Water-filling maximizes the finite parallel Gaussian allocation objective

For $P=0$, feasibility forces every $p_j=0$, so the maximum is $0$. Suppose $P>0$, put $a={\min }_j{\lambda }_j$, and define $g(T)={\sum }_j(T-{\lambda }_j{)}^{+}$. For real $u,v$, considering whether each is nonnegative gives $\mathrm{\mid }{u}^{+}-v^{+}\mathrm{\mid }\le \mathrm{\mid }u-v\mathrm{\mid }$; hence $\mathrm{\mid }g(T)-g(S)\mathrm{\mid }\le L\mathrm{\mid }T-S\mathrm{\mid }$, proving continuity, including at the breakpoints. We have $g(a)=0$ and $g(a+P)\ge P$ because a branch with ${\lambda }_j=a$ contributes $P$. By [F2] on $[a,a+P]$, a solution exists; it lies above $a$ since $P>0$. If $a<S<T$, that same branch grows by $T-S>0$ and all other terms are nondecreasing, so $g(T)>g(S)$ and the solution is unique.

Put $p_j^{\ast }=(T-{\lambda }_j{)}^{+}$. Then ${\sum }_j{p}_j^{\ast }=P$. Extend $f_j(p)=\frac{1}{2}\ln (1+p\mathrm{/}{\lambda }_j)$ to the open interval $p>-{\lambda }_j$. Its affine inner function has positive value and derivative $1\mathrm{/}{\lambda }_j$, so [F1] and the chain rule in [F4] give $f_j^{\mathrm{\prime }}(p)=1\mathrm{/}(2({\lambda }_j+p))$; the quotient rule gives $f_j^{\mathrm{\prime }\mathrm{\prime }}(p)=-1\mathrm{/}(2({\lambda }_j+p{)}^2)<0$. Thus [F3] applied to $-f_j$ proves concavity on this open interval. For any $a,b$ in it and $0<t<1$, concavity gives $f_j(a+t(b-a))\ge (1-t)f_j(a)+t{f}_j(b)$. Subtracting $f_j(a)$, dividing by $t$, and taking $t\downarrow 0$ yields $f_j(b)-f_j(a)\le f_j^{\mathrm{\prime }}(a)(b-a)$. This proves the tangent bound also when $a=0$, which is an interior point of the extended domain. On an active branch $f_j^{\mathrm{\prime }}(p_j^{\ast })=1\mathrm{/}(2T)$; on an inactive branch $p_j^{\ast }=0$ and $f_j^{\mathrm{\prime }}(0)=1\mathrm{/}(2{\lambda }_j)\le 1\mathrm{/}(2T)$. Summing the tangent bounds gives, for every feasible $p$, ${\sum }_j[f_j(p_j)-f_j(p_j^{\ast })]\le (2T{)}^{-1}({\sum }_j{p}_j-P)\le 0$. Thus $p^{\ast }$ is globally optimal.

Substitution of $p_j^{\ast }$ yields $F(p^{\ast })=\frac{1}{2}{\sum }_j{\ln }^{+}(T\mathrm{/}{\lambda }_j)$. The feasible sets are nested as $P$ grows, so $C_L$ is nondecreasing. If $P_0,P_1\ge 0$ and $p^{(0)},p^{(1)}$ are maximizing allocations, then $\theta p^{(0)}+(1-\theta )p^{(1)}$ is feasible at $\theta P_0+(1-\theta )P_1$. Concavity of each $f_j$ gives $C_L(\theta P_0+(1-\theta )P_1)\ge \theta C_L(P_0)+(1-\theta )C_L(P_1)$, including $P_i=0$ by step 1.1.