bound of integrable function I want to prove the following conjecture:
if an integrable function $f(x)$ is continuous on (0,T] and unbounded at $x=0$, then there exists positive $M$ and $\alpha\in(0,1]$ such that
$$
|f(x)|\leq M|x|^{\alpha-1} \quad\mbox{for}\quad x\in[0,T].
$$
First of all, is this statement valid??
If so, how to prove it?
 A: Counterexample: $f(x) = 1/(x\ln^2 (x)), x \in (0,1/2].$
A: The following isn't really rigorous, but it can be made rigorous with more work.
Let $T(x) = \begin{cases}1-|x| & \text{if} \ |x| \le 1 \\ 0 & \text{if} \ |x| > 1\end{cases}$, and define $f(x) = \displaystyle\sum_{n = 1}^{\infty}e^nT(e^{2n}(x-\tfrac{1}{n}))$. 
Graphically, $T(x)$ is a spike centered at $x = 0$ with width $2$ and height $1$. 
Hence, $e^nT(e^{2n}(x-\tfrac{1}{n}))$ is a spike centered at $x = \tfrac{1}{n}$ with width $2e^{-2n}$ and height $e^n$. 
It is easy to see that the functions $\{e^nT(e^{2n}(x-\tfrac{1}{n}))\}_{n = 1}^{\infty}$ have pairwise disjoint supports (i.e. the spikes don't overlap). Then since $e^nT(e^{2n}(x-\tfrac{1}{n}))$ is continuous for each $n$, $f(x)$ is continuous. 
Also, $\displaystyle\int_{0}^{2}f(x)\,dx = \sum_{n = 1}^{\infty}\dfrac{1}{2} \cdot e^n \cdot 2e^{-2n} = \dfrac{1}{e-1} < \infty$, i.e. $f(x)$ is integrable. 
However, $f(\tfrac{1}{n}) = e^n$ for all positive integers $n$. 
You can then show that for any $M$ and any $\alpha \in (0,1]$, there exists a large enough integer $n$ such that $f(\tfrac{1}{n}) = e^{n} > M\left(\tfrac{1}{n}\right)^{\alpha-1}$.
