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!


  • 3
    $\begingroup$ What is the domain of definition $I$ of the function $F$? $\endgroup$ Sep 29 '12 at 20:55
  • $\begingroup$ Is the sequence increasing, or are each $F_{n}$ increasing? $\endgroup$
    – T. Eskin
    Sep 29 '12 at 21:00
  • $\begingroup$ @ThomasE. each $F_n$ is increasing $\endgroup$
    – P. M. O.
    Sep 29 '12 at 21:24
  • $\begingroup$ where have you found this problem? $\endgroup$
    – Exodd
    Jul 29 at 10:08

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.


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

  • $\begingroup$ Your fnction $F_n$ is not increasing: $F_n(-n-1)<F_n(0)>F_n(n+1)$. $\endgroup$ Sep 29 '12 at 21:15
  • $\begingroup$ Oh my bad I thought the sequence of functions was increasing, not the functions themselves.. Thanks I was too hasty.. $\endgroup$
    – clark
    Sep 29 '12 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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