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Here is Prob. 2, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f^\prime(x) > 0$ 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 $$ g^\prime\left( f(x) \right) = \frac{1}{f^\prime(x)} \qquad \qquad \qquad (a < x < b). $$

My effort:

Let $x_1$ and $x_2$ be any two real numbers such that $$ a < x_1 < x_2 < b.$$ Then $f$ is differentiable on the interval $\left[ x_1, x_2 \right]$ and hence on the segment $\left( x_1, x_2 \right)$, and $f$ is of course continuous on the interval $\left[ x_1, x_2 \right]$. So, by the Mean-Value Theorem, we can find a point $x \in \left( x_1, x_2 \right)$, such that $$ f\left( x_2 \right) - f\left( x_1 \right) = \left( x_2 - x_1 \right) f^\prime (x) > 0, $$ and so $$ f\left( x_1 \right) < f\left( x_2 \right) \qquad \qquad \qquad ( a< x_1 < x_2 < b), $$ and therefore $f$ is strictly increasing on $(a, b)$.

As $f$ is strictly increasing on $(a, b)$, so it is injective, and therefore the inverse function $g$ exists; this $g$ is a mapping of $\mathrm{range} f$ into (rather onto) $(a, b)$, and is defined by $$ g \left( y \right) = x \ \mbox{ for all } \ y = f(x) \in \mathrm{range} f.$$ Thus the mapping $h$ of $(a, b)$ into $(a, b)$, given by $$ h(x) = g \left( f(x) \right) \qquad \qquad \qquad ( a< x < b),$$ is the identity mapping. So $$h^\prime(x) = 1 \qquad \qquad \qquad (a < x < b). \tag{1} $$ But by Theorem 5.5 in Baby Rudin (i.e. the Chain Rule), we know that if $g^\prime$ exists at each point in the range of $f$, then we obtain $$h^\prime(x) = g^\prime\left( f(x) \right) f^\prime(x) \qquad \qquad \qquad (a < x < b). \tag{2} $$ From (1) and (2), we can conclude that $$ g^\prime\left( f(x) \right) f^\prime(x) = 1 \qquad \qquad \qquad (a < x < b), $$ and since $f^\prime(x) > 0$ for all $x \in (a, b)$, therefore $$ g^\prime\left( f(x) \right) = \frac{1}{f^\prime(x) } \qquad \qquad \qquad (a < x < b). $$

Is my reasoning so far correct? If so, then how to show that the function $g$ is indeed differentiable (at each point in the range of $f$)?

Or, is there a problem in my reasoning above?

P.S.:

After reading the above answers, I have arrived at this solution to my original problem and would be grateful for the appraisal thereof of the Math SE community.

As $f$ is differentiable at each point in $(a, b)$, so $f$ is continuous in $(a, b)$, and as $f^\prime(x) > 0$ for every $x \in (a, b)$, so $f$ is strictly increasing on $(a, b)$. Thus $f$ is a continuous bijective mapping of $(a, b)$ onto the range of $f$. We now show that the range of $f$ is also an open interval.

Suppose $y_1$ and $y_2$ are any two points in the range of $f$ such that $y_1 < y_2$, and suppose that $y$ is any real number such that $y_1 < y < y_2$. As $y_1$ and $y_2$ are in the range of $f$, so there exist some points $x_1$ and $x_2$ in $(a, b)$ such that $f\left(x_1 \right) = y_1$ and $f \left( x_2 \right) = y_2$; moreover, as $y_1 < y_2$ and as $f$ is strictly increasing, so we must also have $x_1 < x_2$, for otherwise we would obtain $f\left( x_1 \right) \geq f\left( x_2 \right)$.

Now as $f$ is continuous on the interval $\left[ x_1, x_2 \right]$ and as $$y_1 = f \left( x_1 \right) < y < f \left( x_2 \right) = y_2,$$ so by the intermediate-value theorem for continuous functions, there is some point $x \in \left( x_1, x_2 \right)$ such that $y = f(x)$.

Thus we have shown that, for any points $y_1$ and $y_2$ in the range of $f$ such that $y_1 < y_2$, the segment $\left( y_1, y_2 \right)$ is also contained in the range of $f$. That is, the range of $f$ is an interval.

