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Question: 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 $$ g'(f(x)) = \frac{1}{f'(x)}$$ for $(a< x < b)$.

Relevant concepts needed

1) Definition: Let $f$ be defined and real valued on $[a,b]$. For any $x\in [a,b]$ we form the quotient: $$\phi(t) = \frac{f(t) - f(x)}{t-x}$$ for $(a < t < b, t \neq x)$ and define $$f'(x) = \lim_{t \rightarrow x} \phi(t)$$ Provided the limit exists. This is our defintion of the derivative at $x$.

2) Since $g$ is defined as the inverse function of $f$, then $g(x) = f^{-1}(x)$

I've already done the first and second parts of the question: Showing $f$ is strictly increasing and $g$ is differentiable. My issue is with the third portion, showing $ g'(f(x)) = \frac{1}{f'(x)}$.

I made some attempts and was not successful and then I found solutions for the question both of which were similar in style. I'm going to reproduce the solution here but with questions about the reason / intuition behind each step. I've reached a stage in my mathematical journey where I feel I'm hitting a glass ceiling preventing me from making the jump to possess the ability to think more "creatively" about these concepts. So I hope trying to understand how the masters approach these situations would be enlightening. Without further ado the solution:

Solution:

Fix $\epsilon > 0$ and $x \in (a,b)$. $f'(x) > 0$ means that there exists $\eta > 0$ such that $ 0 < |x - t| < \delta_{1}$ implies $$\Bigg|\frac{f(t) - f(x)}{t-x} \Bigg| > \eta$$

$$\Rightarrow \Bigg| \frac{1}{\frac{f(t) - f(x)}{t-x}} \Bigg| < \frac{1}{\eta}$$

$\textit{$\textbf{Comment/Question}$}$: What is the reason behind this step? The idea behind the existence of an $\eta$ makes sense to me by Archimedian properties of the Reals, but I don't see the intuition behind needing to establish this? I get vibes that there is some sort of argument that is going to involve continuity, but still not sure of its placement.

From the definitiuon of $f'(x)$, we also have for $\eta f'(x) \epsilon > 0$ there exists a $\delta_{2} > 0$ such that $ 0 < |x - t| < \delta_{2}$ implies $$\Bigg|f'(x) - \frac{f(t) - f(x)}{t-x} \Bigg| > \eta f'(x) \epsilon$$

$\textit{$\textbf{Comment/Question}$}: Again I have no idea the purpose of this step, but I again get continuity sort of vibes from this portion.

Let $y = f(x)$ and $\delta = min\{\delta_{1},\delta_{2} \}$. Then for any $u \in (f(x-\delta), f(x + \delta)$, let $g(u) = t$ so $t \in (x-\delta, x + \delta)$ and $$\Bigg| \frac{g(u) - g(y)}{u-y} - \frac{1}{f'(x)} \Bigg|= \\"magical\ manipulations" < \frac{ \eta f'(x) \epsilon}{\eta f'(x)} = \epsilon$$

Which shows $ g'(f(x)) = \frac{1}{f'(x)}$.

$\textit{$\textbf{Comment/Question}$}$: I get how to do the mechanical manipulations to arrive at the result, but what was the purpose of defining $u \in (f(x-\delta), f(x + \delta)$, let $g(u) = t$ so $t \in (x-\delta, x + \delta)$ ? I get the purpose of it in terms of trying to align our argument with the definition of continuity, but it still seems uncomfortable that we can arbitrarily just "let this equal this". This probably falls under my inability to think more creatively about the solution.

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  • $\begingroup$ The reason all this seems magical is you are reading it backwards; when doing an argument like this, you already know what the answer is and start from the end and work backwards. You eventually get a sum of $n$ things less than epsilon, and then you divide each by $n$ so that the sum is less than epsilon. Of course, this proof is quite a bit cleaner when not done via $\epsilon-\delta$ arguments, but once one is familiar with them each step makes sense. It's merely a matter of familiarity and knowing what to expect will be needed. $\endgroup$ – Brevan Ellefsen May 1 '19 at 22:27
  • $\begingroup$ I understood the "magical manipulations" at the very end. It was everything else that has me stumped $\endgroup$ – dc3rd May 1 '19 at 22:29
  • $\begingroup$ Ah, but then you miss the point. There is nothing more special about any of the other manipulations than the ones at the end. The first one is just because we know the answer should have $f'(x)$ in the denominator, so we are going to have to flip it at some point. The second step is the definition of a limit, which we know we will have to use in the process since we are doing our argument $\epsilon-\delta$ style. The author just knows ahead of time what will need to be used due to familiarity with the process, so steps are done seemingly "out of nowhere" to someone less experienced. $\endgroup$ – Brevan Ellefsen May 1 '19 at 22:31
  • $\begingroup$ Your explanation does bring some clarity to me. But I am still confused as to why we have to use this $\delta$ - $\epsilon$ process? So for context, in a question like this the first thing I would ask myself is "what does it mean for $g'(f(x)) = \frac{1}{f'(x)}$? And from that I would attempt to construct a solution fulfilling that. But I guess in this case I don't really "know what it means" for $g'(f(x)) = \frac{1}{f'(x)}$. $\endgroup$ – dc3rd May 1 '19 at 23:10
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Intuition: on an interval small enough, and with the origin shifted appropriately, $f(x)$ is close enough to a linear function with positive slope. For such a function, to be denoted $f_{L}$, we have: $$ f_{L}(x) = \alpha x, \quad \alpha > 0. $$ By direct inspection: $$ f_{L}'(x) = \alpha > 0, \quad g_{L}(y) = f_{L}^{-1}(y) = \alpha^{-1} y, \quad g_{L}'(y) = \alpha^{-1} = {1 \over f_{L}'(x)}, $$ where $y = f_{L}(x)$.

So, the key here is that a smooth monotone function is locally equivalent to a linear function with nonzero slope. The rigorous statement of this is one of the versions of the Implicit Function Theorem (see, for example, Arnol'd's Ordinary Differential Equations, any edition after the 1st).

For a proof in your second question, it can be done without "epsilonics".

Let's form the function $\phi(x)$ as follows: $$ \phi(x) = g \circ f (x). $$ Now, note that $$ x = \phi(x) = g \circ f (x) $$ and differentiate using the Chain Rule.

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  • $\begingroup$ I'm still a bit confused because I'm trying to understand what the purpose of the $\delta$ - $\epsilon$ proof. As I mentioned above in a comment, whenever I encounter a question asking to prove a result the first thing I ask myself is "what does it mean for this result to exist?" Then I would attempt to reverse engineer things by way of a definitions and such, but here I guess I don't fully know what it means for $g'(f(x)) = \frac{1}{f'(x)}$ $\endgroup$ – dc3rd May 2 '19 at 0:10
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    $\begingroup$ I see. The intuition I have found helpful and was trying to convey is the relation between the slopes of (i) a linear nonconstant function, and (ii) its inverse. This relation is reciprocity. I don't see why you have to use the $\epsilon-\delta$ approach: are you required to do so by the instructor? $\endgroup$ – avs May 2 '19 at 0:14
  • $\begingroup$ No. I am doing this as personal study to prepare for an upcoming course in Real Analysis. So I am trying to comprehend the thought process used, so hopefully I can apply the creative thinking myself down the line. $\endgroup$ – dc3rd May 2 '19 at 1:28
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    $\begingroup$ Well, if it helps, the $\epsilon-\delta$ construct is not central, but is only a stepping stone to operating with open neighborhoods (generally, in a topological space). In real analysis, the thought process behind the result we are talking about is diffeomorphic equivalence: a function that is smooth enough near a non-critical point can be "smoothly transformed" into a linear function in a neighborhood of that point. And, for linear functions, the reciprocity result you are seeking is obtained by direct computation. $\endgroup$ – avs May 2 '19 at 1:37
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    $\begingroup$ So this broader generalized explanation you just provided does make perfect sense to me. I still can't see how this delta-epsilon proof is arriving at that same thing...... $\endgroup$ – dc3rd May 2 '19 at 2:25
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For geometric intuition (not necessarily help with the proof) use the fact that the graph of the inverse function of $f$ is the reflection of $f$ over the line $y=x$. Then look at the relation between the slopes of the tangents at reflected points.

For example draw the graphs of $y=x^2$ and $y = \sqrt{x}$ and compare the slopes at $(2,4)$ and $(4,2)$.

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Let $A=f((a, b)) $ then by the properties of continuous functions $A$ is an open interval. Let $x\in A$ and $y\in(a, b) $ be such that $f(y) =x, y=g(x) $. Let $h\neq 0$ be such that $x+h\in A$. Then by strict monotone nature of $f$ we have a $k\neq 0$ such that $y+k\in(a, b) $ and $f(y+k) =x+h, g(x+h) =y+k$ and $kh>0$.

The key is to observe that as $h\to 0$ the variable $k$ also tends to $0$. This is more commonly stated as "inverse of a continuous function is also continuous" (you should prove this as this is non-trivial). Now the evaluation of the derivative of $g$ is pretty standard. We have $$\frac{g(x+h) - g(x)} {h} =\frac{y+k-y} {f(y+k) - f(y)} $$ and the right side tends to $1/f'(y)$ as $h, k$ tend to $0$. Thus $g'(f(y)) =1/f'(y)$.

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