Let $(\Omega, \mathcal{F}, \mathbb{P}, \mathcal{F}_t)$ be a given probability space with usual conditions, on which $W$ is a standard Brownian motion. For $x \ge 0$, consider $$X(t) = x + \int_0^t \sigma (X(s)) dW(s)$$ Assume $\sigma \in C^{0,1/2}_{loc}$, $\sigma(0) = 0$, $\sigma>0$ on $(0,\infty)$. By [Karatzas and Shereve 98], there exists a unique strong solution with absorbing state at zero. Denote the running maximum by $X^*(T) = \sup_{s\in [0,T]} X(s)$.
Question: For a fixed $T$, is this possible to show that $\mathbb{P} ( X^*(T) \ge \beta) = o(\beta^{-1})$ as $\beta \to \infty$?
I am trying to use time-changed Brownian motion, i.e. $X(t) = x + B([X]_t)$, where $B$ is BM, and $[X]$ is quadratic variation. There is also density function available for running maximum $B^* (T)$, i.e. $\mathbb{P}(B^*(T) \ge \beta) = 2 - 2 \Phi(\beta/\sqrt{T}) = o(\beta^{-1})$, where $\Phi(\cdot)$ is c.d.f of standard normal distribution. But, I could not succeed using those facts to prove it.