Lemma 1: Let
be defined and differentiable
on a convex open set A. Then for all
, one has
where
is chosen to be an upper bound on all the
for all
,
, and all
.
Proof: Let
. For each
, the Mean Value
Theorem says that there is a
between 0 and 1 such that
By the chain rule,
So,
Theorem 1: (Inverse Function Theorem) Let
be continuously differentiable on an open subset containing
.
If
, then there are open subsets
containing
and
containing
such that
and there is an inverse
which is differentiable with derivative satisfying:
Exercise 1: Show that Theorem 1 is true for linear functions.
Proof: (of Theorem 1) By replacing
with
, we
can assume that
is the identity map (Why?).
Apply Lemma 1 to
:
for
and
in some open rectangle containing
. But then,
Rearranging gives
Since
,
and so by choosing the rectangle
small enough, we can assume that
and so
In particular, it follows that f is one-to-one when restricted to this
rectangle, and the inverse will be continuous if it exists.
Replacing the rectangle with a smaller one, we can assume the
same is true when f is restricted to the closure of the rectangle. Now
the boundary
of the rectangle is compact and so
is also compact
and does not contain
. Let
be the minimum of
for
. Let
be the set
.
Now, for every
, there is at least one
in the rectangle with
. In fact, consider the function
. The
image of
under the closure of the rectangle. The minimum of this
function does not occur on
because
. So it must occur where
the derivative is zero, i.e. one has:
for
. But by taking the rectangle sufficiently small, we can
assume that
for all
in the rectangle. But then
the only solution of this system of linear equations is
.
If
, then
maps the open set
one-to-one and onto
the open set
. It remains to check differentiability of the inverse.
Since
is differentiable, one has for
,
where
. Letting
and
, we get after substitution:
Rearranging gives:
It remains to show that
Since the derivative is just a linear function, it is enough to show that
. As
, we have
because
is continuous. So,
.
But we know that
So, the limit of the product
is zero, as desired.
Note: It follows from the formula for the derivative of the inverse that the inverse is also continuously differentiable.
Corollary 1: (The Implicit Function Theorem) Let
be continuously differentiable in an open set containing the point
which satisfies
. Suppose that
.
Then there is a continuously differentiable function
mapping
an open set
containing
to an open set
containing
such that
for all
.
Proof: We can extend
to a function
by
and apply the Inverse Function Theorem to get
an inverse
defined in an open subset containing
and mapping onto
an open subset containing
. Let
where
is the projection onto the last
coordinates of
.
Then one has
.
Exercise 2: Show that the Inverse Function Theorem is a Corollary of the Implicit Function Theorem.