= Solution
For $a>0$, the <Brownian reflection principle> gives, for $t>0$,
$$
\mathbb P(H_a>t)=2\Phi(a/\sqrt t)-1\longrightarrow0,
$$
where $\Phi$ is the <standard normal distribution function>. Hence $H_a<\infty$ <almost surely>; for $a=0$, $H_0=0$.
For $u\geq0$, use the <Exponential martingale for Brownian motion> $Z_t=\exp(uB_t-u^2t/2)$. At $H_a\wedge t$, it is bounded by $e^{ua}$ because $B_s<a$ before the first hit and $B_{H_a}=a$ by continuity. The bounded <optional stopping theorem> gives $\mathbb EZ_{H_a\wedge t}=1$. Its limit is $e^{ua-u^2H_a/2}$, and the <dominated convergence theorem> yields $\mathbb E e^{ua-u^2H_a/2}=1$. Equivalently, the <Brownian first-passage Laplace transform> is
$$
\boxed{\mathbb E e^{-\lambda H_a}=e^{-a\sqrt{2\lambda}}\qquad(a,\lambda\geq0).}
$$
Taking $\lambda=u^2/2$ proves the displayed exponential identity. The cases $u=0$ and $a=0$ are included. The restriction $u\geq0$ is what gives the upper bound on the stopped exponential.
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