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Suppose $\epsilon =o(1)$ and $\epsilon \ge n^{-1/6}$. The breadth-first exploration of a binomial random graph has deterministic drift $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$ up to $3\epsilon n$ steps. Its weighted martingale has variance $O(\epsilon n)$. For $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$, the Doob L2 maximal inequality makes its maximum smaller than $h{\epsilon }^{2}n/6$ with high probability. Then $A_t>0$ throughout $[h\epsilon n,(2-h)\epsilon n]$, giving a graph component of order $(2-o(1))\epsilon n$.

For the upper expectation bound, let $K=\lceil {\epsilon }^{-3}\rceil$. The dominating binomial branching process has branching survival probability $(2+o(1))\epsilon$ by the binomial branching survival correction. Its branching process conditioned on extinction has mean $1-\epsilon +O({\epsilon }^2+\epsilon /n)$ and total-progeny expected value $O(1/\epsilon )$. Therefore $\mathbb{E}{N}_{\ge K}\le n\rho +O(n/(\epsilon K))$. Since $L_1\le K+N_{\ge K}$ and $K=o(\epsilon n)$, this yields the matching upper bound. Controlling rare large components is necessary to conclude an expected value asymptotic from a typical-size statement.

Write $\mu =1+\epsilon$, $p=\mu /n$, $q=1-p$, and let $T$ be the total progeny of a binomial branching process. The preceding binomial branching survival correction gives $\rho =(2+o(1))\epsilon$. For a branching process conditioned on extinction, the offspring probability generating function is $f((1-\rho )s)/(1-\rho )$, with meanfor sufficiently large $n$. Summing expected generation sizes therefore gives $\mathbb{E}[T;T<\infty ]=(1-\rho )/(1-f^{\prime }(1-\rho ))=O(1/\epsilon )$.

Put $K=\lceil {\epsilon }^{-3}\rceil$ and let $N_{\ge K}$ count vertices in graph components of order at least $K$. A breadth-first exploration of a binomial random graph is dominated by $T$. The Markov inequality applied to finite total progeny givesAlways $L_1\le K+N_{\ge K}$. Since $n{\epsilon }^4\ge n^{1/3}\to \infty$, we have $K=o(\epsilon n)$ and $n/(\epsilon K)=O(n{\epsilon }^2)=o(\epsilon n)$. Hence $\mathbb{E}{L}_1\le (2+o(1))\epsilon n$. This bound controls the expected value directly, including rare large graph components.

For the lower bound, use the breadth-first exploration of a binomial random graph with $U_t$ unseen vertices, $A_t$ active vertices and $t$ explored vertices, starting from $U_0=n$, $A_0=0$. Let $b_t=1_{\{A_{t-1}=0\}}$ indicate a new root. Conditional on the past, the newly discovered count $Z_t$ has a binomial distribution $\mathrm{Bin}(U_{t-1}-b_t,p)$. Put $D_t=Z_t-p(U_{t-1}-b_t)$. The unseen-count recursion givesHere $M_t$ is a martingale. Up to $T_0=\lceil 3\epsilon n\rceil$, martingale-difference orthogonality gives $\mathbb{E}{M}_{T_0}^2\le C\epsilon n$, since each conditional variance is at most $np<2$ and $q^{-2T_0}$ is bounded.

Take $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$. Then $h\to 0$, $\epsilon =o(h)$ and $h^{2}n{\epsilon }^3\to \infty$. The Doob L2 maximal inequality impliesThe Taylor expansion, uniformly for $t\le T_0$, gives $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$. Thus $F(t)\ge h{\epsilon }^{2}n/3$ throughout the integer interval $[\lceil h\epsilon n\rceil ,\lfloor (2-h)\epsilon n\rfloor ]$ for large $n$. The new-root sum is nonnegative, so on the preceding event $A_t>0$ throughout this interval. No graph component finishes there: all these explored vertices belong to one graph component, of order at least $(2-2h)\epsilon n-O(1)$, with high probability. Its expected value is therefore at least $(2-o(1))\epsilon n$. Combining both inequalities provesThe barely-supercritical largest-component expectation uses both a positive exploration window and a finite-progeny bound; a bare convergence-in-probability assertion would not by itself justify this expectation.

For the binomial branching process, the offspring probability generating function is $f(s)=(1-p+ps{)}^n$. The Galton-Watson extinction fixed point gives $1-\rho =f(1-\rho )=(1-p\rho {)}^n$.

The printed finite-binomial upper bound is false without an additional asymptotic qualification. For $n=2$ and $p=(1+\epsilon )/2$, solving the fixed-point equation exactly givesFor instance $\epsilon =1/20$ gives $\rho =80/441>1/10=2\epsilon$. More generally, at every fixed $n>1$ the Taylor expansion gives $\rho =2n\epsilon /(n-1)+O_n({\epsilon }^2)$, again contradicting that upper bound for sufficiently small positive $\epsilon$.

Here is the corrected binomial branching survival correction. The Poisson branching process of mean $1+\epsilon$ has branching survival probability $r$ satisfying $-\log (1-r)=(1+\epsilon )r$. Expanding the logarithm givesThus $2\epsilon /(1+2\epsilon )\le r\le 2\epsilon$, which implies the requested lower estimate $r\ge 2\epsilon -4{\epsilon }^2$. Since $(1-p+ps{)}^n\le e^{np(s-1)}$ on $[0,1]$, iteration of the two offspring probability generating functions gives $\rho \ge r$.

For the actual binomial branching process, expansion of its exact fixed-point equation givesIf $(1+\epsilon {)}^2<n$, all coefficients are nonnegative. Keeping the first term provesIn particular, for $n\to \infty$ and $\epsilon =o(1)$ this gives $\rho =2\epsilon +O({\epsilon }^2+\epsilon /n)$. The printed bounds also hold for the binomial branching process in an explicit large-$n$ regime: $0<\epsilon \le 1/4$ and $n\epsilon \ge 2$. Indeed these conditions give $n{p}^2\le \epsilon$ and $n{p}^3\le 1/4$. In the preceding nonnegative series, evaluation at $x=2\epsilon$ givesThe series is increasing, and its value at the actual branching survival probability $\rho$ is $\epsilon$, so $\rho \le 2\epsilon$. Together with the lower bound already proved, this recovers $2\epsilon -4{\epsilon }^2\le \rho \le 2\epsilon$ in that regime. In particular it applies eventually to part (ii). The counterexample shows why a regime condition is needed for a literal finite-$n$ statement.

For a Galton-Watson process started from one ancestor, this is the probability that every generation is nonempty. It is the complement of the branching extinction probability. The binomial branching survival correction distinguishes finite-binomial reproduction from the Poisson branching process near mean one.