_{r})|<\frac\epsilon9$ for $r=k$ or $r=k+1$).

Per pointwise convergence at $x_0,\ldots,x_m$ there exists $N\in\mathbb N$ such that $|F_n(x_k)-F(x_k)|<\frac{2\epsilon}9$ for all $n>N$ and $0\le k\le m$. Then for $n>N$ and $x\in I$ we find $k$ with $x_k\le x\le x_{k+1}$. Then $$|F_n(x_{k+1})-F_n(x_k)|\\\le |F_n(x_{k+1})-F(x_{k+1})|+|F(x_{k+1})-F(x_k)| +|F(x_{k})-F_n(x_k)|\\ <\frac{2\epsilon}9+\frac{2\epsilon}9+\frac{2\epsilon}9=\frac{2\epsilon}3.$$ By the above remarks we have $|F(x)-F(x_r)|<\frac\epsilon9$ for $r=k$ or $r=k+1$. and therefore (using either $F_n(x_k)=F_n(x_r)\le F_n(x)\le F_n(x_{k+1})$ or $F_n(x_k)\le F_n(x)\le F_n(x_r)=F_n(x_{k+1})$) $$|F_n(x)-F(x)|\le |F_n(x)-F_n(x_r)|+|F_n(x_r)-F(x_r)|+|F(x_r)-F(x)|\\ <|F_n(x_{k+1})-F_n(x_k)|+\frac{2\epsilon}9+\frac\epsilon9\\ <\frac{2\epsilon}3+\frac{2\epsilon}9+\frac\epsilon9=\epsilon. $$ Therefore $\sup |F_n(x)-F(x)|<\epsilon$ for all $n>N$, i.e. the convergence is uniform.

share|improve this answer
add comment

I assume $$I= \mathbb{R}$$ Take $F_n(x)= 1 \forall x \in [-n,\infty] F_n(x)=0\,\, \mathrm{otherwise}$

Then $\lim F_n =F$

Where $F(x)= \forall 1 \in \mathbb{R}$ so $F$ is continuous also $$ | \!| F_n - F| \!| =1 \forall n \in \mathbb{N}$$ So the convergence is not uniform.

You may be interested in Dini's Theorem http://www.math.ubc.ca/~feldman/m321/dini.pdf

share|improve this answer
    
Your fnction $F_n$ is not increasing: $F_n(-n-1)<F_n(0)>F_n(n+1)$. –  Hagen von Eitzen Sep 29 '12 at 21:15
    
Oh my bad I thought the sequence of functions was increasing, not the functions themselves.. Thanks I was too hasty.. –  clark Sep 29 '12 at 21:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.