= Solution
Take \b[$g_n(z)=z^n/n$ and $g(z)=0$], for $n\geq1$. These functions are entire and
$$
\sup_{|z|\leq1}|g_n(z)|=1/n\longrightarrow0,
$$
so convergence is uniform on the closed unit disc. But $g_n'(z)=z^{n-1}$, giving
$$
\boxed{\sup_{|z|\leq1}|g_n'(z)-g'(z)|=1}
$$
for every $n$. In particular $g_n'(1)=1$ never approaches zero. On every strictly smaller disc the <derivative> does converge uniformly, consistent with part (iii). This is <uniform holomorphic convergence without boundary derivative convergence>.
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