1
$\begingroup$

Suppose $f\colon \mathbb{R} \to \mathbb{R}$ is a continuous function. Let $x\in \mathbb{R}$ and let $(x_n) \subseteq \mathbb{R}$ be a sequence converging to $x$. Let $(y_n)$ be a subsequence of $(x_n)$.

Is the following true?

$$\lim_{n\to \infty} \frac{|f(y_n)-f(x)|}{|y_n-x|}=\lim_{n\to \infty} \frac{|f(x_n)-f(x)|}{|x_n-x|} \, $$ provided $\lim_{n\to \infty} \frac{|f(y_n)-f(x)|}{|y_n-x|}$ exists or is $\infty$.

I think so but can't use algebra of limits because of the denominator.

I am asking this because I am trying to show that if for every sequence $(x_n)$ converging to $x$ there is a subsequence $(y_n)$ such that $\lim_{n\to \infty} \frac{|f(y_n)-f(x)|}{|y_n-x|}=\infty$, then $\liminf_{z\to x}\frac{|f(y)-f(x)|}{|y-x|}=\infty$.

$\endgroup$
1
  • $\begingroup$ Every subsequence of a convergent sequence converges to the same limit as the original sequence. $\endgroup$ – Viktor Vaughn Jul 21 '14 at 13:13
1
$\begingroup$

You can consider both these expressions as sequences themselves, i.e. \[a_n = \frac{|f(y_n)-f(x)|}{|y_n - x|},\quad b_n=\frac{|f(x_n)-f(x)|}{|x_n - x|}.\]

Now, it shouldn't be so hard to see that $y_n$ being a subsequence of $x_n$ means that $a_n$ is a subsequence of $b_n$, and exactly the same conclusions apply: if $b_n$ converges, then $a_n$ converges with the same limit. Note that this means that if you can prove both sequences converge, you can find their common limit by calculating the limit of the subsequence. On the other hand, it is possible for $a_n$ to converge even if $b_n$ does not.

There are functions $f$ that are continuous but not differentiable, and those functions will have sequences like $b_n$ that do not converge with subsequences $a_n$ that do.

$\endgroup$
2
  • $\begingroup$ Thank you for your help. However, would the continuity of $f$ not help at all to establish any conclusions in the case $a_n \to \infty$ ? I know that for arbitrary sequences you could have $b_n= 1$ if $n$ is odd and $b_n=n$ if $n$ is even and then the equality would fail. But for this definition of $a_n$ and $b_n$, and given the continuity of $f$ is it possible to cook up an example such that $a_n \to \infty$ but $b_n$ has a finite limit or no limit at all? $\endgroup$ – Student Jul 21 '14 at 13:33
  • 1
    $\begingroup$ If $a_n \to \infty$, then $b_n$ either does not converge, or it converges to $\infty$. Both of these are possible regardless of the continuity of $f$. $\endgroup$ – Ben Millwood Jul 21 '14 at 14:24

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.