I had a question regarding locally and globally Lipschitz functions.
Suppose there exist some real $\delta > 0$ and $M > 0$ such that $f : \mathbb{R} \to \mathbb{R}$ satisfies $|f(x) – f(y)| \leq M|x – y|$ for all $x, y \in \mathbb{R}$ satisfying $|x – y| < \delta$. That is, suppose $f$ is locally Lipschitz, and the same Lipschitz constant works for all small enough neighborhoods of $\mathbb{R}$. Then is the function globally Lipschitz too? If not, what is a counterexample?
I know that if the function is everywhere differentiable, the above condition implies its derivative is bounded, and thus it must be globally Lipschitz, so a counterexample must not be differentiable. Also, the image of a counterexample would have to be unbounded in $\mathbb{R}$, since if for all $M' \geq M$ we have $|f(x) – f(y)| > M'|x – y|$ for some $x, y \in \mathbb{R}$, we know this is only possible if $|x – y| \geq \delta$ by the condition above, so we find $|f(x) – f(y)| > M'\delta$ for all $M' \geq M$ and some $x, y \in \mathbb{R}$, and it follows from here that the image must be unbounded since $M'\delta \to \infty$ as $M' \to \infty$. I also know that if the domain was a bounded subset of $\mathbb{R}$ then we could just recursively take bigger Lipschitz constants to connect the neighborhoods and cover the whole domain with a single Lipschitz constant, but I don't see how this would apply to an unbounded domain like all of $\mathbb{R}$.
Is there more to this condition/argument that I am missing?
Thanks.
Edit: Also, as a follow-up question, if instead $\delta$ depended on $y$ (or equivalently $x$) then would this even imply uniform continuity, or would it just be imply local continuity?
Best Answer
Commenters have already answered in the affirmative, but let me give a rigorous answer to your follow-up as well.
If $\delta$ depends on $x$, we still obtain a global Lipschitz condition. That is, as long as each $x$ has a $\delta_x$ for which $|f(x)-f(y)|\leq M|x-y|$ whenever $|x-y|<\delta_x$, then $f$ is globally $M$-Lipschitz.
To see this, let $x,y\in \mathbb R$, and define $S:=\{z\in[x,y]\mid |f(x)-f(z)|\leq M|x-z|\}$.
Let $z=\sup(S)$, with $z\in S$ since $S$ is closed by continuity of $f$.*
If $y\neq z$, then choosing $z'\in (z,y)$ with $|z'-z|<\delta_z$, we have
\begin{align*} |f(x) - f(z')| &\leq |f(x)-f(z)| + |f(z)-f(z')|\\ &\leq M|x-z|+M|z-z'|=M|x-z'|, \end{align*} so that $z'\in S$, contradicting that $z'>z=\sup(S)$.
Therefore $y=z\in S$, so the Lipschitz condition is satisfied.
Remark
Though the question was stated for $\mathbb R$, the proof immediately extends to the setting where $f\colon X\to Y$ is a map between metric spaces, and $X$ is a geodesic metric space (every two points can be joined by an isometric image of an interval). In particular, it holds whenever the domain is any convex subset of a normed vector space.
*Clarification
To see $S$ is closed, note that it is the inverse image of the closed set $(-\infty,0]$ under the continuous map $z\mapsto |f(x)-f(z)|-M|x-z|$, defined on $[x,y]$. If you haven't had point-set topology, you can also just observe that if $|f(x)-f(z_i)|\leq M|x-z_i|$ for a sequence $z_i\to z$, then from sequential definition of continuity, $$|f(x)-f(z)|=\lim_{i\to\infty} |f(x)-f(z_i)|\leq\lim_{i\to \infty} M|x-z_i|=M|x-z|.$$