 Solution

Put $Y=\sqrt{\tau }$ and $V=X_{\tau }^{\ast }$. Splitting according to whether $V>\delta \lambda$, the good-lambda bound for Brownian stopping time and maximum givesIntegrate against the positive measure $d\Phi (\lambda )$. The tail integral formula for expectation in Stieltjes form yieldsBy moderate growth, choose $K_{\beta }$ with $\Phi (\beta x)\le K_{\beta }\Phi (x)$. ThenFix $\beta >1$ and choose $\delta$ small enough that $K_{\beta }\eta <1/2$. Absorb the last term and apply moderate growth once more to the fixed dilation $1/\delta$. This provesTo avoid subtracting infinities, perform the argument first for $\tau \wedge n$, where Brownian maxima have all finite polynomial moments and moderate growth is at most polynomial, and then use monotone convergence. Applying the Dambis-Dubins-Schwarz theorem transfers this Brownian estimate to continuous local martingales, giving the lower-maximum side of the Burkholder-Davis-Gundy inequalities.