The definition of locally Lipschitz

Question

Marsden's Elementary Classical Analysis seems to indicate this definition:

• A function $f:A{\subset}\mathbb R^n\to\mathbb R^m$ is locally Lipschitz if

  for each $x_0{\in}A,$ there exist constants $M{>}0$ and $\delta_0{>}0$ such that $$||x-x_0||<\delta_0\implies||f(x)-f(x_0)||\leq M||x-x_0||.$$

Here's a scan of the first edition of the text, where the only change from the second (latest) edition referred to above is that the last sentence now reads "This is called the local Lipschitz property" (emphasis mine).

Two questions:

1. Is the correct inequality $M\geq 0$ or $M>0$?

2. Does $M$ depend on $x_0$, like $\delta_0$ seems to?

---

EDIT: After pondering further, I've revised the definition to this:

A function $f:A{\subset}\mathbb R^n\to\mathbb R^m$ is locally Lipschitz at $x_0{\in}A$ if

there exist constants $\delta{>}0$ and $M{\in}\mathbb R$ such that for each $x{\in}A,$ $$||x-x_0||<\delta\implies||f(x)-f(x_0)||\leq M||x-x_0||.$$

Unlike regular/global Lipschitz, local Lipschitz can be defined at a point, and implies pointwise continuity.

Answer 1

A function $f\colon A \to B$ is locally xyz if every point $x \in A$ has a neighbourhood $U$ such that $f\lvert_U$ is xyz. (For things like local homeomorphisms, one has to also consider a neighbourhood of $f(x)$, but for locally Lipschitz, only the restriction of the domain is relevant.)

$f\colon A \to \mathbb{R}^m$, where $A\subset \mathbb{R}^n$, is (globally) Lipschitz-continuous, if there is a constant $L \geqslant 0$ with

$$\bigl(\forall y,z \in A\bigr)\bigl(\lVert f(y)-f(z)\rVert \leqslant L\cdot \lVert y-z\rVert\bigr).$$

$L$ is called a Lipschitz constant for $f$. An equivalent definition is that

$$L_f := \sup \left(\left\lbrace \frac{\lVert f(y)-f(z)\rVert}{\lVert y-z\rVert} : y,z \in A, y \neq z\right\rbrace \cup \{0\}\right) < \infty.$$

$L_f = 0$ if and only if $f$ is constant.

Thus $f \colon A \to \mathbb{R}^m$ is locally Lipschitz-continuous, if every point $x_0 \in A$ has a neighbourhood $U$ such that $f\lvert_U$ is Lipschitz-continuous. Since the open balls with centre $x_0$ form a neighbourhood basis at $x_0$, that is equivalent to

$$\bigl(\forall x_0\in A\bigr) \bigl(\exists \delta_0 > 0\bigr)\left(L_{f;x_0,\delta_0} < \infty\right),$$

where

$$L_{f;x_0,\delta_0} := \sup \left(\left\lbrace \frac{\lVert f(y)-f(z)\rVert}{\lVert y-z\rVert} : y,z \in A, \lVert y-x_0\rVert < \delta_0, \lVert z-x_0\rVert < \delta_0, y\neq z\right\rbrace \cup \{0\}\right);$$

or that for all $x_0 \in A$ there exist constants $\delta_0 > 0$ and $M_0 \geqslant 0$ ($\delta_0$ depending on $x_0$, and $M_0$ depending on $x_0$ and $\delta_0$) such that

$$\bigl(y,z\in A, \lVert y-x_0\rVert < \delta_0, \lVert z-x_0\rVert < \delta_0 \Rightarrow \lVert f(y)-f(z)\rVert \leqslant M_0\cdot \lVert y-z\rVert\bigr).$$

We can allow or disallow the choice $M_0 = 0$ in the definition, that doesn't change the definition. For most functions and points, $M_0 = 0$ would not be a possible choice, since the inequality with $M_0 = 0$ says that $f$ is constant in a neighbourhood of $x_0$, but there is no reason to disallow it a priori. With the notation above, any $M \geqslant L_{f;x_0,\delta_0}$ satisfies the condition and would be a possible choice for $x_0$ and $\delta_0$.