Inverse and Implicit Function Theorems - Differential Calculus, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Handout 4. The Inverse and Implicit Function Theorems
Recall that a linear map L : R
n
? R
n
with detL 6= 0 is one-to-one. By the
next theorem, a continuously di?erentiable map between regions in R
n
is locally
one-to-one near any point where its di?erential has nonzero determinant.
Inverse Function Theorem. Suppose U is open in R
n
and F : U ? R
n
is a
continuously di?erentiable mapping, p ? U, and the di?erential at p, dF
p
, is an
isomorphism. Then there exist neighborhoods V of p in U and W of F(p) in R
n
so
that F :V ?W has a continuously di?erentiable inverse F
-1
:W ?V with
d(F
-1
)
y
=

dF
F
-1
{y}

-1
for y?W .
Moreover, F
-1
is smooth (in?nitely di?erentiable) whenever F is smooth.
Thus, the equation y = F(x), written in component form as a system of n
equations,
y
i
= F
i
(x
1
,...,x
n
) for i = 1,...,n ,
can be solved for x
1
,...,x
n
in terms of y
1
,...,y
n
provided we restrict the points x
and y to small enough neighborhoods of p and F(p). The solutions are then unique
and continuously di?erentaible.
Proof : Let L =dF
p
, and note that the number
? =
1
2
inf
|v|=1
|L(v)| =
1
2sup
|w|=1
|L
-1
(w)|
is positive. Since dF
x
is continuous in x at x =p, we have the inequality
sup
|v|=1
|dF
x
(v)-L(v)| = ?
true for all x in some su?ciently small ball V about p in U. Thus, by linearity,
|dF
x
(v)-L(v)| = ?|v| for all v?R
n
and x?V .
With each y?R
n
, we associate the function
A
y
(x) = x + L
-1

y - F(x)


.
Then
F(x) =y if and only if x is a fixed point of A
y
.
1
Page 2


Handout 4. The Inverse and Implicit Function Theorems
Recall that a linear map L : R
n
? R
n
with detL 6= 0 is one-to-one. By the
next theorem, a continuously di?erentiable map between regions in R
n
is locally
one-to-one near any point where its di?erential has nonzero determinant.
Inverse Function Theorem. Suppose U is open in R
n
and F : U ? R
n
is a
continuously di?erentiable mapping, p ? U, and the di?erential at p, dF
p
, is an
isomorphism. Then there exist neighborhoods V of p in U and W of F(p) in R
n
so
that F :V ?W has a continuously di?erentiable inverse F
-1
:W ?V with
d(F
-1
)
y
=

dF
F
-1
{y}

-1
for y?W .
Moreover, F
-1
is smooth (in?nitely di?erentiable) whenever F is smooth.
Thus, the equation y = F(x), written in component form as a system of n
equations,
y
i
= F
i
(x
1
,...,x
n
) for i = 1,...,n ,
can be solved for x
1
,...,x
n
in terms of y
1
,...,y
n
provided we restrict the points x
and y to small enough neighborhoods of p and F(p). The solutions are then unique
and continuously di?erentaible.
Proof : Let L =dF
p
, and note that the number
? =
1
2
inf
|v|=1
|L(v)| =
1
2sup
|w|=1
|L
-1
(w)|
is positive. Since dF
x
is continuous in x at x =p, we have the inequality
sup
|v|=1
|dF
x
(v)-L(v)| = ?
true for all x in some su?ciently small ball V about p in U. Thus, by linearity,
|dF
x
(v)-L(v)| = ?|v| for all v?R
n
and x?V .
With each y?R
n
, we associate the function
A
y
(x) = x + L
-1

y - F(x)


.
Then
F(x) =y if and only if x is a fixed point of A
y
.
1
Since dA
y
= Id -L
-1

dF
x


=L
-1

L-dF
x


, the above inequalities imply that
|dA
y
x
(v)| =
1
2
|v| for x?V and v?R
n
.
Thus, for w,x?V,
|A
y
(w)-A
y
(x)| = |
Z
1
0
d
dt
A
y

x+t(w-x)


