Lipschitz Conditions for Inverses: Sqrt Example
Suppose we start with positive real numbers
.
Let
on the interval
:

We
know that
satisfies the following Lipschitz conditions on the interval
:
so,
by the Inverse Function Theorem, its inverse, the square-root,
satisfies:
on
the interval
.
If we let
and
,
then this Lipschitz condition
becomes:
Suppose, now, that we had started with the interval
,
where
and numbers
in this interval. We'd like to say the same Lipschitz conditions hold, but how
do we know that there are numbers
and
whose squares are
and
respectively. In other words, how do we know that
and
have square roots? Clearly it suffices to show that
has a square root, since the same argument will show that
has one. The idea is to show that there are numbers
such that
since then we can apply the Inverse function theorem to
on the interval
(with
).
We will use the inequality
easily
proved by induction. When
this becomes
.
We will let
and choose
so that
.
This will happen when
,
so we can let
,
which
gives
Letting
(
)
now gives
.
To construct
such that
,
use the procedure just described to find a number
such that
.
Then
;
thus, we let
.
Exercise 1. Find numbers
and
such that
.
Explain in detail why we can now conclude that
has a square root.
Exercise 2. If
,
prove that the square root function satisfies the following Lipschitz
condition on the interval
:
Exercise 3. Explain how to generalize this discussion to find
Lipschitz conditions for the
th
root function
on a positive interval
.
Exercise 4. Suppose
.
For any
,
use
to find an exponent
such that
.
Also, find an exponent
such that
.
(Hint: Let
,
so that
;
imitate what we did for squaring.)
Exercise 5. Deduce from Exercise 4 that any number
has a logarithm to base
(
).
You'll need the inverse function theorem applied to
.
Exercise 6. Find Lipschitz conditions for
on the positive interval
.