= Solution
{c}
Put $Y=\sqrt\tau$ and $V=X_\tau^*$. Splitting according to whether $V>\delta\lambda$, the <good-lambda bound for Brownian stopping time and maximum> gives
$$
\mathbb P(Y>\beta\lambda)\leq
\mathbb P(V>\delta\lambda)+\eta\mathbb P(Y>\lambda),
\qquad \eta=\frac{\delta^2}{\beta^2-1}.
$$
Integrate against the positive measure $d\Phi(\lambda)$. The <tail integral formula for expectation> in Stieltjes form yields
$$
\mathbb E\Phi(Y/\beta)
\leq\mathbb E\Phi(V/\delta)+\eta\mathbb E\Phi(Y).
$$
By <moderate growth>, choose $K_\beta$ with $\Phi(\beta x)\leq K_\beta\Phi(x)$. Then
$$
\mathbb E\Phi(Y)
\leq K_\beta\mathbb E\Phi(V/\delta)
+K_\beta\eta\mathbb E\Phi(Y).
$$
Fix $\beta>1$ 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 proves
$$
\boxed{\mathbb E\Phi(\sqrt\tau)\leq C_\Phi\mathbb E\Phi(X_\tau^*).}
$$
To 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>.
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