The extension appeared in the previous post and is very simple: given a closed set and a bounded function
, we define
for all
. It is easy to see that
as
. Indeed, the boundedness of
implies that the infimum is actually taken only over the set
where
. As the distance
tends to 0, the set
shrinks at the same rate: indeed,
is contained in the ball of radius
with center
. Thus,
furnishes a continuous extension of
whenever
is continuous on
.
Now suppose that is Lipschitz on
. Will the extension be Lipschitz? Let’s consider a one-dimensional example with
,
, and
on
. When
, the extension is
When , the infimum is at
, hence
. When
, it shifts to
, which yields
. Here is the plot:

While the original function had Lipschitz constant 1, the extension has Lipschitz constant 5 (attained to the right of the transition point
). We can make this progressively worse by replacing 5 with larger numbers. So, the Lipschitz constant of the extension does not admit a bound in terms of
alone. The oscillation
enters the game as well… although it’s not entirely clear that it must. In the example above, there is no apparent reason for
to decrease so quickly (or at all) when
.
Should the a_{x,M} have M = 1 + sup f – inf f?
You are right. And then M+1 can be replaced by M below. Edited.
Thank. These notes definitely helped me to get better sense of the construction.