 Solution

For this concave recruitment map, the extinction state is invariant because $g(0)=0$. Near extinction, $N_{t+1}=g^{\prime }(0)N_t+O(N_t^2)$ with multiplier above one, so small positive populations initially grow. The strict concavity $g^{\prime \prime }<0$ means marginal recruitment decreases as population increases. More precisely, for $N>0$,since $g(N)={\int }_0^N{g}^{\prime }(s)ds>N{g}^{\prime }(N)$. Thus per-capita recruitment decreases and the graph bends downwards from an initial slope greater than that of the identity line.

A positive fixed point, if present, is the unique positive intersection with $g(N)=N$: the strictly decreasing ratio can cross one at most once. The sketch shows the usual saturating example $g(N)=3N/(1+N)$, which crosses at $N=2$.

The three assumptions alone do not guarantee this crossing. For example $g(N)=2N+1-e^{-N}$ satisfies all three and remains strictly above $N$ for every $N>0$. An extra large-population restriction, such as $g(N)<N$ somewhere, supplies existence. Keeping this distinction avoids inventing a carrying capacity that is not implied by the printed hypotheses.