Let $f(x)$ be a smooth function for all $x \geq 0$ and suppose that it satisfies $$f'(x) = o(x^\alpha)$$ as $x \to 0$ and for some $\alpha > 0$. Can we conclude that $$ f(x) = c + o ( x^{1+\alpha}) ? $$ I can prove this if $\alpha = 0$ almost trivially. In this case, we have $f'(x) = o(1)$ which implies that $f'(0) = 0$. By definition of derivative, we have $$ f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x} = 0 $$ But now, by definition of little-$o$ notation, $$ f(x) - f(0) = o(x) \implies f(x) = f(0) + o(x). $$ How do I extend such an argument for all $\alpha$? In fact, I will also be happy if there is a proof for $\alpha \in {\mathbb N}$.
Here is my attempt at a proof for the case $\alpha = 1$. I think if you agree that this proof is correct, then I can easily generalize this to all integer values of $\alpha$.
For $\alpha=1$, we have $f'(x) = o(x)$ so $$ \lim_{x \to 0} \frac{f'(x)}{x} = 0 $$ Since $f'(0) = 0$, we can use L'Hôpital's rule which implies $f''(0) = 0$. We now use the following definition the second derivative $$ f''(0) = \lim_{x \to 0} \frac{2[f(x) - f(0)-xf'(0)]}{x^2} = 0 $$ That this definition of $f''(0)$ is true follows from L'Hôpital's rule. Now using the definition of little-$o$ notation and using $f'(0) = 0$ again, we find $$ f(x) - f(0) = o(x^2) \implies f(x) = f(0) + o(x^2) . $$
Assuming I haven't made any serious error above, this type of argument clearly generalizes trivially for all $\alpha \in {\mathbb N}$.
The statement $f'(x)=o(x^{\alpha})$ as $x\to 0$ is equivalent to saying that for all $\epsilon>0$, there is a $\delta>0$ such that $|f'(x)|<\epsilon x^{\alpha}$ for all $x < \delta$. This means that for all $x\in[0, \delta)$
$$|f(x)-f(0)| = \left|\int_0^x{f'(t)~dt}\right| \le \int_0^x{\epsilon t^{\alpha}~dt} \le \epsilon x^{1+\alpha}$$
But since $\epsilon$ can be arbitrarily small, this shows that $f(x)=f(0)+o(x^{1+\alpha})$ as $x\to0$.