= Solution
Fix $a<b<c<1$ and choose
$$
\lambda=\operatorname{arctanh}b.
$$
Under $\mathbb P_\lambda$, the increments remain independent and identically distributed, with mean $\psi'(\lambda)=b$. The <strong law of large numbers> therefore gives
$$
\mathbb P_\lambda(an\leq S_n\leq cn)\longrightarrow1.
$$
Part (b) implies
$$
\liminf_{n\to\infty}\frac1n\log\mathbb P(S_n\geq an)
\geq-\lambda c+\psi(\lambda).
$$
Let $c\downarrow b$ and then $b\downarrow a$. Since $\lambda b-\psi(\lambda)=\psi^*(b)$ and $\psi^*$ is continuous on $[0,1)$,
$$
\boxed{\liminf_{n\to\infty}\frac1n
\log\mathbb P(S_n\geq an)\geq-\psi^*(a).}
$$
The same argument includes $a=0$ by taking $b\downarrow0$.
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