dt|
=
Z
1
0
|dA
y
x+t(w-x)
(w-x)|dt =
1
2
|w-x| . (*)
ItfollowsthatA
y
hasatmostone?xedpointinV,andthereisatmostonesolution
x?V for F(x) =y.
Next we verify that W = F(V) is open. To do this, we choose, for any point
˜ w =F(˜ x)?W with ˜ x?V,asu?cientlysmallpositiver,sothattheballB =B
r
(˜ x)
has closure B ?V. We will show that B
?r
(˜ w)?W. This will give the openness of
W.
For any y?B
?r
(˜ w), and A
y
as above,
|A
y
(˜ x)- ˜ x| = |L
-1
(y- ˜ w)| <
1
2?
?r =
r
2
.
For x?B it follows that
|A
y
(x)- ˜ x| = |A
y
(x)-A
y
(˜ x)|+|A
y
(˜ x)- ˜ x| <
1
2
|x- ˜ x|+
r
2
= r .
So A
y
(x)?B. By (*) A
y
thus gives a contraction of B. So A
y
has ?xed point x in
B, and y =F(x)?F(B)?F(V) =W. Thus B
?r
(˜ w)?W.
Next we show that F
-1
: W ? V is di?erentiable at each point y ? W and
that
d(F
-1
)
y
= M
-1
where M =dF
x
with x =F
-1
(y)?V .
Supposey+k?W andx+h =F
-1
(y+k)?V. Then, withourpreviousnotations,
|h-L
-1
(k)| = |h-L
-1

F(x+h)-F(x)


| = |A
y
(x+h)-A
y
(x)|=
1
2
|h| ,
which implies that
1
2
|h| = |L
-1
(k)| =

1
2?


|k| .
2
Page 3


Handout 4. The Inverse and Implicit Function Theorems
Recall that a linear map L : R
n
? R
n
with detL 6= 0 is one-to-one. By the
next theorem, a continuously di?erentiable map between regions in R
n
is locally
one-to-one near any point where its di?erential has nonzero determinant.
Inverse Function Theorem. Suppose U is open in R
n
and F : U ? R
n
is a
continuously di?erentiable mapping, p ? U, and the di?erential at p, dF
p
, is an
isomorphism. Then there exist neighborhoods V of p in U and W of F(p) in R
n
so
that F :V ?W has a continuously di?erentiable inverse F
-1
:W ?V with
d(F
-1
)
y
=

dF
F
-1
{y}

-1
for y?W .
Moreover, F
-1
is smooth (in?nitely di?erentiable) whenever F is smooth.
Thus, the equation y = F(x), written in component form as a system of n
equations,
y
i
= F
i
(x
1
,...,x
n
) for i = 1,...,n ,
can be solved for x
1
,...,x
n
in terms of y
1
,...,y
n
provided we restrict the points x
and y to small enough neighborhoods of p and F(p). The solutions are then unique
and continuously di?erentaible.
Proof : Let L =dF
p
, and note that the number
? =
1
2
inf
|v|=1
|L(v)| =
1
2sup
|w|=1
|L
-1
(w)|
is positive. Since dF
x
is continuous in x at x =p, we have the inequality
sup
|v|=1
|dF
x
(v)-L(v)| = ?
true for all x in some su?ciently small ball V about p in U. Thus, by linearity,
|dF
x
(v)-L(v)| = ?|v| for all v?R
n
and x?V .
With each y?R
n
, we associate the function
A
y
(x) = x + L
-1

y - F(x)


.
Then
F(x) =y if and only if x is a fixed point of A
y
.
1
Since dA
y
= Id -L
-1

dF
x


=L
-1

L-dF
x


, the above inequalities imply that
|dA
y
x
(v)| =
1
2
|v| for x?V and v?R
n
.
Thus, for w,x?V,
|A
y
(w)-A
y
(x)| = |
Z
1
0
d
dt
A
y

x+t(w-x)


dt|
=
Z
1
0
|dA
y
x+t(w-x)
(w-x)|dt =
1
2
|w-x| . (*)
ItfollowsthatA
y
hasatmostone?xedpointinV,andthereisatmostonesolution
x?V for F(x) =y.
Next we verify that W = F(V) is open. To do this, we choose, for any point
˜ w =F(˜ x)?W with ˜ x?V,asu?cientlysmallpositiver,sothattheballB =B
r
(˜ x)
has closure B ?V. We will show that B
?r
(˜ w)?W. This will give the openness of
W.
For any y?B
?r
(˜ w), and A
y
as above,
|A
y
(˜ x)- ˜ x| = |L
-1
(y- ˜ w)| <
1
2?
?r =
r
2
.
For x?B it follows that
|A
y
(x)- ˜ x| = |A
y
(x)-A
y
(˜ x)|+|A
y
(˜ x)- ˜ x| <
1
2
|x- ˜ x|+
r
2
= r .
So A
y
(x)?B. By (*) A
y
thus gives a contraction of B. So A
y
has ?xed point x in
B, and y =F(x)?F(B)?F(V) =W. Thus B
?r
(˜ w)?W.
Next we show that F
-1
: W ? V is di?erentiable at each point y ? W and
that
d(F
-1
)
y
= M
-1
where M =dF
x
with x =F
-1
(y)?V .
Supposey+k?W andx+h =F
-1
(y+k)?V. Then, withourpreviousnotations,
|h-L
-1
(k)| = |h-L
-1

