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Yesterday I did a post about this problem but, I didn't get very far in the problem and I really didn't understand the solution in this post. I have read and tried the problem more and I'm making a proof on my own. I'd appreciate any suggestions, and a little bit of help in the last part.

Here's the problem: Suppose that the function $f > 0$ has the property that $$(f')^2=f-{1 \over f^2}.$$ Find a formula for $f''$ in terms of $f$. (Why is this problem in this chapter?)

Now obviously, with some calculations if $f \neq 0$ then

$$ \begin{align} (f')^2 &= f-{1 \over f^2} \\ 2f'f'' &= f'+{2 \over f^3}f' \\ 2f'' &= 1+{2 \over f^3} \\ f'' &= {1 \over 2} + {1 \over f^3} = g(x) \end{align} $$

, and with $(f')^2=f-{1 \over f^2} \geq 0, f > 0$ you can easily get $f \geq 1$ So,as I was reading the solution by Paul Frost, 'my idea' was

Suppose there exists some $x_1$ such that $f(x_1) = 1$ , so $f'(x_1) = 0$

Then ( This is like ' case 1 ' ) if there exist a neigborhood around $x_1$ in which $f \neq 1$, and hence $f' \neq 0$, then for all x in the neigborhood, we apply the Mean Value Teorem and find that $$\frac{f'(x) - f'(x_1)}{x - x_1} = f''(y(x)) = g(y(x))$$ for some $y(x)$ between $x$ and $x_1$. If $x \to x_1$, then clearly $y(x) \to x_1$. Hence $$f''(x_1) = \lim_{x \to x_1} \frac{f'(x) - f'(x_1)}{x - x_1} = \lim_{y(x) \to x_1} g(y(x)) = g(x_1) $$

So we are 'done' with the first case

The second case then would be that for every neigborhood around $x_1$ there always exist some $x$ such that $ f'(x) = 0 $. In this part I really didn't understood the solution by Paul Frost, but i think I came up with: $$f''(x_1) = \lim_{x \to x_1}\frac{f'(x) - f'(x_1)}{x - x_1} $$ and since in every neigborhood around $x_1$ there is always $x$ with $f(x) = 0$ Then $$\lim_{x \to x_1}\frac{f'(x) - f'(x_1)}{x - x_1} = 0$$ obviously so $f''(x_1) = 0$. This is missing an $\epsilon, \delta $ proof. Now since in every interval around $x_1$ there exist x such that $f(x) = 1$, then f is constant ( can I just say this?, my idea is that if in every interval around $x_1$, there is some x with $f(x) = 1$, then if f was not constant, f should be increasing in some other interval since $f \geq 1$, but this is a contradiction )

Is my proof correct? I know there are some observations to be made and some basic $\epsilon, \delta $ arguments, but I don't know if I'm not proving something that has to be proven or things like that. I'd appreciate any help, thanks for advance.

Paco Antonio
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  • This should be a comment, but for some dumb reason you need 50 reputation to comment, well guess I'll write it here then much better aye? I'm doing this problem now myself and I am having a lot of annoying difficulties. There's one that appears in the first case of your proof, when there exists a neighbourhood around $x_1$ in which $f \neq 1$. In order to apply MVT to $f'$ in the neighbourhood, we need $f'$ to be continuous at $x_1$. But it's not obvious that it is so. –  Oct 19 '20 at 14:41

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Your proof is correct. For the second case you do not need a formal $\varepsilon$-$\delta$-argument. We know that $$f''(x_1) = \lim_{x \to x_1}\frac{f'(x) - f'(x_1)}{x - x_1} .$$ Therefore for any sequence $(\xi_n)$ with $\xi_n \ne x_1$ and $\xi_n\to x_1$ we have $$\lim_{n \to \infty}\frac{f'(\xi_n) - f'(x_1)}{\xi_n - x_1} = f''(x_1) .$$ Now for each $n$ there exists $\xi_n \ne x_1$ such that $\lvert x_1 - \xi_n \lvert < 1/n$ and $f'(\xi_n) = 0$. But then clearly $\xi_n \to x_1$ and thus $$\lim_{n \to \infty}\frac{f'(\xi_n) - f'(x_1)}{\xi_n - x_1} = \lim_{n \to \infty}\frac{0 - 0}{\xi_n - x_1} = 0 .$$

Paul Frost
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  • Did you see my "answer" below? I don't understand why the 1st half is valid since in order to apply MVT, you need continuity of $f'$ at $x_1$. Can you explain? –  Oct 20 '20 at 03:10
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    The basic assumption is that $f''$ exists, i.e. that $f$ is twice differentiable. If we do not assume that, I don't have idea how to proceed. In fact, if we only know that $f'$ exists and satisfies the functional equation in your question https://math.stackexchange.com/q/3871836, I do not know whether we can prove that $f''$ exists in all points. We certainly know that $(f'(x))^2$ is differentiable, but that does not allow to conclude $f'(x)$ is. If we assume that $f'(x)$ is continuous, it is easy to see that $f''(x)$ exists in all points with $f'(x) \ne 0$. – Paul Frost Oct 20 '20 at 08:49
  • @SenZen And, by the way, I do not know any example for $f$ except the trivial one $f \equiv 1$. But at least it is easy to see that $f'(x)$ is continous at $x_0$ if $f'(x_0) = 0$. – Paul Frost Oct 20 '20 at 08:54