= Solution
Let $M_n=\max_{1\leq i\leq n}X_i$. For $0<s<1/c$, the exponential of a maximum is bounded by a sum, giving
$$
e^{sM_n}\leq\sum_{i=1}^n e^{sX_i},\qquad
\mathbb E e^{sM_n}\leq n\exp\!\left(\frac{\nu s^2}{2(1-cs)}\right).
$$
Applying <Jensen's inequality> to the <logarithm> yields
$$
\mathbb EM_n\leq\frac1s\log\mathbb E e^{sM_n}
\leq\frac{\log n}{s}+\frac{\nu s}{2(1-cs)}.
$$
For $n>1$, put $a=\sqrt{2\log n/\nu}$ and choose $s=a/(1+ca)$. This gives $1-cs=1/(1+ca)$ and therefore
$$
\frac{\log n}{s}+\frac{\nu s}{2(1-cs)}
=\frac{\log n}{a}+c\log n+\frac{\nu a}{2}
=\sqrt{2\nu\log n}+c\log n.
$$
For $n=1$, let $s\downarrow0$ in the preceding upper bound, obtaining $\mathbb EX_1\leq0$. Consequently
$$
\boxed{\mathbb E\max_{1\leq i\leq n}X_i\leq\sqrt{2\nu\log n}+c\log n\quad(n\geq1).}
$$
The proof uses only the individual <moment-generating function> bounds for <Sub-Gamma random variables in the right tail>; it requires no <independence>. Under the paper's literal one-sided definition, the negative part need not be integrable. The positive part of $M_n$ is integrable by the displayed exponential bound; if $\mathbb EM_n=-\infty$, the conclusion holds in the extended sense. Otherwise the application of <Jensen's inequality> is ordinary. Equivalently, truncate $M_n$ below and pass to the limit.
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