Notes - Analysis II HT23, Continuous inverse function theorem


Can you state the continuous inverse function theorem?


Let $f : I \to \mathbb{R}$ be a strictly monotonic and continuous function on an interval $I$. Then $f^{-1} : f(I) \to I$ is also strictly monotonic and continuous.

If $f : I \to \mathbb{R}$ is strictly monotonic and continuous and $I$ is an interval, then what is true about its inverse?


$f^{-1} : f(I) \to I$ is also strictly monotonic and continuous.

What are the two seperate parts to proving the continuous inverse function theorem, i.e.

Let $f : I \to \mathbb{R}$ be a strictly monotonic and continuous function on an interval $I$. Then $f^{-1} : f(I) \to I$ is also strictly monotonic and continuous.


  • Proving that $f^{-1}$ is well-defined and strictly monotonic
  • Proving that $f^{-1}$ is continuous

When proving the second part of the continuous inverse function theorem,

Let $f : I \to \mathbb{R}$ be a strictly monotonic and continuous function on an interval $I$. Then $f^{-1} : f(I) \to I$ is also strictly monotonic and continuous.

You start by fixing $y _ 0 = f(x _ 0)$ and $\varepsilon > 0$ and then what do you want to be true about any $y$ between $y _ 0 - \delta$ and $y _ 0 + \delta$ for some magical value of $\delta$?


\[x_0 - \varepsilon < f^{-1}(y) < x_0 + \varepsilon\]

When proving the second part of the continuous inverse function theorem,

Let $f : I \to \mathbb{R}$ be a strictly monotonic and continuous function on an interval $I$. Then $f^{-1} : f(I) \to I$ is also strictly monotonic and continuous.

You start by fixing $y _ 0 = f(x _ 0)$ and $\varepsilon > 0$ and then you construct some magical $\delta$ such that for any $ \vert y - y _ 0 \vert < \delta$,

\[x_0 - \varepsilon < f^{-1}(y) < x_0 + \varepsilon\]

Can you draw a picture to represent this? It’s like the standard diagram of continuity but the axes are swapped.


Todo, page 32 of the Analysis II lecture notes if you really want to check.

At the end of the continuous inverse function theorem, you have

\[\delta = \min\\{y_0 - f(x_0-\varepsilon), f(x_0 + \varepsilon) - y_0\\}\]

and are considering

\[|y - y_0| < \delta\]

Can you finish the proof from here?


\[\begin{aligned} &|y - y_0| < \delta \\\\ \implies& y_0 - \delta < y < y_0 + \delta \\\\ \implies&f(x_0 - \varepsilon) < y < f(x_0 + \varepsilon) \\\\ \implies&x_0 - \varepsilon < f^{-1}(y) < x_0 + \varepsilon \\\\ \implies&f^{-1}(y_0) - \varepsilon < f^{-1}(y) <f^{-1}(y_0) + \varepsilon \\\\ \implies& |f^{-1}(y) - f^{-1}(y_0)| < \varepsilon \end{aligned}\]

Proofs

Prove the continuous inverse function theorem:

Let $f : I \to \mathbb{R}$ be a strictly monotonic and continuous function on an interval $I$. Then $f^{-1} : f(I) \to I$ is also strictly monotonic and continuous.


Todo.




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