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For $n\ge1$ and $0<x<H_n/2$, the unique optimizer in definitions is

$$p_m=\frac1{1+e^{cn/m}},\qquad F_n(c):=\sum_{m=1}^n\frac1{m(1+e^{cn/m})}=x,$$

with a unique $c=c_{x,n}>0$. At $x=0$ the optimizer is all zero; for $x\ge H_n/2$ it is all $1/2$.

Fix $x_0>0$ and $0<\delta<1$. Uniformly for

$$x_0\le x\le\frac{1-\delta}{2}\log n$$

and sufficiently large $n=n(x_0,\delta)$,

$$n^{-1+\delta/2}\le c\le C(x_0),\qquad c\asymp_{x_0,\delta}e^{-2x},\qquad cn\ge n^{\delta/2},$$

and, with the continuous multiplier and exponent,

$$\left|\log(c/\lambda_x)\right|=O_{x_0}((cn)^{-1}),\qquad \frac{\mathcal H_n(x)}n=c_x+O_{x_0}((cn)^{-1}).$$

Source: published PDF, pp. 3–4, Lemma 2. The proof uses strict concavity, correcting the printed “strongly convex.” The printed multiplier comparison is not uniform down to $x=0$: for fixed $n$, $c_{x,n}\to\infty$ there. The positive lower bound $x_0$ is essential for the estimates, and is available in Theorem 4. The growing-range Riemann estimates below expand the source's compact-range argument; no discrete slab is assumed nonempty.

Bears on. Problem 297.

Proof

$F_n$ is continuous and strictly decreasing from $H_n/2$ to 0 on $[0,\infty)$. This proves the existence and uniqueness of $c$ in the stated open range. For the corresponding $p_m$,

$$h'(p_m)=\log_2\frac{1-p_m}{p_m}=\frac{cn}{m\log2}.$$

For any feasible vector $(r_m)$, the supporting-tangent inequality for strictly concave $h$ gives

$$\sum_mh(r_m)\le\sum_mh(p_m) +\frac{cn}{\log2}\left(\sum_m r_m/m-x\right) \le\sum_mh(p_m).$$

Strict concavity forces equality only at the displayed vector, including competitors on the boundary of the cube. The two endpoint regimes follow directly from positivity of $1/m$ and the unique maximum $h(1/2)=1$.

Put $\phi_c(y)=1/[y(1+e^{c/y})]$ for $y>0$, with $\phi_c(0)=0$. As a function of $t=c/y$, it is $c^{-1}t/(1+e^t)$. The derivative of $t/(1+e^t)$ has numerator $1+e^t(1-t)$, strictly decreasing for $t>0$ from 2 to $-\infty$. Thus this function is unimodal with bounded maximum; the total variation of $\phi_c$ on $[0,1]$ is $O(1/c)$. For a bounded-variation function, comparing the value at each right endpoint of an interval of length $1/n$ with its integral gives error at most its variation on that interval divided by $n$. Summing gives

$$|F_n(c)-F(c)|\le C/(cn). \tag{1}$$

Take $c_*=n^{-1+\delta/2}$. The formula for $F$ in entropy_exponent gives $F_n(c_*)=(1/2-\delta/4)\log n+O(1)$, which exceeds the allowed upper value of $x$ for large $n$. Monotonicity implies $c\ge c_*$. Choose a fixed $C_0=C_0(x_0)$ with $F(C_0)<x_0/2$. Equation (1) makes $F_n(C_0)<x_0$ for large $n$, so $c\le C_0$. Now (1) has error $O(n^{-\delta/2})$.

On $0<c\le C_0$, the quantity $F(c)-\frac12\log(1/c)$ is bounded: use its expansion near zero and continuity away from zero. Hence $x=\frac12\log(1/c)+O_{x_0}(1)$ and $c\asymp_{x_0,\delta}e^{-2x}$.

Both $c$ and $\lambda_x$ have a common fixed upper bound depending on $x_0$. Since

$$\frac{dF(c)}{d\log c}=-\frac1{1+e^c},$$

its magnitude is bounded away from zero on their intervening range. The mean value theorem and (1) give $|\log(c/\lambda_x)|=O_{x_0}((cn)^{-1})$.

Finally $g_c(y)=h((1+e^{c/y})^{-1})$, with $g_c(0)=0$, increases in $y$ and lies in $[0,1]$. Its Riemann error is at most $1/n$. For $t=c/y$,

$$\frac{\partial g_c(y)}{\partial\log c} =-\frac{t^2e^t}{(1+e^t)^2\log2},$$

which is uniformly bounded for $t\ge0$. Comparing its integrals at $c$ and $\lambda_x$ therefore costs $O_{x_0}((cn)^{-1})$. The error $1/n$ is of the same order because $c\le C_0$. This proves the entropy approximation uniformly.