Finite Gaussian water-level allocation

Define $S(\theta )={\sum }_i\min ({\lambda }_i,\theta )$ for $0\le \theta \le {\max }_i{\lambda }_i$. It is continuous, $S(0)=0$, and $S({\max }_i{\lambda }_i)={\sum }_i{\lambda }_i$. If $0\le {\theta }_1<{\theta }_2<{\max }_i{\lambda }_i$, some ${\lambda }_j>{\theta }_2$, so the $j$-th summand increases strictly and the others do not decrease. Thus for $0<D<{\sum }_i{\lambda }_i$, the intermediate-value theorem gives a unique $\theta$ with $S(\theta )=D$.

For $t>0$, $\ln t\le t-1$: if $t\ge 1$, integrate $1\mathrm{/}u\le 1$ on $[1,t]$; if $0<t<1$, use $\ln t=-{\int }_t^{1}du\mathrm{/}u$ and $1\mathrm{/}u\ge 1$ on $[t,1]$. Hence $-\ln t+t-1\ge 0$.

Fix $0<D<{\sum }_i{\lambda }_i$, its $\theta$, and $c=(2\theta \ln 2{)}^{-1}$. If ${\lambda }_i>\theta$, then for $0<{\delta }_i\le {\lambda }_i$, ${\varphi }_i({\delta }_i)-{\varphi }_i(\theta )+c({\delta }_i-\theta )=\frac{-\ln ({\delta }_i\mathrm{/}\theta )+{\delta }_i\mathrm{/}\theta -1}{2\ln 2}\ge 0$ by step 1.2; the inequality also holds at ${\delta }_i=0$ because ${\varphi }_i(0)=+\mathrm{\infty }$. If $0<{\lambda }_i\le \theta$, then ${\varphi }_i({\lambda }_i)=0$, and for $0<{\delta }_i\le {\lambda }_i$, step 1.2 gives ${\varphi }_i({\delta }_i)\ge \frac{1-{\delta }_i\mathrm{/}{\lambda }_i}{2\ln 2}\ge \frac{{\lambda }_i-{\delta }_i}{2\theta \ln 2}=-c({\delta }_i-{\lambda }_i)\mathrm{.}$ It holds at ${\delta }_i=0$ by the infinite-value convention. A zero eigenvalue has only ${\delta }_i=0$ and contributes zero. Thus for every feasible coordinate, ${\varphi }_i({\delta }_i)\ge {\varphi }_i({\delta }_i^{\ast })-c({\delta }_i-{\delta }_i^{\ast })$, with ${\delta }_i^{\ast }=\min ({\lambda }_i,\theta )$.

Summing the supporting inequalities in step 2.1 and using ${\sum }_i{\delta }_i\le D={\sum }_i{\delta }_i^{\ast }$ proves ${\sum }_i{\varphi }_i({\delta }_i)\ge {\sum }_i{\varphi }_i({\delta }_i^{\ast })$. The latter sum is exactly $\frac{1}{2}{\sum }_i{\log }_2^{+}({\lambda }_i\mathrm{/}\theta )$; the allocation is feasible, so it attains the minimum.

If $D=0$, nonnegativity forces every ${\delta }_i=0$, giving the two stated cases. If $D\ge {\sum }_i{\lambda }_i$, choose ${\delta }_i={\lambda }_i$ for every i; all rates are zero, and no feasible objective can be negative. These cases complete the endpoint and degenerate branches.