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From Pugh's analysis book, prelim problem 57 from Chapter 4:

Let $f$ and $f_n$ be functions from $\Bbb R$ to $\Bbb R$. Assume that $f_n(x_n)\to f(x)$ as $n\to\infty$ whenever $x_n\to x$. Prove that $f$ is continuous. (Note: the functions $f_n$ are not assumed to be continuous.)

here's my attempt: assume $x_n \to x$. we want to show that $f(x_n) \to f(x)$. so $|f(x_n) - f(x)| \leq |f(x_n)-f_n(x_n)| + |f_n(x_n)-f(x)|$. The second term can be made to be less than any $\varepsilon > 0$ for $n$ sufficiently large. i'm having trouble with the first term. can anyone help? thank you!

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2  
need some uniform property? –  Yimin Feb 22 '13 at 4:00
1  
this is from Pugh's real mathematical analysis. –  Aden Dong Feb 22 '13 at 4:03
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specifically, question 57 of the prelim problems in chapter 4. –  Aden Dong Feb 22 '13 at 4:04
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I think the question may be: prove the function $\,f\,$ is continuous at $\,x\,$ if $$x_n\xrightarrow[n\to\infty]{} x\Longrightarrow f(x_n)\xrightarrow[n\to\infty]{} f(x)$$ –  DonAntonio Feb 22 '13 at 4:08
2  
This is called continuous convergence. Carathéodory advertised its use in complex analysis, but (unfortunately?) it never really caught on. –  Martin Feb 22 '13 at 4:58

4 Answers 4

up vote 5 down vote accepted

Suppose $(x_n)$ converges to $x$ and let $\epsilon>0$.

Note that the hypotheses imply $(f_n)$ converges to $f$ pointwise. From this, choose a subsequence $(f_{n_k})$ of $(f_n)$ such that $$ |f_{n_k} (x_k) -f(x_k)|<\epsilon$$ for every $k$.

Claim: $f_{n_k}(x_k)$ converges to $f(x)$.

Proof of Claim: Consider the sequence $(y_n)$ $$(y_n)= (\underbrace{x_1,x_1,\ldots,x_1}_{n_1\text{-terms}}\,,\, \underbrace{x_2,x_2,\ldots,x_2}_{n_2-n_1\text{-terms}}\,,\, \underbrace{x_3,x_3,\ldots,x_3}_{n_3- n_2 \text{-terms}},\ldots). $$ Since this sequence converges to $x$, we have $f_n(y_n)\rightarrow f(x)$. Thus $f_{n_k}(y_{n_k})=f_{n_k}(x_k)\rightarrow f(x)$.


Now use the Claim and the inequality: $$ |f(x_k)-f(x)| \le |f(x_k)-f_{n_k}(x_k)|+|f_{n_k}(x_k)-f(x)|. $$

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Nice proof, +1. You have a typo: it should be $f(x)$ and not $x$ in your proof of the claim. –  1015 Feb 22 '13 at 13:09
    
@julien Thanks, corrected. –  David Mitra Feb 22 '13 at 13:10
    
There were two. It remains one. –  1015 Feb 22 '13 at 13:12
    
@julien I need more coffee... –  David Mitra Feb 22 '13 at 13:13
    
Could you explain to me why we can choose the sequence you write about in line 4? Why is it true that the diestance is less than $\epsilon$? –  Hagrid Mar 18 '13 at 11:10

Here's a hint: fix $n$ and consider the constant sequence $x_m = x_n$ for all $m$. This converges to $x_n$, so your assumption tells you that $f_m(x_n)$ converges to $f(x_n)$ as $m \to \infty$. Now look at building inequalities involving things like $f_m(x_n)$, and don't forget that convergent sequences are also Cauchy.

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ahh great hint. thank you! –  Aden Dong Feb 22 '13 at 4:53

Nice Problem. $\newcommand\abs[1]{\left\lvert#1\right\rvert}$There's another approach:

Suppose $\{x_n\}$ is an arbitrary sequence which converges to $x$. It suffices to prove that $$f(x_n)\to f(x)\tag{*}$$

The key is to construct an increasing integer sequence $0<N_1<N_2<\dotsb$ such that $$\abs{f_{N_k}(x_k+1/N_k)-f(x_k)}<1/k\tag1$$ We'll construct terms inductively.

Put $N_0=0$, and suppose $N_0,\dotsc,N_{k-1}$ are constructed. Now consider sequence $\{f_m(x_k+1/m)\}$. Since $x_k+1/m\to x_k$ as $m\to\infty$, we have $$\lim_{m\to\infty}f_m(x_k+1/m)=f(x_k)$$ So there's some $N_k>N_{k-1}$ such that $\abs{f_{N_k}(x_k+1/N_k)-f(x_k)}<1/k$.

Now let's consider sequence, namely $\{a_n\}$: $x_1+1/N_1,\dotsc,x_1+1/N_1,x_2+1/N_2,\dotsc,x_2+1/N_2,\dotsc$. $x_1+1/N_1$ appears $N_1-N_0$ times, followed by $x_2+1/N_2$ with $N_2-N_1$ times, then $x_3+1/N_3$ with $N_3-N_2$ times, and so forth, we have $a_n\to x$, so $f_n(a_n)\to f(x)$ as $n\to\infty$

Notice that $a_{N_k}=x_k+1/N_k$, and $f_{N_k}(a_{N_k})=f_{N_k}(x_k+1/N_k)$, therefore $$\lim_{k\to\infty}f_{N_k}(x_k+1/N_k)=f(x)\tag2$$

(*) follow from (1) and (2)

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Note that $f_{n}(x) \to f(x)$ for all $x$.

Assume that f is discontinuous at a.This means that

there is a sequence $x_n \to a$ while

$f(x_n)\notin [f(a)-2\epsilon ,f(a)+2\epsilon] $ for some $\epsilon>0 $.

Then construct another sequence $y_n $ as follows:

$y_1=y_2=...=y_{N_1}=x_1 $ where $ |f_{N_1}(x_1) -f(a)|>\epsilon$

$y_{N_1+1}=y_{N_1+2}=...=y_{N_{2}}=x_2$ where $ |f_{N_2}(x_2)-f(a)|>\epsilon$ and $N_2>N_1$

...

Now , $y_n\to a$ but $f_n(y_n)\to\neq f(a)$ !

So , we get a contradiction.

f is continuous.

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