= Solution
Let $G=E\mathbin\square F$. To prove <convexity>, take $u_1,u_2$ with finite $G$ values, $0<t<1$, and $\varepsilon>0$. By the definition of <infimum>, choose $v_i$ with
$$
E(v_i)+F(u_i-v_i)\leq G(u_i)+\varepsilon.
$$
Put $u_t=tu_1+(1-t)u_2$ and $v_t=tv_1+(1-t)v_2$. <Convexity> of $E$ and $F$ gives
$$
\begin{aligned}
G(u_t)&\leq E(v_t)+F(u_t-v_t)\\
&\leq t[E(v_1)+F(u_1-v_1)]
+(1-t)[E(v_2)+F(u_2-v_2)]\\
&\leq tG(u_1)+(1-t)G(u_2)+\varepsilon.
\end{aligned}
$$
Let $\varepsilon\downarrow0$. If either endpoint value is infinite, the convexity inequality is automatic, and $t=0,1$ gives equality. Thus \b[the <infimal convolution> of two <convex functions> is <convex>] under the stated properness assumption. The proof uses approximate minimizing splits, so no attainment of the inner <infimum> is assumed.
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