Strict concavity and uniqueness of a finite stationary maximizer

Fix distinct ${\beta }_0,{\beta }_1\in \mathrm{\Theta }$, put $d={\beta }_1-{\beta }_0$, and define $q(t)=\mathrm{\ell }({\beta }_0+td;y,\varphi )$ for $0\le t\le 1$. Convexity keeps this segment in $\mathrm{\Theta }$. By [F1], $q^{\mathrm{\prime }\mathrm{\prime }}(t)=d^{\mathsf{T}}{\mathrm{\nabla }}^2\mathrm{\ell }({\beta }_0+td)d=-a(\varphi {)}^{-1}{\sum }_i{b}^{\mathrm{\prime }\mathrm{\prime }}(x_i^{\mathsf{T}}({\beta }_0+td))(x_i^{\mathsf{T}}d{)}^2<0$, because $Xd\mathrm{\ne }0$ by full column rank and every displayed $b^{\mathrm{\prime }\mathrm{\prime }}$ is positive. The mean value theorem [F2] applied to $q^{\mathrm{\prime }}$ makes $q^{\mathrm{\prime }}$ strictly decreasing. For any $0<t<1$, another application of [F2] gives $(q(t)-q(0))\mathrm{/}t=q^{\mathrm{\prime }}(u)$ for some $u\in (0,t)$ and $(q(1)-q(t))\mathrm{/}(1-t)=q^{\mathrm{\prime }}(v)$ for some $v\in (t,1)$; strict decrease gives the first secant slope larger than the second, which rearranges to $q(t)>(1-t)q(0)+tq(1)$. Thus every nonconstant segment is strictly concave.

Two distinct stationary points would give $q^{\mathrm{\prime }}(0)=q^{\mathrm{\prime }}(1)=0$ on their connecting segment, contradicting strict decrease. If ${\beta }_0$ is stationary and ${\beta }_1\mathrm{\ne }{\beta }_0$, then $q^{\mathrm{\prime }}(0)=0$ and strict decrease gives $q^{\mathrm{\prime }}(t)<0$ for $0<t\le 1$; [F2] applied to $q$ yields $\mathrm{\ell }({\beta }_1;y,\varphi )-\mathrm{\ell }({\beta }_0;y,\varphi )=q^{\mathrm{\prime }}(c)<0$ for some $c\in (0,1)$. Thus any stationary point is the unique global maximizer when it exists.