[proofplan]
We compute the stability function by applying Backward Euler to the scalar test equation and solving the resulting one-step recurrence. Once the rational stability function is identified, A-stability reduces to the elementary geometric inequality $|1-z|\geq 1$ on the closed left half-plane. The L-stability decay condition then follows because the denominator tends to infinity along every sequence in the closed left half-plane with $|z|\to\infty$.
[/proofplan]
[step:Derive the Backward Euler stability function from the test equation]
Let $\lambda\in\mathbb{C}$, let $k>0$ be the time step, and define $z:=k\lambda\in\mathbb{C}$. Let $(y_n)_{n=0}^{\infty}\subset\mathbb{C}$ denote the sequence of numerical approximations generated by the method, where $y_n$ approximates the exact solution at time $nk$. For the scalar test equation $y'(t)=\lambda y(t)$, the Backward Euler update is
\begin{align*}
y_{n+1}=y_n+k\lambda y_{n+1}.
\end{align*}
Rearranging gives
\begin{align*}
(1-k\lambda)y_{n+1}=y_n.
\end{align*}
Equivalently, whenever $1-z\neq 0$,
\begin{align*}
y_{n+1}=\frac{1}{1-z}y_n.
\end{align*}
Therefore the stability function of Backward Euler is the map
\begin{align*}
R:\mathbb{C}\setminus\{1\}\to\mathbb{C},\quad R(z)=\frac{1}{1-z}.
\end{align*}
[guided]
We start from the definition of the stability function: apply the one-step method to the scalar test equation and record the multiplier taking $y_n$ to $y_{n+1}$. Fix $\lambda\in\mathbb{C}$ and a time step $k>0$, and define the dimensionless stability variable $z:=k\lambda$. Let $(y_n)_{n=0}^{\infty}\subset\mathbb{C}$ be the numerical sequence produced by the method, with $y_n$ approximating the exact solution at time $nk$. The Backward Euler method evaluates the right-hand side at the new time level, so for $y'(t)=\lambda y(t)$ its update equation is
\begin{align*}
y_{n+1}=y_n+k\lambda y_{n+1}.
\end{align*}
The unknown is $y_{n+1}$, so we collect the terms involving $y_{n+1}$ on the left:
\begin{align*}
(1-k\lambda)y_{n+1}=y_n.
\end{align*}
Since $z=k\lambda$, this is
\begin{align*}
(1-z)y_{n+1}=y_n.
\end{align*}
If $1-z\neq 0$, division by $1-z$ is valid and gives
\begin{align*}
y_{n+1}=\frac{1}{1-z}y_n.
\end{align*}
Thus the scalar amplification factor is $1/(1-z)$. Hence the stability function is
\begin{align*}
R:\mathbb{C}\setminus\{1\}\to\mathbb{C},\quad R(z)=\frac{1}{1-z}.
\end{align*}
The exclusion of $z=1$ is exactly the point where the implicit linear equation has coefficient $0$ and the rational formula has a pole. This point lies in the open right half-plane, so it will not affect A-stability on the closed left half-plane.
[/guided]
[/step]
[step:Prove the closed left half-plane lies in the stability region]
Let $z\in\mathbb{C}$ satisfy $\operatorname{Re}(z)\leq 0$. Write $z=x+iy$ with $x,y\in\mathbb{R}$. Then
\begin{align*}
|1-z|^2=|1-x-iy|^2=(1-x)^2+y^2.
\end{align*}
Since $x\leq 0$, we have $1-x\geq 1$, and therefore
\begin{align*}
|1-z|^2=(1-x)^2+y^2\geq 1.
\end{align*}
Thus $1-z\neq 0$ and
\begin{align*}
|R(z)|=\frac{1}{|1-z|}\leq 1.
\end{align*}
This proves A-stability.
[/step]
[step:Verify the stiff-decay limit in the left half-plane]
Let $(z_n)_{n=1}^{\infty}$ be any sequence in $\mathbb{C}$ such that $\operatorname{Re}(z_n)\leq 0$ for every $n\in\mathbb{N}$ and $|z_n|\to\infty$. For each $n\in\mathbb{N}$, the previous step gives $1-z_n\neq 0$. Also, by the [reverse triangle inequality](/theorems/2300),
\begin{align*}
|1-z_n|\geq ||z_n|-1|.
\end{align*}
Since $|z_n|\to\infty$, the right-hand side tends to $\infty$, so $|1-z_n|\to\infty$. Hence
\begin{align*}
|R(z_n)|=\frac{1}{|1-z_n|}\to 0.
\end{align*}
Because the sequence $(z_n)_{n=1}^{\infty}$ was arbitrary, this proves
\begin{align*}
\lim_{\substack{|z|\to\infty, \operatorname{Re}(z)\leq 0}} R(z)=0.
\end{align*}
[/step]
[step:Conclude that Backward Euler is L-stable]
The Backward Euler stability function is
\begin{align*}
R:\mathbb{C}\setminus\{1\}\to\mathbb{C},\quad R(z)=\frac{1}{1-z}.
\end{align*}
The closed left half-plane is contained in its domain, and $|R(z)|\leq 1$ for every $z$ with $\operatorname{Re}(z)\leq 0$, so the method is A-stable. The same stability function also satisfies $R(z)\to 0$ as $|z|\to\infty$ within the closed left half-plane. These are precisely the two required conditions for L-stability, so Backward Euler is L-stable.
[/step]
