= Solution
Write $\phi(g)=a+ib$ with $a,b\in\mathbb R$, where $g$ is real valued. The normalization and the <operator norm> bound imply, for every real $t$,
$$
|1+it\phi(g)|^2\leq\|1+itg\|_\infty^2=1+t^2\|g\|_\infty^2.
$$
The left side is $(1-tb)^2+t^2a^2$, so
$$
-2tb+t^2(a^2+b^2-\|g\|_\infty^2)\leq0.
$$
Divide by $t>0$ and let $t\downarrow0$ to obtain $b\geq0$. Divide by $t<0$, reversing the inequality, and let $t\uparrow0$ to obtain $b\leq0$. Thus \b[$\boxed{\operatorname{Im}\phi(g)=0}$] for every real-valued $g$. The first-order term in $t$ is what forces reality; this is the complex-functional step in the <unital contraction positivity criterion>.
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