We now show that the range of $f$ is an open interval. For any point $y_0$ in the range of $f$, we have a unique point $x_0 \in (a, b)$ such that $y_0 = f\left(x_0\right)$. But as $a < x_0 < b$, so we also have $$ a < \frac{a+x_0}{2} < x_0 < \frac{x_0+b}{2} < b,$$ and therefore $$ f \left( \frac{a+x_0}{2} \right) < f\left( x_0 \right) < f \left( \frac{x_0+b}{2} \right);$$ that is, for any point $y_0$ in the range of $f$, we have points $f\left(\frac{a+x_0}{2} \right)$ and $f \left( \frac{x_0+b}{2} \right)$ in the range of $f$ such that $$ f\left(\frac{a+x_0}{2} \right) < y_0 < f \left( \frac{x_0+b}{2} \right).$$ So the range of $f$ has no maximum element and no minimum element.

Hence the range of $f$ is a possibly infinite (on either side) open interval, say $(c, d)$.

Let the function $g \colon (c, d) \to (a, b)$ be the inverse of the function $f \colon (a, b) \to (c, d)$. We show that $g$ is continuous by showing that the inverse image under $g$ of every open set in $(a, b)$ is open in $(c, d)$. For this it suffices to show that the inverse image under $g$ of every open interval $(u, v) \subset (a, b)$ is open in $(c, d)$.

But as $f$ and $g$ are bijective and are the inverses of each other, so $$g^{-1} \left[ (u, v) \right] = f \left[ (u, v) \right]. $$ We now show that $$f \left[ (u, v) \right] = \left( f(u), f(v) \right).$$ Suppose $y \in f \left[ (u, v) \right]$. Then $y = f(x)$ for a (unique) point $x \in (u, v)$. As $a < u < x < v < b$ and as $f$ is strictly increasing on $(a, b)$, so we must have $f(u) < f(x) < f(v)$, that is, $y \in \left( f(u), f(v) \right)$.

Conversely, suppose that $y \in \left( f(u), f(v) \right)$. Then $f(u) < y < f(v)$, and $f$ is continuous on the closed interval $[u, v]$; so (by the intermediate-value theorem for continuous functions) there is some (unique) point $x \in (u, v)$ such that $y = f(x)$, which implies that $y \in f\left[ (u, v) \right]$.

Therefore, $f \left[ (u, v) \right] = \left( f(u), f(v) \right)$, which is open in $(c, d)$, as required.

Thus the inverse function $g \colon (c, d) \to (a, b)$ is continuous whenever $f^\prime(x) > 0$ for all $x \in (a, b)$.

Now let $y_0 \in (c, d)$. Then there is a unique point $x_0 \in (a, b)$ such that $y_0 = f\left( x_0 \right)$, which implies that $g \left( y_0 \right) = x_0$.

Since $g$ is continuous at $y_0$, so $$ \lim_{y \to y_0} g(y) = g\left( y_0 \right); $$ that is, $$\lim_{y \to y_0} x = x_0, \ \mbox{ where } \ x = g(y); $$ thus, as $y \to y_0$ in $(c, d)$, the point $x = g(y) \to x_0$ in $(a, b)$.

Now as $f^\prime\left( x_0 \right) \neq 0$, so we find that $$g^\prime \left( y_0 \right) = \lim_{y \to y_0 } \frac{ g(y) - g\left( y_0 \right) }{ y - y_0 } = \lim_{x \to x_0 } \frac{ x - x_0 }{ f(x) - f \left( x_0 \right) } = \lim_{x \to x_0 } \frac{ 1 }{ \frac{ f(x) - f\left( x_0 \right) }{ x - x_0 } } = \frac{1}{ f^\prime \left( x_0 \right) },$$ which shows that $g^\prime$ exists at each point $y_0 \in (c, d)$, and also that $$ g^\prime \left( y_0 \right) = \frac{1}{ f^\prime \left( x_0 \right) },$$ as required.

Have I finally managed to get this solution correct?

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  • $\begingroup$ You are assuming g is differentiable…btw I also did this as a homework yesterday hahaXD I used caratheodory thm twice…see Bartle thm 6.1.8 for details $\endgroup$ – Li Chun Min Apr 9 '17 at 12:36
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The problem in getting $g'(f(x))=1/f'(x)$ from $g(f(x))=x$ is exactly that you have first to prove that $g$ is differentiable.

Lemma. If $f$ is strictly increasing and continuous on $(a,b)$, then the range of $f$ is an open interval and the inverse function of $f$ is continuous.

You can try your hand in proving the lemma. With it, the proof of the inverse function theorem on derivatives is easy. Let $y_0$ belong to the range of $f$, with $y_0=f(x_0)$; then $$ \lim_{y\to y_0}\frac{g(y)-g(y_0)}{y-y_0}= \lim_{x\to x_0}\frac{x-x_0}{f(x)-f(y_0)} $$ by using the substitution $x=g(y)$ in the limit, which is possible because $g$ is continuous.


