bounded by the convex envelope. To prove this statement, it is sufficient to show that any linear function that lower bounds all points in $\mathcal{S}_{\textup{Lower}}\cup \{({0,J})\}$, also lower bounds the point $(M,R^*)$. We prove this for any such linear function, denoted by $A+BM$, by first finding a converse bound of $R^*$ using Theorem \[th:general\] with certain parameters $s$ and $\alpha$, and then proving that this converse bound is lower bounded by the linear function. We consider the following $2$ possible cases:

If $A\geq 0$, note that $(0,J)$ should be lower bounded by the linear function, so we have $A\leq J$. Thus, we can choose $s=\lceil A\rceil$, $\alpha=A-s+1$, and let $\ell$ be the minimum value in $\{1,...,s\}$ such that (\[eq:sml\]) holds. Because $\left(\frac{N-\ell+1}{s},\frac{s-1}{2}+\frac{\ell(\ell-1)}{2s}\right)\in\mathcal{S}_{\textup{Lower}}$, we have $$\begin{aligned}
            A+B\frac{N-\ell+1}{s}\leq \frac{s-1}{2}+\frac{\ell(\ell-1)}{2s}.
        \end{aligned}$$ By the definition of $\alpha$, we have $A=s-1+\alpha$. Consequently, the slope $B$ can be upper bounded as follows: $$\begin{aligned}
            B&\leq \frac{s(s-1)+\ell(\ell-1)-2As}{2(N-\ell+1)}\nonumber\\
            &=-\frac{s(s-1)-\ell(\ell-1)+2\alpha s}{2(N-\ell+1)}.
        \end{aligned}$$ Thus, for any $M\geq0$, we have $$\begin{aligned}
            A+BM\leq s-1+\alpha-\frac{s(s-1)-\ell(\ell-1)+2\alpha s}{2(N-\ell+1)}M.
        \end{aligned}$$ Note that the RHS of the above inequality is exactly the lower bound provided in Theorem \[th:general\]. Hence, $A+BM\leq R^*$.

If $A<0$, let $s=\ell=1$, we have $(N,0)\in\mathcal{S}_{\textup{Lower}}$ from (\[def:sl\]). Hence, $A+BN\leq 0$, and for any $M\in[0,N]$ we 