[proofplan] The proof is an unpacking of the definitions. In the forward direction, A-stability gives the bound $|R(z)| \leq 1$ on the closed left half-plane, and multiplying a left-half-plane number $\lambda$ by a positive real time step $k$ keeps it in that half-plane. In the reverse direction, the stated scalar stability condition recovers the A-stability bound by taking $k = 1$ and $\lambda = z$. The equivalence with non-asymptotic linear stability is exactly the scalar criterion for the recurrence $y_{n+1} = qy_n$. [/proofplan]

[step:Use A-stability to bound every Dahlquist amplification factor]Assume the method is A-stable. By the closed-half-plane convention, $\mathbb{C}_{\leq 0} \subset U$ and $|R(z)| \leq 1$ for every $z \in \mathbb{C}_{\leq 0}$. Let $\lambda \in \mathbb{C}$ satisfy $\operatorname{Re}(\lambda) \leq 0$, and let $k > 0$. Since $k$ is a positive real number, $\operatorname{Re}(k\lambda) = k\operatorname{Re}(\lambda) \leq 0$. Thus $k\lambda \in \mathbb{C}_{\leq 0}$. Since $\mathbb{C}_{\leq 0} \subset U$, the value $R(k\lambda)$ is defined, and the A-stability bound gives $|R(k\lambda)| \leq 1$. By the definition of the stability function, applying the one-step method to $y' = \lambda y$ with time step $k$ gives $y_{n+1} = R(k\lambda)y_n$. The scalar non-asymptotic linear stability criterion says that the fixed point $0$ of the recurrence $y_{n+1} = qy_n$ is stable precisely when $|q| \leq 1$. Applying this criterion with $q := R(k\lambda)$ proves that the fixed point $0$ is non-asymptotically linearly stable.[/step]

[guided]Assume the method is A-stable. Under the convention in the statement, this means two separate facts: first, $\mathbb{C}_{\leq 0} \subset U$, so the stability function is defined on the entire closed left half-plane; second, $|R(z)| \leq 1$ for every $z \in \mathbb{C}_{\leq 0}$. Now choose an arbitrary $\lambda \in \mathbb{C}$ with $\operatorname{Re}(\lambda) \leq 0$ and an arbitrary time step $k > 0$. We must show that the amplification factor produced by the method is bounded in modulus by $1$. The only point to check is that $k\lambda$ is still in the closed left half-plane. Since $k$ is a positive real scalar, \begin{align*} \operatorname{Re}(k\lambda) = k\operatorname{Re}(\lambda). \end{align*} Because $\operatorname{Re}(\lambda) \leq 0$ and $k > 0$, this gives \begin{align*} \operatorname{Re}(k\lambda) \leq 0. \end{align*} Hence $k\lambda \in \mathbb{C}_{\leq 0}$. The inclusion $\mathbb{C}_{\leq 0} \subset U$ guarantees that $R(k\lambda)$ is defined, and the A-stability modulus bound gives \begin{align*} |R(k\lambda)| \leq 1. \end{align*} By the defining property of the stability function, applying the method to the Dahlquist test equation $y' = \lambda y$ with time step $k$ produces the scalar recurrence $y_{n+1} = R(k\lambda)y_n$. For a scalar recurrence $y_{n+1} = qy_n$, non-asymptotic linear stability of the fixed point $0$ means exactly that one step does not increase perturbations in modulus, which is the condition $|q| \leq 1$. With $q := R(k\lambda)$, the bound just proved is precisely this condition. Therefore the fixed point $0$ is linearly stable in the non-asymptotic sense.[/guided]

[step:Recover the A-stability bound from the scalar fixed-point condition] Conversely, assume that $\mathbb{C}_{\leq 0} \subset U$ and that for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}(\lambda) \leq 0$ and every $k > 0$, the recurrence $y_{n+1} = R(k\lambda)y_n$ has the fixed point $0$ linearly stable in the non-asymptotic sense, equivalently $|R(k\lambda)| \leq 1$. Let $z \in \mathbb{C}_{\leq 0}$ be arbitrary. Set $\lambda := z$ and $k := 1$. Then $\operatorname{Re}(\lambda) = \operatorname{Re}(z) \leq 0$, and the assumed scalar stability condition gives \begin{align*} |R(z)| = |R(1 \cdot z)| \leq 1. \end{align*} Since $z \in \mathbb{C}_{\leq 0}$ was arbitrary, $|R(z)| \leq 1$ for every $z \in \mathbb{C}_{\leq 0}$. Together with the assumed inclusion $\mathbb{C}_{\leq 0} \subset U$, this is exactly A-stability under the closed-half-plane convention. [/step]

[step:Conclude the equivalence] The forward implication shows that A-stability implies the stated non-[asymptotic stability](/page/Asymptotic%20Stability) of every Dahlquist fixed point with $\operatorname{Re}(\lambda) \leq 0$ and $k > 0$. The reverse implication shows that this scalar fixed-point stability condition, together with $\mathbb{C}_{\leq 0} \subset U$, implies the defining A-stability bound on the whole closed left half-plane. Therefore the two conditions are equivalent. [/step]