= Solution
<Convergence of a numerical method> on $[0,T]$ means that its grid error obeys $\max_{0\leq nh\leq T}|y_n-y(nh)|\to0$ as $h\to0$, with consistent initial values.
First, the implicit step of <Backward Euler method> is well-defined for $hL<1$: the map $z\mapsto y_n+hf(t_{n+1},z)$ is a <contraction mapping> on $\mathbb R$, so the <Banach fixed-point theorem> gives a unique solution. Let $M=\max_{[0,T]}|y''|$, finite under the stated smoothness. The exact solution's step defect is
$$
d_n=y(t_{n+1})-y(t_n)-hy'(t_{n+1})=\int_{t_n}^{t_{n+1}}[y'(s)-y'(t_{n+1})]ds,
$$
so $|d_n|\leq Mh^2/2$. Subtract the exact and numerical step equations. The <Lipschitz condition> gives
$$
|e_{n+1}|\leq\frac{|e_n|+Mh^2/2}{1-hL},\qquad e_n=y_n-y(t_n).
$$
With exact initial data, sum this <geometric progression> to obtain
$$
|e_n|\leq\frac{Mh}{2L}\big[(1-hL)^{-n}-1\big].
$$
For $hL\leq1/2$ and $nh\leq T$, $-\log(1-hL)\leq2hL$, so the bracket is at most $e^{2LT}-1$. Thus \b[the global error is $O(h)$, uniformly on $[0,T]$, and <Backward Euler method> converges with order one]. A vanishing initial error adds only $(1-hL)^{-n}|e_0|$.
Order one is sharp: for $y'=y$, $y(0)=1$ and $nh=T$, $y_n=(1-h)^{-n}=e^T(1+Th/2+O(h^2))$.
For the <Dahlquist test equation> $y'=\lambda y$, the <stability function> is $R(z)=1/(1-z)$, $z=h\lambda$. Therefore its <linear stability domain> is
$$
\boxed{\{z\in\mathbb C:|1-z|\geq1\}.}
$$
The strict inequality gives decay; equality gives bounded amplification. The whole closed left half-plane lies in this region, so <Backward Euler method> is <A-stable>.
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