The negative reciprocal-power function is convex

Put $\alpha =1\mathrm{/}s$, so $0<\alpha \le 1$. If $\alpha =1$, then ${\phi }_s(t)=-t$ is affine and hence convex. If $0<\alpha <1$, on $(0,\mathrm{\infty })$ the derivative formula [F1] gives ${\phi }_s^{\mathrm{\prime }\mathrm{\prime }}(t)=\alpha (1-\alpha )t^{\alpha -2}\ge 0$. Thus [F2] makes ${\phi }_s$ convex on $(0,\mathrm{\infty })$.

In the second case $t^{\alpha }\to 0$ as $t\downarrow 0$. Indeed, for $\epsilon >0$ put $\delta ={\epsilon }^{1\mathrm{/}\alpha }$. The derivative $\alpha t^{\alpha -1}$ is positive on $(0,\mathrm{\infty })$, so [F5] makes $t\mapsto t^{\alpha }$ increasing there; if $0<t<\delta$, then $t^{\alpha }<{\delta }^{\alpha }=\epsilon$ by [F4]. Thus ${\phi }_s$ is continuous at zero with the stated value. For $a,b\ge 0$ and $0\le \lambda \le 1$, apply the convexity inequality on $(0,\mathrm{\infty })$ to $a+\epsilon$ and $b+\epsilon$, then let $\epsilon \downarrow 0$; continuity at zero and on positive inputs gives the inequality on $[0,\mathrm{\infty })$. In the affine case the same inequality holds directly.