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.
Best Answer
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 $$\lim_{n \to \infty}\frac{f'(\xi_n) - f'(x_1)}{\xi_n - x_1} = \lim_{n \to \infty}\frac{0 - 0}{\xi_n - x_1} = 0 .$$