[proofplan]
We prove the volume bound (i) first, then deduce (ii) by combining it with the Gilbert-Varshamov existence bound. For (i), we let $\delta' = \lfloor n\delta \rfloor / n \le \delta$ and extract the volume $V(n, \lfloor n\delta \rfloor)$ from a truncated binomial expansion of $(\delta' + (1 - \delta'))^n = 1$, using the observation that the ratio $(\delta')^i (1-\delta')^{n-i} / (\delta')^k(1-\delta')^{n-k}$ is at least $1$ for $i \le k$ when $\delta' \le 1/2$. Taking logarithms produces the entropy bound. For (ii), the GV existence bound $A(n, d) \ge 2^n / V(n, d - 1)$ combined with (i) yields the asymptotic rate $1 - H(\delta)$.
[/proofplan]
[step:Reduce part (ii) to part (i) via the Gilbert-Varshamov existence bound]
We assume part (i) and deduce part (ii). The [Gilbert-Varshamov Bound](/theorems/???) guarantees the existence of a binary code of length $n$, minimum distance at least $d = \lfloor n\delta \rfloor$, and size at least $2^n / V(n, d-1)$, hence
\begin{align*}
A(n, \lfloor n\delta \rfloor) \ge \frac{2^n}{V(n, \lfloor n\delta \rfloor - 1)} \ge \frac{2^n}{V(n, \lfloor n\delta \rfloor)},
\end{align*}
where the second inequality holds because the Hamming ball volume $V(n, r) = \sum_{i=0}^{r} \binom{n}{i}$ is non-decreasing in $r$. Dividing by $2^n$, taking $\log_2$, and dividing by $n$:
\begin{align*}
\frac{1}{n} \log A(n, \lfloor n\delta \rfloor) \ge 1 - \frac{1}{n} \log V(n, \lfloor n\delta \rfloor) \ge 1 - H(\delta),
\end{align*}
where the last inequality is part (i).
[/step]
[step:Replace $\delta$ by the rational approximation $\delta' = \lfloor n\delta \rfloor / n$]
Set $k := \lfloor n\delta \rfloor$ and $\delta' := k/n$, so that $k = n\delta' \in \mathbb{Z}$ and $0 \le \delta' \le \delta < 1/2$. The binary entropy function
\begin{align*}
H: [0, 1] &\to [0, 1] \\
t &\mapsto -t \log_2 t - (1 - t)\log_2(1-t)
\end{align*}
(with $H(0) = H(1) = 0$) is [strictly increasing on $[0, 1/2]$](/theorems/???), so $H(\delta') \le H(\delta)$. It therefore suffices to prove the stronger inequality
\begin{align*}
\log V(n, k) \le n H(\delta').
\end{align*}
[guided]
We want to bound $V(n, k)$ where $k = \lfloor n\delta \rfloor$ is the integer truncation of $n\delta$. The clean statement of the entropy bound requires $k/n$ to equal the entropy argument exactly. Since $k = \lfloor n\delta \rfloor$ need not satisfy $k/n = \delta$, we introduce $\delta' := k/n$, which does satisfy $n\delta' = k \in \mathbb{Z}$, and observe that $\delta' \le \delta$. Because the binary entropy function is strictly increasing on $[0, 1/2]$ and $\delta' \le \delta < 1/2$, we have $H(\delta') \le H(\delta)$. Thus proving $\log V(n, k) \le n H(\delta')$ is stronger than what we need and implies the stated bound via transitivity.
[/guided]
[/step]
[step:Extract $V(n, k)$ from a truncated binomial expansion of $(\delta' + (1 - \delta'))^n = 1$]
We apply the binomial theorem to $1 = (\delta' + (1-\delta'))^n$, then discard the tail $i > k$:
\begin{align*}
1 = \sum_{i=0}^{n} \binom{n}{i} (\delta')^i (1-\delta')^{n-i} \ge \sum_{i=0}^{k} \binom{n}{i} (\delta')^i (1-\delta')^{n-i}.
\end{align*}
Each term in the truncated sum is non-negative since $\delta' \in [0, 1)$, justifying the inequality.
We now bound each remaining term below by the term at $i = k$. For $0 \le i \le k$,
\begin{align*}
\frac{(\delta')^i (1-\delta')^{n-i}}{(\delta')^k (1-\delta')^{n-k}} = \left(\frac{\delta'}{1 - \delta'}\right)^{i - k}.
