Analysis II HT23, Continuous inverse function theorem

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@State the continuous inverse function theorem.

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?

monotonic-continuous-inverse-is-continuous

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

What are the two separate 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.

continuous-inverse-proof-two-parts
  • 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 $\delta$?

continuous-inverse-proof-target
\[x _ 0 - \varepsilon < f^{-1}(y) < x _ 0 + \varepsilon\]

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

\[ \vert y - y _ 0 \vert < \delta\]

Can you finish the proof from here?

continuous-inverse-proof-conclusion
\[\begin{aligned} & \vert y - y _ 0 \vert < \delta \\ \implies& y _ 0 - \delta < y < y _ 0 + \delta \\ \implies&f(x _ 0 - \varepsilon) < y < f(x _ 0 + \varepsilon) \\ \implies&x _ 0 - \varepsilon < f^{-1}(y) < x _ 0 + \varepsilon \\ \implies&f^{-1}(y _ 0) - \varepsilon < f^{-1}(y) <f^{-1}(y _ 0) + \varepsilon \\ \implies& \vert f^{-1}(y) - f^{-1}(y _ 0) \vert < \varepsilon \end{aligned}\]

Proofs



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