Let $(X_{t})_{t\ge0}$ be adapted to $(\mathcal{F}_{t})_{t\ge0}$ with continuous trajectories. Assume that $X_{0} = 0$ and $X_{t}^{4} - 3t^{2}$ is a martingale with respect to $(\mathcal{F}_{t})$. Let $$\tau = \inf\left\{t>0 : |X_{t}| = \sqrt[4]{(t^2+9)}\right\}.$$ Prove that $\mathbb{E}\tau^2 < \infty$ and compute $\mathbb{E}\tau^2$.
I know that if $\mathbb{E}\tau^2 < \infty$ then by Doob we get $\mathbb{E}\tau^2 = 4.5$. The problem is that I cannot prove $\mathbb{E}\tau^2 < \infty$. I was thinking of a function $g(t) = |X_{t}| - \sqrt[4]{(t^2+9)}$ which is continuous a.s. and $g(0) < 0$. Then if I find any $t_{0}$ such that $g(t_{0}) > 0$ it will be proved. Is there any other way of doing that?
The optional stopping theorem shows that $$M_t :=X_{t \wedge \tau}^4-3 (t \wedge \tau)^2, \qquad t \geq 0; $$ is a martingale; in particular, $\mathbb{E}(M_t)=0$ for all $t \geq 0$, i.e. $$\mathbb{E}(X_{t \wedge \tau}^4) = 3 \mathbb{E}((t \wedge \tau)^2).$$ Since $X_{t \wedge \tau}^4 \leq (t \wedge \tau)^2+9$ for all $t \geq 0$, this gives $$2 \mathbb{E}((t \wedge \tau)^2) \leq 9.$$ Applying the monotone convergence theorem, we conclude that $\mathbb{E}(\tau^2) \leq 9/2< \infty$.