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.

This is actually a generalized version I wrote of a homework question that intrigued me:

Let $f$ be continuous in $[0,1]$ and $\forall x\in [0,1], \ f(x)=f(1-x)$. If $r\in [0,1]$ and $1-r$ are the only sub limits of $a_{n}$ then $f(a_{n})$ converges.

(the original question states $r=\frac{1}{3}$)

I solved the original question using a method shown in class and tried to implement it to prove the generalized version (see my own answer). I was wondering if there's a better way.

share|improve this question
    
Again: please do not use a half sentence as your title. –  Arturo Magidin Nov 13 '10 at 22:08
1  
@Arturo: I'm not a native English speaker and I have no idea how to write the title differently. I can't write the entire question because it's too long so I wrote the important part. I wrote it the same way you edited my previous question, feel free to edit it again... –  daniel.jackson Nov 14 '10 at 7:34
    
I'm not a native speaker either. I don't want to come in and rewrite your posts, which is why I only added ellipses in the previous one. But perhaps something like "Convergence of $f(a_n)$ when $(a_n)$ has only two limit points" or the like. The general gist, but not the particulars, would be the idea. –  Arturo Magidin Nov 14 '10 at 23:00
add comment

3 Answers

up vote 0 down vote accepted

We claim that $f(a_n) \to \xi = f(r) = f(1-r)$. If $r = \tfrac{1}{2}$ then $(a_n)$ must be convergent (why?). Since $f$ is continuous $f(a_n)$ converges too.

So assume that $r \neq \tfrac{1}{2}$. WLOG we can assume that $r \lt \tfrac{1}{2}$. Consider the subsequence $(a_{n_k})$ consisting of the elements of $(a_n)$ that are less than $\tfrac{1}{2}$ and the subsequence $(a_{n_l})$ consisting of the elements of $(a_n)$ that are greater than or equal to $\tfrac{1}{2}$ and convince yourself that they are in fact subsequences and that they exhaust the $a_n$. In particular $a_{n_k} \to r$ and $a_{n_l} \to 1-r$. Now all that's left to do is to observe that since $f$ is continuous, $f(a_{n_k})$ and $f(a_{n_l})$ are eventually in any open interval around $\xi$. Since they exhaust $(a_n)$ we must have $f(a_n) \to \xi$.

share|improve this answer
add comment

Here's my own attempt at solving this (comments are appreciated!):

We denote $b_{n}=f(a_{n})$ and by Bolzano-Weistrauss, $b_{n_{k}}\to b$. The subsequence $a_{n_{k}}$ might not converge, but again by Bolzano-Weistrauss there exists $a_{n_{k_{m}}}$ that converges to either $r$ or $1-r$ (because those are the only sublimits of $a_{n}$).

  1. if $a_{n_{k_{m}}}\to r$ then $b_{n_{k_{m}}}=f(a_{n_{k_{m}}})\to f(r)$ (because $f$ is continuous in $[0,1]$).
  2. Similarly, if $a_{n_{k_{m}}}\to 1-r$ then $b_{n_{k_{m}}}=f(a_{n_{k_{m}}})\to f(1-r)$.

But since $\forall x\in [0,1], \ f(x)=f(1-x)$, we get $f(r)=f(1-r)$.

To finish the argument we note that $b_{n_{k_{m}}}$ is a subsequence of $b_{n_{k}}\to b$, thus $b_{n_{k_{m}}}\to b$ and we conclude that $b=f(r)=f(1-r)$.

Since $b$ was chosen arbitrarily we get that $b_{n}$ has a single sublimit which implies $b_{n}\to f(r)$

share|improve this answer
add comment

It is quite obvious that if $r$ and $1-r$ are the only sublimits of $\{a_n\}$ in the interval $[0,l]$, then for any $\delta \gt 0$, $(r-\delta, r+\delta)\cup (1-r-\delta,1-r+\delta)$ contains all but finitely many terms of the sequence. Let $\epsilon>0$. Choose $\delta>0$ such that $|f(x)-f(r)|\lt \epsilon$ for all $x\in (r-\delta, r+\delta)$. Then by symmetry of $f$, also $|f(x)-f(r)|\lt \epsilon$ for all $x\in (1-r-\delta, 1-r+\delta)$. It follows that $$|f(a_n)-f(r)|\lt \epsilon$$ for all but finitely many $n$'s. This proves that $\lim_{n\to\infty} f(a_n)=f(r)$.

share|improve this answer
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.