In my differential equations class, our Professor gave the following exercise:
(1) Suppose that the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ satisfies the Lipschitz Condition on the whole space $\mathbb{R}^2$. Give an example that shows that this does not guarantee that $f$ is continuous on $\mathbb{R}$.
Lipschitz Condition: Let $D \subseteq \mathbb{R}^{n+1}$ be some domain in which $(t_0,x_0)$ is an interior point. Then the function $f(t,x)$ defined on $D$ satisfies the Lipschitz condition on D if there exists a constant $L$ such that: $$||f(t,x)-f(t,y)|| \leq L||x-y|| \quad(\forall (t,x), (t,y) \in D)$$
I've got a couple questions about this.
Why does this definition have the same $t$ in each case? If the domain of $f$ is $\mathbb{R}^2$, how come that we only care about $x,y \in \mathbb{R}$? Surely we would want $f(t,x)$ to be close to $f(s,y) \quad \forall (x,t), (y,s) \in D$?
Am I right to generally think about the Lipschitz condition as meaning that the function $f$ has a bounded derivative? Are there any good intuitive sources as to why a bounded derivative implies a unique solution?
Finally, to answer (1), my professor defines $f(t,x):$ \begin{cases} x & \text{if } t \geq 0 \\ x + 1 & \text{if } t < 0 \\ \end{cases}
This answer seems to follow from the fact that the RHS of the Lipschitz condition doesn't involve t (So we can "create" discontinuities by defining an $f$ piecewise in $t$. I don't really get how we can do this though (Question above).
I also thought Lipschitz was a stronger form that simple continuity. I.e. Lipschitz implies continuity, especially when it is globally Lipschitz. How can we have a Lipschitz function that's not continuous? Am I misunderstanding these properties and in what context they apply? Thanks.
For completeness, my Professor's answer to (1): "But since the variable $t$ plays no real role in this definition, it's easy to have a non-continuous function that satisfies this [Lipschitz] condition. For instance, define $f(t,x)$ [As above]. It is easy to show that this function satisfies the Lipschitz Condition (With any $L$ at least 1). But the function is clearly not continuous in any point of the form $(0,x)$."
