11
$\begingroup$

Let $f:\mathbb R^d \to \mathbb R^m$ be a map of class $C^1$. That is, $f$ is continuous and its derivative exists and is also continuous. Why is $f$ locally Lipschitz?

Remark

Such $f$ will not be globally Lipschitz in general, as the one-dimensional example $f(x)=x^2$ shows: for this example, $|f(x+1)-f(x)| = |2x+1|$ is unbounded.

  • 3
    Well, $x \mapsto x^2$ is not particularly Lipschitz, is it? You seem to miss either a *locally* or that $df$ must be bounded.2011-10-17
  • 0
    @t.b. In some contexts $C^1$ implies that $\sup |f| + \sup |df| < \infty$. (At least, when *I* write $C^1(M,\mathbb{R}^k)$ with $M$ being non-compact, that's what I would mean.) Of course, this is not what the OP wrote in the parenthetical.2011-10-17
  • 0
    yes you're right. I meant locally Lipschitz.2011-10-17
  • 0
    @bass: use that continuous functions are locally bounded.2011-10-17
  • 0
    First reduce to the scalar-valued case (this is easy). Then note that $\left|f\left(x\right)-f\left(y\right)\right|=\left|\int_{0}^{1}\frac{d}{dt}f\left(tx+\left(1-t\right)y\right)dt\right|$. Then use the chain rule and...2011-10-17

3 Answers 3