L-stability and Stiff decay In my Numerical Methods for PDEs textbook by Ari Uscher L-stability and stiff decay are introduced by considering a generalized test equation:
$y' = \lambda (y - g(t)), 0 < t < b$
where g(t) is a bounded but otherwise arbitrary function. As $Re(\lambda) \rightarrow -\infty$ the exact solution satisfies $y(t) \rightarrow g(t), 0< t < b$. My first question (1) is that this is not obvious to me and no explanation is given. How can one conclude that? 
It goes on to define stiff decar for a method as $|y_n - g(t_n)| \rightarrow 0$ as $\Delta t Re(\lambda) \rightarrow 0$. My second question (2) is how this is found in practice. If I want to determine if an ODE is stiff, how do I find/choose a g(t) to satisfy the above requirement? Or does any arbitrary bounded g satisfy it if my ODE is stiff as the test equation supposedly shows (which I don't understand and is listed as question 1).
 A: Let me answer your questions in turn.
(1) Let me provide an "engineering explanation". For $\text{Re}(\lambda) \ll 0$, a  function $y(\cdot)$ satisfying $y'(t)=-\lambda y(t)$ decays to $0$ very fast. (More precisely, noting the true solution is $Ce^{-\lambda t}$, one can show $y(t)$ converges uniformly to $0$ on any interval $[a,b]$ as $\text{Re} (\lambda )\to -\infty$.)
Now consider the equation $y' = \lambda(y-g(t))$. Consider the quantity, $e(t) = y(t) - g(t)$, which characterizes how close $y(t)$ is to $g(t)$. Then
$$
e'(t) = y' - g'(t) = \lambda(y-g(t)) - g'(t) = \lambda e(t) - g'(t).
$$
If $\text{Re}(\lambda) \ll 0$, $g'(t)$ looks tiny compared to $\lambda e(t)$. Thus,
$$
e'(t) = \lambda e(t) - g'(t) \approx \lambda e(t).
$$
Thus, as we argued above, $e(t)$ will decay to $0$ exceptionally fast, and thus $y(t)$ will decay to $g(t)$ quite fast. To make this precise, you can use any number of perturbation theorems for ODEs (the Existence and Uniqueness theorem in Ascher and Petzold's book will suffice. Hint: $g'(t)$ is the $r(t,y)$.)
(2) When talking about numerical methods for ODEs, it is important to distinguish between properties of the ODES and of the methods we use to solve the ODEs. Stiffness is a property of an ODE, and usually having corresponds large eigenvalues in the Jacobian matrix of the right-hand side function compared to the interval over which the ODE is being solved. The problem $y' = \lambda(y-g(t))$ is stiff for $\text{Re}(\lambda) \ll 0$, because the eigenvalue (here just $\lambda$) becomes very large. Stiff decay is a property of a method to solve ODEs, which a method either does or does not possess. Stiff decay refers to whether the method correctly decays $y' = \lambda y$ (taking $g \equiv 0$) to $0$ in one step with large $\Delta t\cdot \text{Re}(\lambda)$. The backwards Euler method has stiff-decay, whereas the trapezoid method does not.
