Let $f:\Bbb R\to \Bbb R$ be a continuous function which is strictly increasing, let $g:\Bbb R\to \Bbb R$ be a function such that $f\circ g$ is continuous. Show that $g$ is continuous.

My try:

Let $a\in \Bbb R$. To show continuity of $g$ at $a$, let $g$ be not continuous at $a$. Then for each $n\in \Bbb N$ we have some $(x_n)_n$ such that $|x_n-a|<\dfrac{1}{n}$ but $$\left|g(x_n)-g(a)\right|\ge \dfrac{1}{n}$$

  • Since $f\circ g$ is continuous then for every sequence $x_n\to a $ we have $ (f\circ g)(x_n)\to (f\circ g)(a)$.
  • Also, since $f$ is continuous then for every sequence $x_n\to a $ we have $ f(x_n)\to f (a)$.

But I am failing to understand how to jot these equations to get the required proof.

Please give some hints.

  • $\begingroup$ You seem to begin a contradiction proof. But where is the contradiction ? $\endgroup$ – Peter Oct 17 '16 at 13:24
  • $\begingroup$ For $f$ not strictly increasing this does not hold, so I assume that you mean strictly increasing. Under these assumptions $f$ is invertible, so you may write $$g(x)=f^{-1}(f\circ g)(x)$$ It remains to show that $f^{-1}$ is continuous, but I am not sure how to do it. For example if $f(x)=x$ then $f^{-1}(x)=\frac1x$ with $f^{-1}$ not defined for $x=0$ but continuous everywhere else. $\endgroup$ – Jimmy R. Oct 17 '16 at 13:35

Assume g is not continuous for some $x\in\mathbb{R}$ now let $x_n \rightarrow x$.

Now what happens to $f(g(x_n))$ ? keep in mind that $f$ is increasing and that $max_{k\geq n} \{|g(x_k)-g(x)|\} > \delta >0$. Look for a contradicition

Let me know if this doesn't help you either

  • $\begingroup$ No it does not help me at all $\endgroup$ – Learnmore Oct 17 '16 at 14:45

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.