F(x+h)-F(x)


| = |A
y
(x+h)-A
y
(x)|=
1
2
|h| ,
which implies that
1
2
|h| = |L
-1
(k)| =

1
2?


|k| .
2
We now obtain the desired formula for d(F
-1
)
y
by computing that
|F
-1
(y+k)-F
-1
(y)-M
-1
k|
|k|
=
|h-M
-1
k|
|k|
= |M
-1

F(x+h)-F(x)-Mh
|h|


|
|h|
|k|
=
1
?
|M
-1

F(x+h)-F(x)-Mh
|h|


| ,
which approaches 0 as |k|? 0 because M =dF
x
.
Finally, since the inversion of matrices is, by Cramer’s rule, a continuous, in
fact, smooth, function of the entries, we deduce from our formula that F
-1
is
continuously di?erentiable. Moreover, repeatly di?erentiating the formula shows
that F
-1
is a smooth mapping whenever F is.
Next we turn to the Implicit Function Theorem. This important theorem gives
a condition under which one can locally solve an equation (or, via vector notation,
system of equations)
f(x,y) = 0
for y in terms of x. Geometrically the solution locus of points (x,y) satisfying the
equation is thus represented as the graph of a function y =g(x). For smooth f this
is a smooth manifold.
Let (x,y) =

(x
1
,...,x
m
),(y
1
,...,y
n
)


denote a point in R
m
×R
n
, and, for
an R
n
-valued function f(x,y) = (f
1
,...,f
n
)(x,y) , let d
x
f denote the partial
di?erential represented by the n × m matrix

?f
i
?x
j

and d
y
f denote the partial
di?erential represented by the n×n matrix

?f
i
?y
j

.
ImplicitFunctionTheorem. Supposef(x,y) is a continuously di?erentiableR
n
-
valued function near a point (a,b)? R
m
×R
n
, f(a,b) = 0, and detd
y
f|
(a,b)
6= 0 .
Then
{(x,y)?W : f(x,y) = 0} = {

x,g(x)


: x?X}
for some open neighborhood W of (a,b) in R
m
× R
n
and some continuously
di?erentiable functiong mapping someR
m
neighborhoodX ofa intoR
n
. Moreover,
(d
x
g)
x
= -(d
y
f)
-1
|
(x,g(x))
d
x
f|
(x,g(x))
,
and g is smooth in case f is smooth.
3
Page 4


Handout 4. The Inverse and Implicit Function Theorems
Recall that a linear map L : R
n
? R
n
with detL 6= 0 is one-to-one. By the
next theorem, a continuously di?erentiable map between regions in R
n
is locally
one-to-one near any point where its di?erential has nonzero determinant.
Inverse Function Theorem. Suppose U is open in R
n
and F : U ? R
n
is a
continuously di?erentiable mapping, p ? U, and the di?erential at p, dF
p
, is an
isomorphism. Then there exist neighborhoods V of p in U and W of F(p) in R
n
so
that F :V ?W has a continuously di?erentiable inverse F
-1
:W ?V with
d(F
-1
)
y
=

dF
F
-1
{y}

-1
for y?W .
Moreover, F
-1
is smooth (in?nitely di?erentiable) whenever F is smooth.
Thus, the equation y = F(x), written in component form as a system of n
equations,
y
i
= F
i
(x
1
,...,x
n
) for i = 1,...,n ,
can be solved for x
1
,...,x
n
in terms of y
1
,...,y
n
provided we restrict the points x
and y to small enough neighborhoods of p and F(p). The solutions are then unique
and continuously di?erentaible.
Proof : Let L =dF
p
, and note that the number
? =
1
2
inf
|v|=1
|L(v)| =
1
2sup
|w|=1
|L
-1
(w)|
is positive. Since dF
x
is continuous in x at x =p, we have the inequality
sup
|v|=1
|dF
x
(v)-L(v)| = ?
true for all x in some su?ciently small ball V about p in U. Thus, by linearity,
|dF
x
(v)-L(v)| = ?|v| for all v?R
n
and x?V .
With each y?R
n
, we associate the function
A
y
(x) = x + L
-1

y - F(x)


