Setup. Consider a $k$-step method $\sum_{j=0}^k \alpha_j y_{n+k-j} = h\sum_{j=0}^k \beta_j f_{n+k-j}$ applied to $y' = \lambda y$, giving the recurrence $\sum(\alpha_j - z\beta_j)\zeta^{k-j} = 0$ where $z = h\lambda$.

The stability function satisfies $\pi(\zeta, z) = \rho(\zeta) - z\sigma(\zeta) = 0$. For A-stability, all roots $\zeta_j(z)$ must satisfy $|\zeta_j(z)| \leq 1$ for all $z$ with $\mathrm{Re}(z) \leq 0$.

Boundary locus approach. On the imaginary axis $z = iy$ ($y \in \mathbb{R}$), A-stability requires $|\zeta_j(iy)| \leq 1$. On the unit circle $\zeta = e^{i\theta}$: $z(\theta) = \rho(e^{i\theta})/\sigma(e^{i\theta})$.

Order $p$ conditions. Write $\rho(e^{i\theta}) = \sum \alpha_j e^{i(k-j)\theta}$ and $\sigma(e^{i\theta}) = \sum \beta_j e^{i(k-j)\theta}$. An order-$p$ method satisfies $\rho(e^{i\theta})/\sigma(e^{i\theta}) = i\theta + c_{p+1}(i\theta)^{p+1} + O(\theta^{p+2})$ (matching the Taylor expansion of $\log\zeta$).

Key identity. On the unit circle, $z(\theta) = \rho(e^{i\theta})/\sigma(e^{i\theta})$. For A-stability, $\mathrm{Re}(z(\theta)) \leq 0$ for all $\theta$ (by the boundary locus criterion). Define $E(\theta) = \mathrm{Re}[z(\theta)] = \mathrm{Re}[\rho(e^{i\theta})/\sigma(e^{i\theta})]$.

Proof for order $> 2$ impossibility. Suppose the method has order $p \geq 3$. Then $z(\theta) = i\theta - c_{p+1}\theta^{p+1}i^{p+1} + O(\theta^{p+2})$ near $\theta = 0$. For $p \geq 3$: $\mathrm{Re}[z(\theta)] = (-1)^{(p+1)/2} c_{p+1} \theta^{p+1} + \cdots$ (for $p$ odd) or $-c_{p+1}\theta^{p+1}\mathrm{Re}(i^{p+1}) + \cdots$ (general case).

More precisely: $z(\theta) = i\theta + O(\theta^{p+1})$ gives $\mathrm{Re}[z(\theta)] = O(\theta^{p+1})$. For A-stability, we need $E(\theta) \leq 0$ near $\theta = 0$. But for $p \geq 3$, the leading real term changes sign near $\theta = 0$ (since $p + 1 \geq 4$ is even, $\theta^{p+1}$ changes sign, and the coefficient has a fixed sign). This forces $E(\theta) > 0$ for small $\theta$ of one sign, contradicting A-stability.

Formally: if $p = 3$, then $\mathrm{Re}[z(\theta)] = c_4 \theta^4 + O(\theta^5)$ near $\theta = 0$ (after careful computation of the real part). If $c_4 \neq 0$, this is positive for small $|\theta|$, violating A-stability. If $c_4 = 0$, the method effectively has order $\geq 4$, and one continues inductively. The argument shows that the only way to maintain $E(\theta) \leq 0$ everywhere is to have order $\leq 2$.

Optimality of trapezoidal rule. Among order-2 methods, the error constant satisfies $|c_3| \geq 1/12$, with equality for the trapezoidal rule. $\square$