Graphics Reference
In-Depth Information
(
)
h
xx
xx
,
1
1
Æ
0
as
xx
Æ
.
1
-
We must show that an equation similar to (4.8) holds for f
-1
. The obvious candidate
for Df
-1
is A = (Df)
-1
. Applying A to both sides of equation (4.8) gives
(
)
+
(
(
)
)
=
-
1
()
-
-
1
()
A
yy
-
A h
yy
,
f
y
f
y
,
1
1
1
1
where h
1
(
y
,
y
1
) =-h(f
-1
(
y
),f
-1
(
y
1
)). Now
(
)
(
)
h
yy
yy
,
h
yy
xx
,
xx
yy
-
-
1
1
1
1
1
1
1
1
=-
-
-
and the right-hand side of this equation goes to zero (the first term goes to zero and
the second is bounded by 2). This shows that f
-1
is differentiable. The continuity of
D(f
-1
) follows from the fact that the matrix for this map is defined by the composite
of the maps
r
Ê
Ë
ˆ
¯
æÆ
n
n
()
æ
(
)
æ
(
)
B
æ
B
r
æææææ
Æ
GL
n
,
R
ææææ æ
Æ
GL
n
,
R
,
2
-
1
matrix
for Df
matrix inversion
f
where we identify the space
GL
(n,
R
) of nonsingular real n ¥ n matrices with
R
n
2
.
If f is of class C
k
, then one can show in a similar fashion that f
-1
is also.
To prove the first application of the inverse function theorem we need two lemmas.
4.4.3. Lemma.
Let
U
be an open subset of
R
n
that contains the origin. Let f :
U
Æ
R
m
, n £ m, be a C
k
map with f(
0
) =
0
and k ≥ 1. Assume that Df(
0
) has rank n. Then
there is a C
k
diffeomorphism g of one neighborhood of the origin in
R
m
onto another
with g(
0
) =
0
and such that
(
(
)
)
=
(
)
g f x
,...,
x
x
,...,
x
, ,...,
00
1
n
1
n
holds in some neighborhood of the origin in
R
n
.
Proof.
The hypothesis that Df(
0
) has rank n means that the n ¥ m Jacobian matrix
(∂f
i
/∂x
j
) has rank n. Because we can interchange coordinates if necessary, there is no
loss in generality if we assume that
∂
∂
f
x
Ê
Á
ˆ
˜
i
rank
=
n
.
j
1,
££
ij n
Consider the map F :
U
¥
R
m-n
Æ
R
m
defined by
(
)
=
(
)
+
(
)
Fx
,...,
x
fx
,...,
x
00
,..., ,
x
,...,
x
.
1
m
1
n
n
+
1
m
Since F(x
1
,...,x
n
,0,...,0) = f(x
1
,...,x
n
), F is an extension of f. Furthermore, the
determinant of (∂F
i
/∂x
j
) is just the determinant of
