3
$\begingroup$

Can someone verify whether my attempt to prove this theorem is correct?

Notice that I use a generalized definition of derivative:

Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Let $p$ both a point and a limit point of $E$. Then we define the derivative of $f$ at $p$ as the limit $f'(p) = \lim_{x \to p, x \in E \setminus \{p\}} \frac{f(x)-f(p)}{x-p}$, provided the limit exists.

Theorem: Let $f: E \subseteq \mathbb{R} \to F \subseteq \mathbb{R}$ be an invertible function that is differentiable at $p \in E$. Suppose that $f^{-1}: F \to E$ is continuous at $f(p)$ and that $f'(p) \neq 0$. Then $f^{-1}$ is differentiable at $f(p)$, and we have

$$(f^{-1})'(f(p)) = \frac{1}{f'(p)}$$

Proof: Before proving the theorem, we have to check that differentiation at $f(p)$ makes sense: we have to show that $f(p)$ is a limit point of $Y$.

Let $\epsilon > 0$. Because $f$ is differentiable at $p$, $f$ is continuous at $p$ and it follows that $|f(p)-f(x)| < \epsilon$ whenever $x \in (p- \delta, p + \delta) \cap E \setminus \{p\}$ for some $\delta > 0$. Notice: $f'(p) \neq 0$ implies that $f$ is not constant on $(p- \delta, p + \delta) \cap E \setminus \{p\}$ (if it were constant, we would have $f'(p) = \lim_{x \to p} \frac{f(x)-f(p)}{x-p} = \lim_{x \to p, x \in E \cap (p- \delta, p + \delta)\setminus \{p\}}\frac{f(x)-f(p)}{x-p} = 0)$. Combining these facts, we deduce that $0 < |f(x)-f(p)| < \epsilon$ for some $x \in E$, and $f(p)$ is a limit point of $F = f(E)$.

Define $$F: E \to \mathbb{R}: x \mapsto \begin{cases} \frac{f(x)-f(p)}{x-p} \quad x \neq p \\ f'(p) \quad x = p\end{cases}$$

Clearly, $F$ is continuous at $p$.

The theorem now follows from the following easy calculation:

$$(f^{-1})'(f(p)) = \lim_{y \to f(p)} \frac{f^{-1}(y)- f^{-1}(f(p))}{y-f(p)}$$

$$= \lim_{y \to f(p)}\frac{1}{\frac{f(f^{-1}(y))-f(p)}{f^{-1}(y)- p}}$$

$$ = \lim_{y \to f(p)} \frac{1}{F(f^{-1}(y))}$$

$$= \frac{1}{F( \lim_{y \to f(p)} f^{-1}(y))} = \frac{1}{F(p)} = \frac{1}{f'(p)}$$

However, some equalities need some explanation. The second equality is justified by noticing that $f^{-1}(y) - p = f^{-1}(y) - f^{-1}(f(p))$ is zero only when $y = f(p)$, so there are no trouble with dividing by zero. The fourth equality uses the continuity of $F$ at $p$ and the fifth equality follows from the continuity of $f^{-1}$ at $f(p) \quad \square$

$\endgroup$
0
1
+50
$\begingroup$

This looks fine to me with one exception. You don't want to claim that $f$ isn't constant on $(p-\delta,p+\delta)$. You want to claim that $f(x)\ne f(p)$ for all $x$ with $0<|x-p|<\delta$. (The proof for that is not quite the proof you've supplied.) You need this because you need to know that as $y\to f(p)$ you can make sense of the inverse function (and you never have a division by zero problem). In some sense, the statement of your theorem is flawed, as you're assuming existence of the inverse function. That should be part of what you prove here — namely, that $f$ is one-to-one on some interval around $p$.

$\endgroup$
12
  • $\begingroup$ Thanks for your answer! It made me realise that the constant part needs modification. I don't really get what's wrong with the theorem statement though. It's completely similar to Tao's statement about this theorem in his analysis book, but he provides another proof. $\endgroup$ – user370967 Aug 22 '18 at 21:54
  • $\begingroup$ The whole point (and, even more importantly so in multivariable) is to have a sufficient condition to be guaranteed an inverse function (locally). [When you get to multivariable analysis/calculus, you might find my YouTube lectures of interest. They're linked in my profile.] $\endgroup$ – Ted Shifrin Aug 22 '18 at 21:56
  • $\begingroup$ Can't we just, if an inverse function locally exist, restrict the domain and then apply this theorem? $\endgroup$ – user370967 Aug 22 '18 at 22:00
  • $\begingroup$ But you don't want to assume it. You want to know that if $f'(p)\ne 0$, then you're guaranteed to have an inverse function on an interval around $p$ ... plus continuity, differentiability, and the formula for the derivative. $\endgroup$ – Ted Shifrin Aug 22 '18 at 22:03
  • $\begingroup$ The problem here is first that my domain isn't necessarily an interval. But let's assume for a moment it is. The condition $f'(p) \neq 0$ should imply that we locally have an inverse? Is there an easy proof? I can see that if we have $f'(p) \neq 0$ in a nbh, we can use IVT for derivatives to show that $f$ is monotonic on a nbh and hence invertible. But is this true? $\endgroup$ – user370967 Aug 22 '18 at 22:10
1
$\begingroup$

Your work will be done faster if not make use of the First Principle of Derivative.

We know that $f^{-1}(f(x))=x$, differentiate it combining with Chain Derivative, we have $$\begin{align}(f^{-1}(f(x)))'&=1\\f^{-1}{'}(f(x))f'(x)&=1\\\therefore f^{-1}{'}(f(x))&=\dfrac1{f'(x)}\end{align}$$

I am not too used in rigorous proof in derivative (such like the first principle) but I think your proof is good enough.

$\endgroup$
2
  • 2
    $\begingroup$ I'm sorry, but I think you can't use the chain rule, otherwise you assume that $f^{-1}$ is differentiable at $f(x)$ (and this is to be proven). $\endgroup$ – user370967 Aug 19 '18 at 13:14
  • 2
    $\begingroup$ Yes, this is the typical "sloppy" plug-and-chug derivation one finds in some calculus books. But it indeed assumes the main issue — that the inverse function is differentiable in the first place. $\endgroup$ – Ted Shifrin Aug 22 '18 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy