1
$\begingroup$

Let $f$ be a locally Lipschitz $C^1$ function defined on $\mathbb{R}$. Define $$g_n(x)=\begin{cases}f(x) &: x \in [-n,n]\\ f(n) &: x \in (n, \infty)\\ f(-n) &:x \in (-\infty, -n)\end{cases}$$.

Then $g_n$ pointwise converges to $g$.

Is $g$ globally Lipschitz? I believe it is, except it is not differentiable at the points $-n$ and $n$. So it is differentiable a.e.

$\endgroup$
  • 1
    $\begingroup$ Yes it is. Determine the Lpischitz constant inside $[-n,n]$. It is easy to check that the same constant holds on $(-\infty,\infty)$. $\endgroup$ – Quickbeam2k1 Apr 7 '14 at 9:50
0
$\begingroup$

You don't need to invoke derivatives, actually. Just write $g_n=f\circ h_n$ where $$h_n(x) = \max(-n,\min(n,x))$$ Check that $h_n$ is a $1$-Lipschitz function (taking maximum of minimum of Lipschitz functions does not increase the Lipschtiz constant.) The Lipschitz constant is submultiplicative under composition.

Here's a more general statement.

Suppose $K$ is a convex closed subset of $\mathbb R^n$ and $f:K\to \mathbb R$ is an $L$-Lipschitz function. Extend $f$ to $\mathbb R^n$ by $f(x)=f(a)$ where $a$ is the point of $K$ nearest to $x$. Then the extended map is also $L$-Lipschitz.

Proof: let $\pi:\mathbb R^n\to K$ be the nearest point contraction, i.e., $\pi(x)=a$ in the above notation. This map is $1$-Lipschitz (see second half of the answer). Therefore, $f\circ \pi$ is $L$-Lipschitz, and this is our extension. $\quad \Box$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.