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This question comes from Michael Spivak's Calculus, Chapter 11, Problem 59, for which there are already a few posted questions about this problem:

"Calculus 4th Edition" by Michael Spivak -- Chapter 11 Problem 59

Spivak — Chapter 11 Problem 59 Problems understanding the solution

Find a formula for $f''$ in terms of $f$, where $f\gt 0$ and $(f')^2=f-\frac{1}{f^2}.$

However, all of these questions seem to gloss over a minor technicality that I can't seem to wrap my head around. It occurs in the very first manipulation:

$[f'(x)]^2 = f(x) - \frac{1}{[f(x)]^2}$

$2f'(x)f''(x) = f'(x) + \frac{2f'(x)}{[f(x)]^2}$

Why is the LHS justified? $[f'(x)]^2 = 2f'(x)f''(x)$ is only true if $f'(x)$ and $f''(x)$ are defined. Of course $f'(x)$ is defined, because otherwise the original equation wouldn't make sense, and we can justify the RHS using the fact that $f(x) \geq 1$ (again because if not, the original equation wouldn't make sense). But I can't find a way to justify the LHS. And I don't know anything about diff. equations to come up with a counter-example if it's not.

Does the question assume $f''(x)$ is defined? If not, why is it necessarily? If so, is there a counter-example where $f$ satisfies the above, but $f''(x)$ is not defined?

Update: I've made some progress. First of all, $f''(x)$ is necessarily defined if $f'(x) \neq 0$, but that's kind of trivial. More interesting is when $f'(x) = 0$.

Note that $f(x) = 1$ is a trivial solution to the equation. If you have another solution $g(x)$ such that $g(x) = 1 : \forall x \in (a,b)$, then $g''(x) = 0 : \forall x \in (a, b)$, and in particular, $g''(a)$ and $g''(b)$ are undefined! So the answer to my question is no. But now comes forth a new question...

What if $g(x)$ is a solution such that $g(x) = 1$ is only true for a single $x$? This is Paul Frost's idea of an isolated point. In this case, must $g''(x)$ be defined? I think it is, and should be $2f'(x)f''(x)$ but it's tricky to prove.

  • derivative of $[f'(x)]^2 $ is $2f'(x) f''(x)$ and the relation becomes $$2 f'(x)f''(x)=\frac{2 f'(x)}{f(x)^3}+f'(x)$$ and is $f'(x)\ne 0$ it gives $$f''(x)=\frac{1}{f(x)^3}+\frac{1}{2}$$ – Raffaele Oct 19 '20 at 10:22
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    @Raffaele That's incorrect. Counter-example: $f(x) = x|x|$. Derivative of $[f'(x)]^2$ at $x = 0$ is $0$, but $2f'(x) f''(x)$ is not defined at x = 0. That is the point of my question. –  Oct 19 '20 at 11:44
  • @Raffaele Well technically actually your conclusion is correct that $f''$ exists if $f' \neq 0$, although your reasoning is flawed. But even if $f' = 0$, we can still calculate $f''$ (see the hyperlinks to the original questions above). However, they seem to assume $f''$ exists even when $f'$ = 0. My question is; is it possible for $f''$ to not exist? –  Oct 19 '20 at 12:44
  • @TheSilverDoe No. It only implies that $f'(x)^2$ is differentiable. Note for example that $f(x)=\begin{cases}1&x\ge0\-1&x<0\end{cases}$ is not differentiable, while $f(x)^2$ is. The OP is asking whether that is the case in this particular problem or not. – Mohsen Shahriari Oct 19 '20 at 19:45
  • Why would $g''=0$ on $(a,b)$ imply undefinability of $g''(a)$ and $g''(b)$? As far as I can see, it only implies that $g''$ cannot be continuous at $a$ and $b$, if defined. – Mohsen Shahriari Oct 20 '20 at 03:30
  • @Mohsen Shahriari Using Rolle's Theorem, $g$ must be strictly increasing on $[b, \infty)$ and strictly decreasing on $(-\infty, a]$. That means $\forall x$ outside of $(a, b): g''(x) = \frac{1}{2} + \frac{1}{g^3(x)}$, and $\forall x \in (a, b) : g''(x) = 0$. But at $a$ and $b$, both left and right 2nd derivatives of $g$ exist, but they do not agree (one is $\frac{3}{2}$, the other is $0$), therefore $g''(a)$ and $g''(b)$ do not exist. –  Oct 20 '20 at 13:10
  • In other words, to use your analogy, yes $g$ is not continuous at $a$ and $b$, but the left and rights limits must exist, which means $g''$ has a jump discontinuity, which isn't possible for derivatives. –  Oct 20 '20 at 13:13
  • @SenZen You're right. I wasn't noting that derivatives have the intermediate value property. – Mohsen Shahriari Oct 20 '20 at 14:35
  • For the case $g(x)=1$ for a single $x$, say $x=0$, it seems that $g$ must be an even function, and thus using L'Hôpital's rule, $g''$ must exist at $x=0$ with $g''(0)=\lim_{x\to0}g''(x)=\frac12+\frac1{\lim_{x\to0}g(x)^3}=\frac32$. – Mohsen Shahriari Oct 20 '20 at 14:44
  • @Mohsen Shahriari Yes but in order to apply L'Hopital's rule, we need the limits of both numerator and denominator to approach 0. Let's call the isolated point $x_1$, where $f(x_1) = 1$ and $f'(x_1) = 0$. The numerator will be $\lim_{x \rightarrow x_1}[f'(x) - f'(x_1)] = \lim_{x \rightarrow x_1}[f'(x)]$. Now if this limit exists, it must be 0, and we get the desired result, however I am having difficulty justifying the existence of this limit, i.e. why must $f'$ be continuous at $x_1$? My intuition tells me it must be. –  Oct 21 '20 at 01:14
  • Update: I've proven $f'$ is continuous at 1. The trick is that $|f'(x)| = \sqrt{f(x) - \frac{1}{f^2(x)}}$, and you can prove $\lim_{x \rightarrow x_1}(\sqrt{f(x) - \frac{1}{f^2(x)}}) = 0$, which gets $\lim_{x \rightarrow x_1}(f'(x)) = 0$ using the fact that $f'(x) > 0 : \forall x > x_1$ and $f'(x) < 0 : \forall x < x_1$. –  Oct 21 '20 at 13:47
  • @SenZen It's a simple corollary of L'Hôpital's rule that since $\lim_{x\to x_1}f''(x)$ exists in the case of $f$ being even around $x_1$, $f''(x_1)$ must exist and be equal to that limit. See here – Mohsen Shahriari Oct 21 '20 at 19:20
  • @Mohsen Shahriari, that's not exactly what the corollary says. You've missed a subtle condition... – SenZen Oct 22 '20 at 08:43
  • @SenZen You're right. Fortunately, in this case, that's true, and you've already managed to show it. – Mohsen Shahriari Oct 22 '20 at 16:26

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One week later, I finally solved this damned problem. I'm writing up my conclusions because I think the other solutions missed some key cases. For details of the proofs for each case, ask below.

Case 1: $x_1$ is a point such that $f(x_1) \neq 1$. Then $f''(x_1) = \frac{1}{2} + \frac{1}{f^3(x_1)}$

Case 2: $x_1$ is a point such that $f(x_1) = 1$, and there exists $x_2$ and $x_3$ such that $x_2 < x_1 < x_3$ and $f(x_1) = f(x_2) = f(x_3) = 1$. Then $f''(x_1) = 0$.

Case 3a: $x_1$ is a point such that $f(x_1) = 1$, and there exists $x_2$ such that $x_2 < x_1$ and $f(x_1) = f(x_2) = 1$. But $\forall x > x_1 : f(x) \neq 1$. Then $f''(x_1)$ is undefined!

Case 3b (The mirror image of Case 3a): $x_1$ is a point such that $f(x_1) = 1$, and there exists $x_3$ such that $x_1 < x_3$ and $f(x_1) = f(x_3) = 1$. But $\forall x < x_1 : f(x) \neq 1$. Likewise, $f''(x_1)$ is undefined.

Case 4: $x_1$ is a point such that $f(x_1) = 1$ and $\forall x \neq x_1 : f(x) \neq 1$. Then $f''(x_1) = \frac{1}{2} + \frac{1}{f^3(x_1)}$.

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    I think you still need to show that there is at least one solution of the differential equation with $f(x_1)=1$ at some point $x_1$ and $f(x_2)\ne1$ at some other point $x_2$. Otherwise, while what you've stated is true, in many cases, it will be only vacuously true. – Mohsen Shahriari Oct 22 '20 at 16:38
  • @Mohsen Shahriari You're right, but unfortunately I have no idea how to do that. Worse, never mind proofs, I don't even intuitively know if there exists a solution that isn't $f = 1$. According to Wolfram Alpha, it can be solved using something called a "hypergeometric function" - no idea what that is, but I do know it's way beyond my current knowledge. –  Oct 23 '20 at 01:07