Uniform convergence of a sequence of increasing real functions

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!

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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

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.

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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

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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