Proposition 5.13.12D8BC10548372

Proof. Convolve $g$ on smaller strips with a nonnegative smooth radial approximate identity. The regularizations $g_\varepsilon$ are smooth and subharmonic and decrease pointwise to $g$ by subharmonic regularization. They are still invariant under imaginary translations, hence have the form $g_\varepsilon(z)=f_\varepsilon(\operatorname{Re}z)$. Therefore $f_\varepsilon''=\Delta g_\varepsilon\geq0$, and each $f_\varepsilon$ is convex on its interval of definition.

Passing to the decreasing limit in the convexity inequality proves convexity of $f$ in the extended sense. This limit is finite everywhere: it is bounded above locally because $g$ is subharmonic, and if it were $-\infty$ at an interior point, convexity with a point beyond any other given point would force it to be $-\infty$ throughout $(a,b)$. That would make $g\equiv-\infty$, which is excluded. Thus $f$ is a real-valued convex function.