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(From Rudin Principles of Mathematical Analysis, 5.2)

Suppose $f'(x) > 0$ in ($a, b$). Prove that $f$ is strictly increasing in ($a, b$), and let $g$ be its inverse function.

Prove that $g$ is differentiable, and that $g'(f(x)) = \frac{1}{f′(x)} \quad (a < x < b)$


Here's an answer I found online:

Let $g : f(a, b) → (a, b) $ be the inverse function of $f$, i.e., $ g(f(x)) = x $ for all $x ∈ (a, b)$.

We now show that $g′(y) = \lim\limits_{z→y} \frac {g(z) − g(y)}{z − y}$ exists for all $y ∈ f(a, b)$.

Put $y = f(x)$ and $z = f(t)$, where $x, t ∈ (a, b)$, then since $f$ is continuous (by Theorem 5.2), so is $g$ (by Theorem 4.17), and $z → y$ implies $t → x$.

It follows that

$$\begin{align*} \lim_{z→y} \frac{g(z) − g(y)}{z − y} &= \lim_{t→x}\frac{g(f(t)) − g(f(x))}{f(t) − f(x)} \\ &= \lim_{t→x}\frac{t − x}{f(t) − f(x)} \\ &= \lim_{t→x}\frac{1}{\frac{f(t) − f(x)}{t − x}} \\ &= \frac{1}{f′(x)} \end{align*}$$


Question:

Why it is necessary to for g to be continuous? The only step that uses continuity is the changing of the limit values ( $z → y$ implies $t → x$), but that comes from $f$ I think?

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  • $\begingroup$ $g$ must be continuous at $y = f(x)$ so that $\lim_{y \rightarrow z} g(z) = g(y) = g(f(x)) = x$. We really need to be able to plug in $x$ into the function $f^{-1}(x)$. $\endgroup$
    – D.B.
    Apr 7, 2019 at 15:30
  • $\begingroup$ When we g(y) = g(f(x)), aren't we just plugging values into g? (since we defined y = f(x)). I don't see why the limit is involved to get from g(z)-g(y) to g(f(t)) - g(f(x)) $\endgroup$
    – Peter_Pan
    Apr 7, 2019 at 15:53

3 Answers 3

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You do not need to assume that the inverse function $g\colon f([a,b])\to [a,b]$ is continuous because that actually does come for free. To see this, it suffices to show that if $U\subset [a,b]$ is an relatively open interval, then $g^{-1}(U)$ is open in $f([a,b])$. Now, $g^{-1}(U) = f(U)$, and since $f$ is strictly increasing, the image of a relatively open interval in $[a,b]$ under $f$ is another relatively open interval in $f([a,b])$.

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In fact, we do need continuity of $g$.

If only g is continuous, then $z→y $ implies $f(z)→f(y) $. But this is unhelpful, since we want $t→x $.

Recall the only thing we know is that $y = f(x)$ and $z = f(t)$.

From this, we know $$g(z) = g( f(t)) = t$$ (since $g$ and $f$ are inverses).

Similarly: $$g(y) = g( f(x)) = x.$$

Now, suppose $g$ is continuous.

Then $z→y $ implies $g(z)→g(y) $. (def of continuity).

It follows immediately that $t→x $.

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In some sense you are right that the change of limit values comes from the properties $f$. However, proving this is harder than just using a known result that the inverse function of $f$ is continuous. But let's do it just for completeness.

What we want to show is that, given $f(t) \to f(x)$ as $t \to x_0$ for some $x_0 \in (a,b)$, it must be the case that $x_0 = x$. We argue by contradiction. Suppose $x_0 \neq x$. By the continuity of $f$, we have that $f(x_0) = f(x)$. Assume, without loss of generality, that $x < x_0$. Then the mean value theorem tells us there exists $c \in (x,x_0) \subset (a,b)$ such that $$f'(c) = \frac{f(x_0) - f(x)}{x_0 - x} = \frac{0}{x_0-x}=0.$$ But this contradicts that $f'(x) > 0$ for all $x \in (a,b)$. Thus, $x = x_0$ must hold. Hence, we have shown that for any $t,x \in (a,b)$, $f(t) \to f(x)$ implies that $t \to x$.

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