Are All Continuous Functions Locally Lipschitz

The definition of locally Lipschitz

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 $\delta_0$.

Comments

I am given 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 $||x-x_0||<\delta_0\implies||f(x)-f(x_0)||\leq M||x-x_0||$.

Source: Marsden's Elementary Classical Analysis; note that the scan below is from an old edition of the textbook; in the latest edition, the last sentence has been changed to read "This is called the local Lipschitz property."

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

2) Does $M$ depend on $x_0$, just like $\delta_0$ does?

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 $||x-x_0||<\delta\implies||f(x)-f(x_0)||\leq M||x-x_0||$.

Unlike regular/global Lipschitz, local Lipschitz can be defined at a point, and implies pointwise continuity.
- 1) $M = 0$ would be allowed, then $f$ would be constant in a neighbourhood of $x_0$. 2) $M$ depends on $x_0$, otherwise you'd get a global Lipschitz constant.
- I think it is not important if you use $M \ge 0$ or $M > 0$. About locality, I believe $M$ and $\delta_0$ can both depend on $x_0$.
- @DanielFischer Thanks, I was wondering why the statement indicated the dependence on $x_0$ (via the subscript $0$) only for $\delta$. Could you post the above as the answer?
- @DanielFischer Could I also insert a "$\forall x\in A$" after the "such that" ?
- I see there's a mistake in the condition. It ought to be $(\exists M,\delta > 0)(\forall x,z\in A)(\lVert x-x_0\rVert < \delta \land \lVert z-x_0\rVert < \delta \Rightarrow \lVert f(x)-f(z)\rVert \leqslant M\cdot \lVert x-z\rVert$.
- Indeed, $x \mapsto x^2 \sin(1/x)$ (extended by continuity) is an example of a function that satisfies the property in the question (as any differentiable function must) but is not Lipschitz on any neighbourhood of $0$ (which I agree is what 'locally Lipschitz' should mean). Perhaps this property should be called 'pointwise Lipschitz' or something?
- Oops, $x \mapsto x^2 \sin(1/x)$ is locally Lipschitz; $x \mapsto x^2 \sin(1/x^2)$ is the counterexample that we need.
I'm sorry, that definitely wasn't the intention. Note that the book doesn't (at least in the posted part) say that that is the definition of a locally Lipschitz function (and rightly so, it isn't, for that you need $\lVert f(y)-f(z)\rVert \leqslant M\cdot \lVert y-z\rVert$ for all $y,z$ close enough to $x_0$). Actually, a continuously differentiable function is locally Lipschitz, but since the derivative isn't assumed continuous in the theorem, one has only the weaker property that might be dubbed "pointwise Lipschitz".
However, I can't say much about the part I can see. If in the proof (or a remark) Marsden says that that condition means $f$ is locally Lipschitz, that would be plain wrong.
I downvoted this answer because it does not address the question: is it the case or not that the definition of locally Lipschitz given by the OP is correct.
The main problem of your question is the lack of reference. For me, Lipschitz continuity is defined as en.wikipedia.org/wiki/Lipschitz_continuity#Definitions, hence following your first paragraph, the Lipschitz definition would change a little, but yet, I'm not sure if the definitions are equivalent. I will up vote your question due to your techniques, but not for your answer.

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stonerhatteniou1990.blogspot.com

Source: https://9to5science.com/the-definition-of-locally-lipschitz

Are All Continuous Functions Locally Lipschitz

The definition of locally Lipschitz

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