Let $g : (a, b) → R$ be uniformly continuous on $(a, b)$. Let $\{x_n\}_{n=1}^\infty$ be a sequence in $(a, b)$ converging to $a$. Prove that $\{g(x_n)\}_{n=1}^\infty$ converges.

The general idea here is to use the uniform continuity of g as well as the fact that since the $\{x_n\}$ converges it is Cauchy and to prove that $\{g(x_n)\}_{n=1}^\infty$ is Cauchy.

From uniform continuity we have $\forall\epsilon > 0, \exists\delta > 0$ such that $|x - y| < \delta$ with $x,y \in (a,b)$ then $|g(x) - g(y)| < \epsilon$.

And since $\{x_n\}_{n=1}^\infty$ is Cauchy we have $\forall\epsilon > 0, \exists N \in J$ such that $\forall n,m \ge N$ then $|x_m - x_n| < \epsilon$.

However, I'm not sure how to combine these two given properties in order to get the Cauchy property for $\{g(x_n)\}_{n=1}^\infty$.

Thanks for the help!

  • $\begingroup$ One way to work you way through this kind of problem is to rename everything that's small and positive to $\varepsilon_1$, $\varepsilon_2$, etc. Then, you just have to find out which ones must be equal. $\endgroup$ – xavierm02 Dec 7 '14 at 23:19

Let $\varepsilon > 0$. Because $g$ is uniformly continuous there exists $\delta > 0$ s.t. $$ |x-y| < \delta \implies |g(x)-g(y)| < \varepsilon, \; x,y, \in (a,b) $$ Because $(x_n)_{n=1}^\infty$ is Cauchy sequence, there exists $n_0 \in \Bbb{N}$ s.t. $|x_n - x_k| < \delta$ when $n \geq n_0$ and $k \geq n_0$. So, if $n, k \geq n_0$ holds, we have $$ |g(x_n) - g(x_k)| < \varepsilon $$ so $(g(x_n))_{n=1}^\infty$ is a Cauchy sequence. Because $\Bbb{R}$ is complete, $(g(x_n))_{n=1}^\infty$ converges.

| cite | improve this answer | |
  • $\begingroup$ How do you know $|g(x_n) - g(x_k)| < \epsilon$ though? $\endgroup$ – jlang Dec 7 '14 at 23:54
  • $\begingroup$ We chose $n \geq n_0$ and $k \geq n_0$. Then $|x_n - x_k| < \delta$. The $\delta$ was chosen to be such that $|x-y| < \delta \implies |g(x)-g(y)| < \varepsilon$. So now $|g(x_k) - g(x_n)| < \varepsilon$. $\endgroup$ – desos Dec 8 '14 at 8:59

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.