[proofplan] The backward Euler equation is a fixed-point equation $y_{n+1} = G(y_{n+1})$ where $G(x) = y_n + hf(t_{n+1}, x)$. The Lipschitz condition makes $G$ a contraction when $h\lambda < 1$, and the Banach fixed-point theorem gives convergence with geometric rate $h\lambda$. [/proofplan] [step:Verify $G$ is a contraction and apply the Banach fixed-point theorem] The backward Euler equation $y_{n+1} = y_n + hf(t_{n+1}, y_{n+1})$ is a fixed-point equation $y_{n+1} = G(y_{n+1})$ where $G(x) = y_n + hf(t_{n+1}, x)$. By the Lipschitz condition on $f$: \begin{align*} \|G(x) - G(\hat{x})\| &= h\|f(t_{n+1}, x) - f(t_{n+1}, \hat{x})\| \leq h\lambda\,\|x - \hat{x}\|. \end{align*} When $h\lambda < 1$, $G$ is a contraction. By the Banach fixed-point theorem, $G$ has a unique fixed point and the iterates $y_{n+1}^{(k+1)} = G(y_{n+1}^{(k)})$ converge to it at rate \begin{align*} \|y_{n+1}^{(k)} - y_{n+1}\| &\leq (h\lambda)^k\,\|y_{n+1}^{(0)} - y_{n+1}\|. \end{align*} [/step]