This question is a follow-up of Calculus with little $o$ notation where I asked if it was true that $$ f'(x) = o (x^\alpha) \qquad \implies \qquad f(x) = f(0) + o(x^{\alpha+1}) \quad \forall \quad \alpha > 0 ~ ?? \tag{1} $$ where $f : [0,\infty) \to {\mathbb R}$ is a smooth function and we are taking the limit $x \to 0$. I got two great answers showing that this was indeed true. I have two follow-up questions:
I believe those two proofs can be extended trivially to the region $\alpha > -1$. Is this correct?
Can we generalize (1) to $\alpha < -1$ in some way? In particular, is the following true: $$ f'(x) = o (x^\alpha) \qquad \implies \qquad f(x) = o(x^{\alpha+1}) \quad \forall \quad \alpha \leq -1 ~ ?? \tag{2} $$ where now $f:(0,\infty) \to {\mathbb R}$ is a smooth function.
EDIT: Apparently, if I don't add context for this question, the post will be closed. I'm certain the context won't help at all, but here goes.
I'm trying to solve the BPS equations in a $U(1)^3$ truncation of $N=8$ gauged supergravity in five dimensions and trying to understand the near horizon geometry of supersymmetric black holes with charged scalar hair (so-called hairy black holes). After a significant amount of simplification, I managed to boil my problem down to a set of 5 coupled ordinary differential equations for 5 functions. I am able to make progress towards my goal if equation (2) of the post holds (or some version thereof) and that's why I'm asking if such a theorem could possibly be true.
For example, one of the 5 differential equations is of the form $$ (2 x U )' = H \sqrt{ 4 + \Phi^2 } . $$ I know that $H=r_+^2 + o(1)$ and $\Phi = C x^{-\alpha} + o(x^{-\alpha})$ for some $\alpha,r_+,C > 0$(as $x \to 0$ which is where the horizon of the black hole is located). Given this information, what can I deduce about the near-horizon behaviour of $U(x)$ (i.e. its behaviour as $x \to 0$)?
Yes, Lutz Lehmann's argument still goes through for $\alpha>-1$: use the extended mean value theorem with $f \colon [0,\infty) \to \mathbb{R}$ as in your question and $g \colon [0,\infty) \to \mathbb{R}$ given by $g(x)=x^{\alpha+1}$, noting that for each $x>0$, $f$ and $g$ are both continuous on $[0,x]$ and differentiable on $(0,x)$.
For $\alpha<-1$, a variant of Luca Armstrong's argument works: Suppose for a contradiction that $f'(x)$ is $o(x^\alpha)$ but $f(x)$ is not $o(x^{\alpha+1})$ as $x \to 0$. So we can find $C>0$ and a strictly decreasing sequence of values $x_n \to 0$ such that $|f(x_n)| \geq Cx_n^{\alpha+1}$ for all $n$; and we can then find $\delta>0$ such that $|f'(x)| \leq -\frac{\alpha+1}{2}Cx^\alpha$ for all $x \in (0,\delta]$. For all sufficiently large $n$, $$ |f(x_n)| \leq |f(x_n)-f(\delta)| + |f(\delta)| \leq \int_{x_n}^\delta |f'(x)| \, dx + |f(\delta)| < \tfrac{1}{2}Cx_n^{\alpha+1} + |f(\delta)|. $$ So taking $n$ sufficiently large that $|f(\delta)| < \tfrac{1}{2}Cx_n^{\alpha+1}$, we have $f(x_n)<Cx_n^{\alpha+1}$, contradicting that $|f(x_n)| \geq Cx_n^{\alpha+1}$.
For $\alpha=-1$, your question 2 isn't really formulated correctly: just taking $f$ to be a non-zero constant would be a trivial counter-example. But even if we were to ask whether $$ f'(x) = o (x^{-1}) \qquad \implies \qquad f(x) = O(1) $$ with a big $O$ on the right side of the implication sign (meaning that it should be easier to fulfil), the answer would still be no: a counter-example is given by $f(x)=\log(-\log x)$ for $x$ close to $0$, where we have $f'(x)=\frac{1}{x\log x}$ for $x$ close to $0$.
The "corrected version" of Eq. (2) for $\alpha=-1$ is $$ f'(x) = o (x^{-1}) \qquad \implies \qquad f(x) = o(-\log x). $$ The proof is identical to the above proof for $\alpha<-1$ except replacing $$ |f(x_n)| \geq Cx_n^{\alpha+1} \quad \text{with} \quad |f(x_n)| \geq -C\log x_n $$ and $$ |f'(x)| \leq -\tfrac{\alpha+1}{2}Cx^\alpha \quad \text{with} \quad |f'(x)| \leq \tfrac{C}{2x}. $$