What is the optimal growth of the constant in BDG? Let $X$ be a continuous local martingale, and $\langle X \rangle$ be its quadratic variation process. The "standard" proof of Burkholder-Davis-Gundy inequalities found in books yields $(\mathsf{E} |X|^{p})^{1/p} \le O(p) \cdot (\mathsf{E} \langle X \rangle ^{p/2})^{1/p}$ for large $p$.
Can the growth rate be improved to, say, $O(p^{1/2})$? For example, if $\langle X \rangle$ is bounded, this estimate gives exponential tails for $|X|$, which is clearly suboptimal, since they should be Gaussian.
 A: I know a version which exactly gives the constant $O(p^{1/2})$ for $p\ge 2$. It is contained in a lecture note by D. Khoshnevisan on SPDE.
A: You are correct that for bounded $<X>_T$ the tails of $X_T$ should be Subgaussian. However, the Burkholder-Davis-Gundy inequality gives an upper bound for the $L^p$-norm of the running supremum $X_T^* = \sup_{t \le T} |X_T|$, of $X$ not just for $X_T$ itself.
I do not see a reason why $X_T^*$ should have Subgaussian tails, even if $<X>_T$ is bounded. In fact it cannot always have Subgaussian tails, otherwise the known optimal constant $p-1$ for $p \ge 2$ (see George Lowthers remark) would not be optimal.
A: This comment was too long. 
For martingales of the form $X_{t} = \int_{0}^{t}f(s,\omega)dB_{s}$ where $B_{t}$ is the standard Brownian motion, $f(s, \omega)$ is non-anticipating function such that  $\int_{0}^{\infty}f^{2}(s, \omega)ds <\infty$ a.s.  (I guess they are called $L^{2}$ integrable continuous  martingales)  one can obtain the bound of order $\sqrt{p}$ for large $p$. This is due to Burgess Davis, (see Section 3) of the reference.  The result essentially follows from the Brownian case with arbitrary  $L^p$ integrable stopping times. 
However, for discrete time martingales, this is not true, and the best possible bound is of order $p$ for large $p$, for example, see paper of G. Wang, Remark 2. On the other hand, there are very special ``conditionally symmetric'' discrete martingales for which one can obtain the bound of order $\sqrt{p}$. And it looks like continuous $L^{2}$ integrable martingales in this sense are similar to conditionally symmetric ones. 
Wang, Gang, Sharp inequalities for the conditional square function of a martingale, Ann. Probab. 19, No.4, 1679-1688 (1991). ZBL0744.60046.
Davis, Burgess, On the $L^p$ norms of stochastic integrals and other martingales, Duke Math. J. 43, 697-704 (1976). ZBL0349.60061.     
