Let $U \subset \mathbb{C}$, and let $R: U \to \mathbb{C}$ be the stability function of a one-step method, meaning that applying the method with time step $k > 0$ to the Dahlquist test equation $y' = \lambda y$ produces the scalar recurrence $y_{n+1} = R(k\lambda)y_n$ whenever $k\lambda \in U$. Define the closed left half-plane by

\begin{align*} \mathbb{C}_{\leq 0} := \{z \in \mathbb{C} : \operatorname{Re}(z) \leq 0\}. \end{align*}

Use the convention that the method is A-stable precisely when $\mathbb{C}_{\leq 0} \subset U$ and $|R(z)| \leq 1$ for every $z \in \mathbb{C}_{\leq 0}$.

Then the method is A-stable if and only if $\mathbb{C}_{\leq 0} \subset U$ and, for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}(\lambda) \leq 0$ and every $k > 0$, the scalar recurrence obtained from the Dahlquist test equation,

\begin{align*} y_{n+1} = R(k\lambda)y_n, \end{align*}

has the fixed point $0$ linearly stable in the non-asymptotic sense; equivalently, $|R(k\lambda)| \leq 1$.