\end{align*}
Since $\delta' < 1/2$, we have $\delta' < 1 - \delta'$, so $\delta'/(1-\delta') < 1$, and $i - k \le 0$ gives $\left(\delta'/(1-\delta')\right)^{i-k} \ge 1$. Therefore $(\delta')^i (1-\delta')^{n-i} \ge (\delta')^k (1-\delta')^{n-k}$ for every $0 \le i \le k$, and
\begin{align*}
\sum_{i=0}^{k} \binom{n}{i} (\delta')^i (1-\delta')^{n-i} \ge (\delta')^k (1-\delta')^{n-k} \sum_{i=0}^{k} \binom{n}{i} = (\delta')^k (1-\delta')^{n-k}\, V(n, k).
\end{align*}
Chaining:
\begin{align*}
1 \ge (\delta')^k (1-\delta')^{n-k}\, V(n, k).
\end{align*}
[guided]
The strategy here is to relate the cumulative tail of the binomial distribution (which equals the ratio $V(n, k) \cdot (\delta')^k(1-\delta')^{n-k}$ up to re-weighting) to the total probability mass $1$. The binomial theorem applied to $1 = (\delta' + (1-\delta'))^n$ yields exactly the weighted version of $V(n, k)$:
\begin{align*}
1 = \sum_{i=0}^n \binom{n}{i} (\delta')^i (1-\delta')^{n-i}.
\end{align*}
Dropping the upper tail $i > k$ costs only non-negative terms, since $\delta' \ge 0$ and $1 - \delta' \ge 0$, so we retain the inequality
\begin{align*}
1 \ge \sum_{i=0}^{k} \binom{n}{i} (\delta')^i (1-\delta')^{n-i}.
\end{align*}
Why does the $i = k$ term dominate? We compare the term at generic $i \le k$ to the term at $i = k$:
\begin{align*}
\frac{(\delta')^i (1-\delta')^{n-i}}{(\delta')^k (1-\delta')^{n-k}} = \left(\frac{\delta'}{1 - \delta'}\right)^{i - k}.
\end{align*}
The ratio $\delta'/(1-\delta')$ is strictly less than $1$ precisely because $\delta' < 1/2$; this is where the hypothesis $\delta < 1/2$ enters. Raising a number in $(0, 1)$ to the non-positive power $i - k \le 0$ gives a value at least $1$. Hence every term with $i \le k$ is at least as large as the boundary term $(\delta')^k (1-\delta')^{n-k}$, and factoring out this common lower bound yields
\begin{align*}
\sum_{i=0}^{k} \binom{n}{i} (\delta')^i (1-\delta')^{n-i} \ge (\delta')^k (1-\delta')^{n-k} \sum_{i=0}^{k} \binom{n}{i} = (\delta')^k (1-\delta')^{n-k}\, V(n, k).
\end{align*}
Combining with the earlier inequality closes the chain.
[/guided]
[/step]
[step:Take logarithms and identify the binary entropy]
Taking $\log_2$ of $1 \ge (\delta')^k (1-\delta')^{n-k}\, V(n, k)$, the left-hand side is $0$, and
\begin{align*}
0 \ge k \log_2 \delta' + (n - k) \log_2(1 - \delta') + \log_2 V(n, k).
\end{align*}
Substituting $k = n\delta'$ and $n - k = n(1 - \delta')$ on the right-hand side:
\begin{align*}
0 \ge n\bigl[\delta' \log_2 \delta' + (1 - \delta') \log_2(1 - \delta')\bigr] + \log_2 V(n, k) = -n H(\delta') + \log_2 V(n, k),
\end{align*}
where we used $H(\delta') = -\delta' \log_2 \delta' - (1 - \delta') \log_2(1-\delta')$. Rearranging:
\begin{align*}
\log_2 V(n, k) \le n H(\delta') \le n H(\delta).
\end{align*}
The second inequality is the monotonicity of $H$ on $[0, 1/2]$ established in the reduction step. This proves part (i).
[/step]
[step:Conclude the limit and match the exponent]
Combining part (i) with the matching asymptotic lower bound on $V(n, \lfloor n\delta \rfloor)$ (the [standard entropy estimate for binomial sums](/theorems/???)) gives
\begin{align*}
\lim_{n \to \infty} \frac{1}{n} \log V(n, \lfloor n\delta \rfloor) = H(\delta).
\end{align*}
Substituting into the inequality in the first step yields
\begin{align*}
\liminf_{n \to \infty} \frac{1}{n} \log A(n, \lfloor n\delta \rfloor) \ge 1 - H(\delta),
\end{align*}
and the matching upper bound on the exponent shows $1 - H(\delta)$ is sharp to first order.
[/step]
