Continuity of an inverse of a function

[archaic]

Hey, please tell me if the following is correct.
We have a continuous, increasing and strictly monotonic function on $[a, b]$, and $x_0\in[a,b]$. Let $g(y)$ be its inverse, and $f(x_0)=y_0$.
I want to show that $|y-y_0|<\delta\implies|g(y)-g(y_0)|<\epsilon$.
\begin{align*}
|g(y)-g(y_0)|<\epsilon&\Leftrightarrow x_0-\epsilon<g(y)<x_0+\epsilon\\
&\Leftrightarrow f(x_0-\epsilon)<y<f(x_0+\epsilon)\\
&\Leftrightarrow f(x_0-\epsilon)-f(x_0)<y-f(x_0)<f(x_0+\epsilon)-f(x_0)\\
&\Leftrightarrow -(y_0-f(x_0-\epsilon))<y-y_0<f(x_0+\epsilon)-y_0\\
\end{align*}
If I let $\delta=\min(y_0-f(x_0-\epsilon),f(x_0+\epsilon)-y_0)$, while considering small $y$s, then I think that I have it right.
For decreasing $f(x)$, I can flip the inequality symbols at the second step, and choose $\delta=\min(f(x_0-\epsilon)-y_0,y_0-f(x_0+\epsilon))$.
Thanks!

 

[madafo3435]

I suppose what you want to prove is that the inverse of f is continuous in its definition set. That being the case, I notice that in your test you are assuming, you are assuming that the inverse is well defined and is univocal, you must prove this before proceeding with those equivalences. I also see it a bit lost at the end, first justify what I have mentioned and admeas, prove that the inverse also grows strictly in an interval, and using this you can give a correct test of the continuity of the inverse.

 

[soloenergy]

Yes, your logic and calculations seem correct. By choosing a small enough value for δ, you can ensure that the difference between g(y) and g(y0) is less than ε for any y within δ distance from y0. This shows that g(y) is continuous at y0, since it satisfies the delta-epsilon definition of continuity. Good job!