Primitive and derivative of little $o$

Question

For $x\in {\rm I\!R}$ in a neighborhood of $0$ and $n>1$, we define the function $f(x) = o(x^n)$ and one of its primitives $F(x) = \int_0^x f(t)dt$.

Is it true that:

1. $F(x) = o(x^{n+1})$

2. $f'(x) = o(x^{n-1})$

How to prove it? Here are my attempts:

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1. I need to prove that:

$$ \lim_{x\rightarrow 0} \frac{F(x)}{x^{n+1}} = 0 $$

I tried to develop the fraction :

$$ \frac{F(x)}{x^{n+1}} = \int_0^x \frac{f(t)}{x^{n+1}}dt $$

But I couldn't go further... A related question seems to add a condition to get this true but I coundn't really understand the comments.

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2. I would use L'Hôpital's rule :

$$ \lim_{x\rightarrow 0} \frac{f'(x)}{x^{n-1}} = \lim_{x\rightarrow 0} \frac{f(x)}{\frac{x^n}{n}} = 0 $$

Is it correct ?

Answer 1

In general the integration of little $o$ is okay, but differentiation is not.

Differentiation Consider the function $$f(x) = \begin{cases} x^3\sin{(1/x^4)}\,, & x\neq0\,;\\ 0\,, & x = 0\,; \end{cases} $$ which is $o(x^2)$ as $x\to0$ (so it satisfies your condition that $n>1$), but $f'(x) = 3x^2\sin{(1/x^4)} - 4x^{-1}\cos{(1/x^4)}\,,$ which is clearly not $o(x)$ as $x\to0\,$.

The moral is that the asymptotic bounds cannot control the wild oscillations.

Integration Suppose $f(x)=o(g(x))$ as $x\to0\,,$ and define $F(x)=\int_0^x f(t) \,\mathrm{d}t\,$, and $G(x)=\int_0^x g(t) \,\mathrm{d}t\,$. By definition of $f(x)=o(g(x))\,$, given any $\epsilon >0\,$, we have that there exists some $\delta>0$ such that $$ \big|f(x)\big| \leq \epsilon \,g(x) $$ whenever $0<x<\delta\,$. We can then estimate $$ \big|F(x)\big| \leq \int_0^x \big|f(t)\big| \, \mathrm{d}t \leq \epsilon\int_0^x g(t)\,\mathrm{d}t = \epsilon \,G(x)\,. $$ As such we have proved that $F(x) = o(G(x))$ as $x\to0\,$.

Answer 2

In differntiation you misused L'Hopital's rule because: the limit of the fraction of the derivatives first has to exist, if you can prove it's existance only then will it be equal to the limit of the fraction of the main functions.