I wanna find a counter example to prove that "if $f(x)\in C^0,f(0)=0$,and for $x\neq 0(f(x)>0)$, then $\int^x_0f(t)dt=o(f(x)),$ as $x\to 0$" is not always true even if $f(x)\in C^p,p\geq1$.
First I found $f(x)=x^n|\sin(\frac1x)|+x^m\in C^0,m>n$, and by using an auxiliary function $\phi(x)$ which is a broken line connecting the point where $|sin(\frac1x)|=0,1$. So for the sequence $\{x_k=\frac1{k\pi}\},\int_0^{x_k}t^n\sin(\frac1t)dt\geq\int_0^{x_k}t^n\sin\phi(t)dt\geq\frac{nx_k^{n+1}}{2(n+1)}-\frac{(x_k)^{n+1}}{n\pi}$ , which means $\lim\limits_{k\to\infty} \frac{\int^{x_k}_0f(t)dt}{f(x)}$ doesn't exist.
Then I found that $f(x)=x^n(\sin(\frac1x)+1)+x^m\in C^{n-2},m>n$, and by using $\varphi(x)=$$\begin{cases}1,sin(\frac1x)>0\\0,sin(\frac1x)\leq0\end{cases}$ and $\int^{x_k}_0t^n\varphi(x)dt\geq\int^{\frac{x_k}2}_0t^ndt$, we can prove that for $\{x_k=\frac{1}{2k\pi-\frac{\pi}2}\}, \lim\limits_{k\to\infty}\frac{\int^{x_k}_0f(t)dt}{f(x)}$ doesn't exist.
As you can see, these two functions are pretty complicated, so I'm wondering if there are better (simpler) counter examples for the problem.
Best Answer
My counterexample: $$ f(x) = \exp\left(\frac{-1}{x}\left(2+\sin\frac{1}{x}\right) \right),\qquad x>0 $$ Then $f(x) \to 0$ as $x \to 0^+$, but $\int_0^x f(t)\;dt$ is not $o(f(x))$.
How did I find this?
First, $\int_0^x f(t)\;dt = o(f(x))$ as $x \to 0^+$ means $$ \lim_{x\to 0^+} \frac{\int_0^x f(t)\;dt}{f(x)} = 0 $$ Note $f(x) \to 0$ and $\int_0^x f(t)\;dt \to 0$, so by l'Hopital we should attempt the limit of $$ \frac{\frac{d}{dx}\int_0^x f(t)\;dt}{f'(x)} = \frac{f(x)}{f'(x)} = \frac{1}{\frac{d}{dx}\log(f(x))} \tag1$$ For our counterexample, we want $\frac{d}{dx}\log(f(x))$ does not go to $\pm \infty$. We do have $f(x) \to 0$, so $\log(f(x)) \to -\infty$. So I found a function that goes to $-\infty$, but oscillates so that its derivative does not go to $\infty$. I used $$ g(x) := \frac{-1}{x}\left(2+\sin\frac{1}{x}\right) $$ with graph like this:
Computation. Define $x_n = \frac{1}{(2 n-1) \pi}$, so that $x_n \to 0^+$, and define interval $$ J_n = \left[\frac{1}{(2n-\frac16)\pi},\frac{1}{(2n-\frac56)\pi}\right] $$ so that $\sin\frac1x \le -\frac12$ for $x \in J_n$.
Then $$ \int_0^{x_n} f(x)\;dx > \int_{J_n} f(x)\;dx > \frac{24}{(12n-1)(12n-5)\pi}\;\exp\left(-\frac32 (2n-1)\pi\right) $$ and $$ f(x_n) = \exp\big(-(2n-1)\pi\cdot (2+0)\big) $$ Divide $$ \frac{1}{f(x_n)}\int_0^{x_n} f(t)\;dt > \frac{24}{(12n-1)(12n-5)\pi}\;\exp\left(\pi n - \frac12 \pi\right) \to \infty \text{ as }n \to \infty . $$ So $$ \frac{1}{f(x)}\int_0^{x} f(t)\;dt \not\to 0\quad\text{ as } x \to 0^+ . $$