Why are these problems well posed? I've the following definition:

The problem $(f, x)$ is well-posed in $E_\delta(x)$ if a constant $L_{abs}\geq 0$ exists such that $$|f(\widetilde{x}) - f(x)| \leq L_{abs}|\widetilde{x} - x|$$ If such a constant $L_{abs}$ does not exist, the problem is called ill-posed.

Here's the measurement is done in absolute error, but there's also a version in terms of relative error.
Then I've the following examples:


*

*The constant mapping $f : \mathbb{R} \mapsto \mathbb{R}$, $f (x) := 0$ is clearly well posed. It is even "destroying" information


*The linear mapping $f : \mathbb{R} \mapsto \mathbb{R}$, $f(x) := ax$, $0 \neq a \in \mathbb{R}$ is also well-posed $\forall x \in \mathbb{R}$. Obviously, we have $L_{abs} =a$.


*The problem $f : \mathbb{R} \mapsto \mathbb{R}$, $f(x):= |x|$ is ill-posed at $x=0$.

I've a few questions regarding these examples.
For all 3 cases, I don't understand why one can claim that the problem is well posed based on the definition.
Regarding the second example, why $L_{abs} = a$?
 A: For the first case:
$\forall x, y \; |f(x) - f(y) | = |0 -  0| = 0$ so taking $L_{abs}  > 0$ it is always true that $|f(x) - f(y)| \leq L_{abs}  |x -  y|$. 
For the linear mapping $x \mapsto ax$:
$\forall x, y \; |f(x) - f(y)| = |ax -  ay| = |a(x-y)| = |a||x-y|$
So $L_{abs}  = |a|$. Note that the definition is with $L_{abs}  \geq 0$ while the example states $a \in \mathbb{R}^*$ so I think there's an issue and that it should be $|a|$ and not $a$. 
For the third case:
Let $x = 0$ and $\tilde{x}  = x + \epsilon = \epsilon$ with $|\epsilon| < \delta$. 
If $\epsilon > 0$ then $|\epsilon| = \epsilon$ and $\epsilon \leq L_{abs}  \epsilon$ for any $L_{abs} \geq 1$. 
If $\epsilon < 0$ then $|\epsilon| = - \epsilon$ and the definition gives $-\epsilon \leq L_{abs} (-\epsilon) $ which is true for $L_{abs} \leq 1$.
So this leaves us with $L_{abs} = 1$ for $x = 0$ and $\tilde{x}$ in the open ball of center $x$ and radius $\delta$, so not sure why the example says the problem is ill-posed  at $0$.
Note however that the absolute value function is not differentiable at $0$.
A: For $2)$, setting $f(x):=ax$ gives : $|f(x)-f(x')| = |a||x-x'|$ and then you can choose $L_{abs} = |a|$. So by your definition, the problem is well-posed.
For $1)$, it's even better, every constant $\lambda >0$ can be chosen for $L_{abs}$ since : $|f(x)-f(x')| = 0 \leq \lambda |x-x'|$.
For $3)$, with $x=0$, we have $|f(x)-f(x')| = |x'| = |x'-0|$ so you can choose $L_{abs} = 1$.