A sketch of the proof of the lemma.

Assume $f$ is strictly increasing and continuous on $(a,b)$. Then, by the intermediate value theorem its range is an interval, which cannot contain its end points (if bounded). Indeed, suppose the range is upper bounded and let $s$ be the supremum. If $s=f(x)$, for some $x\in(a,b)$, then there exists $y\in(x,b)$ and $s=f(x)<f(y)$: a contradiction. Similarly for the infimum.

Let $g$ be the inverse function of $f$ (it exists because $f$ is injective). Then $g$ is increasing as well. Consider $c$ in the domain $I$ of $g$ (which is an open interval). Then it is easy to prove that $$ \lim_{x\to c^-}g(x)=\sup\{g(t):t\in I, t<c\} $$ The supremum exists, because $g(c)$ is an upper bound.

Similarly, $$ \lim_{x\to c^+}g(x)=\inf\{g(t):t\in I, t>c\} $$ and $g(c)$ is a lower bound. Now prove that both limits must be equal to $g(c)$ (hint: both limits belong to $(a,b)$), so $g$ is continuous.

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  • $\begingroup$ can you please also supply a proof of this lemma? I think I can figure out how the range of $f$ is an interval, but I'm not able to figure out if that range is an open interval and how to prove the continuity of the inverse. So I'd appreciate if you could please write out a proof of the lemma you've stated in your answer. $\endgroup$ – Saaqib Mahmood May 3 '17 at 18:51
  • $\begingroup$ @SaaqibMahmuud I added some notes about it $\endgroup$ – egreg May 3 '17 at 21:52
  • $\begingroup$ yes, I can figure out that these two limits are the supremum and the infimum, respectively, but I'm afraid I'm unable to figure out from your hint how these two limits are equal. And, can you please have a look at my original post and read through my solution in the P.S. and comment on whether or not you find it to be satisfactory enough? $\endgroup$ – Saaqib Mahmood May 4 '17 at 5:54
  • $\begingroup$ @SaaqibMahmuud Sorry, but you're not proving that $g$ is continuous. $\endgroup$ – egreg May 4 '17 at 6:03
  • $\begingroup$ @egreq please have a look at my solution again. I'm showing the continuity of $g$ by showing that the image under $f$ of every open interval is also open (in fact an open interval). Don't you accept that argument as correct? $\endgroup$ – Saaqib Mahmood May 4 '17 at 6:21
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Fix $y$ in the range of $f$. To show that $g'(y)$ exists we need to show that $$\lim_{h\to0}\frac{g(y+h)-g(y)}{h}$$ exists. For $h$ sufficiently small so that $y+h$ is still in the range of $f$, we can use the fact that $f$ is strictly increasing to obtain unique numbers $x_{y+h}$ and $x_y$ in the domain of $f$ such that $f(x_{y+h})=y+h$ and $f(x_y)=y$. The limit we need to show exists is now $$\lim_{h\to0}\frac{g(f(x_{y+h}))-g(f(x_y))}{h},$$ or since $g$ is the inverse of $f$, $$\lim_{h\to0}\frac{x_{y+h}-x_y}{h}$$ needs to be shown to exist. Write $$\frac{x_{y+h}-x_y}{h}=\frac{x_{y+h}-x_y}{f(x_{y+h})-f(x_y)}\frac{f(x_{y+h})-f(x_y)}{h}.$$ The right hand side is a product whose individual limits exist and are equal to $1/f'(x_y)$ and $1$ respectively. Therefore $$\lim_{h\to0}\frac{x_{y+h}-x_y}{h}=\frac{1}{f'(x_y)}.$$

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  • $\begingroup$ how do we know that $$\lim_{h \to 0} \frac{x_{y+h}-x_y}{f(x_{y+h})-f(x_y)} = \frac{1}{f^\prime(x)}?$$ $\endgroup$ – Saaqib Mahmood Apr 9 '17 at 12:26
  • $\begingroup$ @SaaqibMahmuud That's from the fact that the limit of an inverse is the inverse of the limit provided that limit is nonzero. $\endgroup$ – user375366 Apr 9 '17 at 12:33
  • $\begingroup$ can you please rigorously state and prove the result that the limit of an inverse is the inverse of the limit, provided that limit is non-zero? I'm afraid I can't figure this out for myself. Neither has Rudin discussed this phenomenon, as far as I can recall. $\endgroup$ – Saaqib Mahmood May 3 '17 at 18:47

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