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Suppose $f$ is a real-valued function, and $a_n,b_n$ are real sequences.

Prove that $f$ is uniformly continuous $\iff (\lim a_n=\lim b_n \implies \lim f(a_n)=\lim f(b_n))$


  • Suppose $f$ is uniformly continuous. Then, in particular, $f$ is continuous. Suppose $a_n\to x$ as $n\to\infty$. We know by continuity of $f$ that: $$\lim a_n=x\implies \lim f(a_n)=f(x)$$ We know that $\lim a_n=\lim b_n=x$ so that also: $$\lim b_n=x\implies \lim f(b_n)=f(x)$$

    Together this gives that if $f$ is uniformly continuous, that $\lim f(a_n)=\lim f(b_n)$.


  • Suppose $f$ is not uniformly continuous. We then know that: $$\forall_{\delta>0}\exists_{\epsilon>0}:|x-a|<\delta \not \Rightarrow |f(x)-f(a)|<\epsilon$$

    Suppose we know that $\lim a_n=a=\lim b_n=b$, we want to show that $\lim f(a_n)\neq f(b_n)$.

    As the sequences are arbitrary in $\mathbb{R}$, we know that: $a-b=0\implies|a-b|<\delta$.

    As $f$ is not uniformly continuous, we know that for a particular $\delta>0$ we pick, the following holds: $$|a-b|<\delta \not \Rightarrow |f(a)-f(b)|<\epsilon \ \ \ (*)$$ Now suppose that $\lim f(a_n)=\lim f(b_n)$ so that $f(a)=f(b)$. Then for any $\epsilon>0$: $$|a-b|=0<\delta\implies|f(a)-f(b)|=0<\epsilon$$ This is contradictory with the statement $(*)$; non-uniform continuity. Thus: $f(a)$ cannot equal $f(b)$ if $a=b$ so that if $f$ is not uniformly continuous, we know that: $$\lim a_n = \lim b_n \not \Rightarrow \lim f(a_n) = \lim f(b_n)$$ $\tag*{$\Box$}$

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  • $\begingroup$ "Now suppose that $\lim f(a_n) = \lim f(b_n)$ so that $f(a) = f(b)$". As I read it you denote $a = \lim a_n$ and $b = \lim b_n$ and since $\lim a_n = \lim b_n$ implies $\lim f(a_n) = \lim f(b_n)$ you (incorrectly) conclude that $f(a) = \lim f(a_n)$ and $f(b) = \lim f(b_n)$. This is it not, true, since you assumed that $f$ is not continuous. And that follows precisely from continuity. You see it? $\endgroup$
    – Olba12
    Mar 8 '18 at 16:02
  • $\begingroup$ @Olba12 Yes I see what is wrong now, but at this point I've tried so many things that I guess I'm just done with this assignment. Proving from right to left seems near impossible to me $\endgroup$
    – Marc
    Mar 8 '18 at 16:11
  • $\begingroup$ What is the domain of $f?$ $\endgroup$
    – zhw.
    Mar 8 '18 at 16:26
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There are various things wrong here.

In the title of your question you require the existence of two sequences; but in the body it becomes clear that you mean: for all pairs of sequences satisfying $\lim a_n=\lim b_n$ something happens. In reality the criterion outlined in skin tone just describes ordinary continuity: Take all $b_n=b$, and you obtain the standard definition.

When it comes to your proof the first big mistake is your description of "not uniformly continuous". It does not mean that for all $\delta>0$ there is an $\epsilon>0$ such that something disagreable happens. Instead it means that there is an $\epsilon_0>0$ such for all $\delta>0$ you can find two points $x$, $y$ with $|x-y|<\delta$ and at the same time $|f(x)-f(y)|\geq\epsilon_0$.

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If you go for contradiction, it would mean whatever $\delta$ you take, we have that $\forall$ $\epsilon$ we have $|x-a_n|< \delta \not \Rightarrow |f(x) - f(a_n)| < \epsilon$. Now take $x=b_n$, and since you can take whatever $\delta$ you want, we have that $\lim a_n = \lim b_n$. Now what does that imply?

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  • $\begingroup$ We know that $\lim a_n - \lim b_n = 0 < \delta$, but as the implication is negated, you don't know whether the right-hand side might be true or not. How to proceed? I have $\lim(a_n-b_n)=0 \not\rightarrow \lim(f(a_n)-f(b_n))=f(0)$ as $f$ is not (uniformly?) continuous. Can we use that $\lim f(a_n-b_n)=f(0)$? $\endgroup$
    – Marc
    Mar 8 '18 at 14:23
  • $\begingroup$ We knot that $\lim a_n = \lim b_n$ but this implied $\lim f(a_n) = \lim f(b_n)$, contradiction. @Marc $\endgroup$
    – Olba12
    Mar 8 '18 at 14:56
  • $\begingroup$ If you negate the proposition, you get that you want to prove that $f$ is not continuous $\implies \lim a_n=\lim b_n \not \Rightarrow \lim f(a_n) = \lim f(b_n)$, or not? How does in the negation, the equality of limits of $a_n,b_n$ imply the equality of limits of $f(a_n),f(b_n)$? $\endgroup$
    – Marc
    Mar 8 '18 at 15:02
  • $\begingroup$ No, we prove by contradiction, not neglection. We assumed that $f$ was not uniform, but still $ \lim a_n = \lim b_n \Rightarrow \lim f(a_n) = \lim f(b_n)$ holds. And then we concluded that in order for $ \lim a_n = \lim b_n \Rightarrow \lim f(a_n) = \lim f(b_n)$ to hold, our first assumption (not uniform) must be wrong, i.e a contradiction. @Marc $\endgroup$
    – Olba12
    Mar 8 '18 at 15:13
  • $\begingroup$ Thank you for elaborating. I just wrote something out and I think I might have it! I'm going to update my proof; could you check if it's any good? $\endgroup$
    – Marc
    Mar 8 '18 at 15:19
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$\impliedby$ fails. On $\mathbb R,$ let $f(x) = x^2.$ Suppose $\lim a_n = \lim b_n.$ Denote the common limit by $L.$ Then by continuity of $f,\lim f(a_n) = f(L) = \lim f(b_n).$ But $f$ is not uniformly continuous on $\mathbb R.$

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