Necessity of consistency. If the method is convergent, it must reproduce constant solutions exactly. For $y' = 0$, $y \equiv C$: $\sum \alpha_j C = 0$, so $\rho(1) = \sum \alpha_j = 0$. For $y' = 1$, $y(t) = t + C$: $\sum \alpha_j (t_{n+k-j}) = h \sum \beta_j$, giving $\rho'(1) = \sigma(1)$ after algebra.

Necessity of zero-stability. Consider $y' = 0$, $y(0) = 0$. The multistep recurrence becomes $\sum \alpha_j y_{n+k-j} = 0$ (homogeneous). The general solution is a linear combination of $\zeta_i^n$ (for simple roots) and $n^m \zeta_i^n$ (for roots of multiplicity $m+1$). If a root $|\zeta_i| > 1$, the solution grows exponentially even with zero initial data perturbed by rounding. If $|\zeta_i| = 1$ with multiplicity $> 1$, $n^m \zeta_i^n$ grows polynomially. Either case prevents convergence.

Sufficiency (outline). Let $e_n = y(t_n) - y_n$ be the global error. Subtracting the numerical scheme from the exact equation yields: $\sum \alpha_j e_{n+k-j} = h \sum \beta_j [f(t_{n+k-j}, y(t_{n+k-j})) - f(t_{n+k-j}, y_{n+k-j})] + h^{p+1}\tau_n$ where $\tau_n = O(1)$ is the local truncation error.

By the Lipschitz condition $|f(t,u) - f(t,v)| \leq L|u-v|$: $\sum \alpha_j e_{n+k-j} = h \sum \beta_j [G_{n+k-j} e_{n+k-j}] + O(h^{p+1})$ where $|G_j| \leq L$.

This is a perturbed linear recurrence. Zero-stability ensures the homogeneous recurrence $\sum \alpha_j e_{n+k-j} = 0$ has bounded solutions. The perturbation theory for difference equations (discrete Gronwall inequality) then gives: $\max_n |e_n| \leq C \cdot \left(\max_j |e_j|_{j<k} + h^p\right)$ where $C$ depends on $L$, $T = t_{\text{final}} - t_0$, and the method coefficients.

If the starting values satisfy $|e_j| = O(h^p)$ for $j = 0, \ldots, k-1$ (e.g., computed by a one-step method of order $p$), then $\max_n|e_n| = O(h^p)$. $\square$