.
Then
F(x) =y if and only if x is a fixed point of A
y
.
1
Since dA
y
= Id -L
-1

dF
x


=L
-1

L-dF
x


, the above inequalities imply that
|dA
y
x
(v)| =
1
2
|v| for x?V and v?R
n
.
Thus, for w,x?V,
|A
y
(w)-A
y
(x)| = |
Z
1
0
d
dt
A
y

x+t(w-x)


dt|
=
Z
1
0
|dA
y
x+t(w-x)
(w-x)|dt =
1
2
|w-x| . (*)
ItfollowsthatA
y
hasatmostone?xedpointinV,andthereisatmostonesolution
x?V for F(x) =y.
Next we verify that W = F(V) is open. To do this, we choose, for any point
˜ w =F(˜ x)?W with ˜ x?V,asu?cientlysmallpositiver,sothattheballB =B
r
(˜ x)
has closure B ?V. We will show that B
?r
(˜ w)?W. This will give the openness of
W.
For any y?B
?r
(˜ w), and A
y
as above,
|A
y
(˜ x)- ˜ x| = |L
-1
(y- ˜ w)| <
1
2?
?r =
r
2
.
For x?B it follows that
|A
y
(x)- ˜ x| = |A
y
(x)-A
y
(˜ x)|+|A
y
(˜ x)- ˜ x| <
1
2
|x- ˜ x|+
r
2
= r .
So A
y
(x)?B. By (*) A
y
thus gives a contraction of B. So A
y
has ?xed point x in
B, and y =F(x)?F(B)?F(V) =W. Thus B
?r
(˜ w)?W.
Next we show that F
-1
: W ? V is di?erentiable at each point y ? W and
that
d(F
-1
)
y
= M
-1
where M =dF
x
with x =F
-1
(y)?V .
Supposey+k?W andx+h =F
-1
(y+k)?V. Then, withourpreviousnotations,
|h-L
-1
(k)| = |h-L
-1

F(x+h)-F(x)


| = |A
y
(x+h)-A
y
(x)|=
1
2
|h| ,
which implies that
1
2
|h| = |L
-1
(k)| =

1
2?


|k| .
2
We now obtain the desired formula for d(F
-1
)
y
by computing that
|F
-1
(y+k)-F
-1
(y)-M
-1
k|
|k|
=
|h-M
-1
k|
|k|
= |M
-1

F(x+h)-F(x)-Mh
|h|


|
|h|
|k|
=
1
?
|M
-1

F(x+h)-F(x)-Mh
|h|


| ,
which approaches 0 as |k|? 0 because M =dF
x
.
Finally, since the inversion of matrices is, by Cramer’s rule, a continuous, in
fact, smooth, function of the entries, we deduce from our formula that F
-1
is
continuously di?erentiable. Moreover, repeatly di?erentiating the formula shows
that F
-1
is a smooth mapping whenever F is.
Next we turn to the Implicit Function Theorem. This important theorem gives
a condition under which one can locally solve an equation (or, via vector notation,
system of equations)
f(x,y) = 0
for y in terms of x. Geometrically the solution locus of points (x,y) satisfying the
equation is thus represented as the graph of a function y =g(x). For smooth f this
is a smooth manifold.
Let (x,y) =

(x
1
,...,x
m
),(y
1
,...,y
n
)


denote a point in R
m
×R
n
, and, for
an R
n
-valued function f(x,y) = (f
1
,...,f
n
)(x,y) , let d
x
f denote the partial
di?erential represented by the n × m matrix

?f
i
?x
j

and d
y
f denote the partial
di?erential represented by the n×n matrix

?f
i
?y
j

.
ImplicitFunctionTheorem. Supposef(x,y) is a continuously di?erentiableR
n
-
valued function near a point (a,b)? R
m
×R
n
, f(a,b) = 0, and detd
y
f|
(a,b)
6= 0 .
Then
{(x,y)?W : f(x,y) = 0} = {

x,g(x)


: x?X}
for some open neighborhood W of (a,b) in R
m
× R
n
and some continuously
di?erentiable functiong mapping someR
m
neighborhoodX ofa intoR
n
. Moreover,
(d
x
g)
x
= -(d
y
f)
-1
|
(x,g(x))
d
x
f|
(x,g(x))
,
and g is smooth in case f is smooth.
3
Proof : De?ne F(x,y) = (x,f(x,y)


, and compute that
detdF
(a,b)
= det(d
y
f)
(a,b)
6= 0 .
The Inverse Function Theorem thus gives a continuously di?erentiable inverse
F
-1
:W ?V forsomeopenneighborhoodsV of(a,b)andW of(a,0)inR
m
×R
n
.
The set X = {x ? R
m
: (x,0) ? W} is open in R
m
, and, for each point
x?X, F
-1
(x,0) =

x,g(x)


for some point g(x)?R
n
. Moreover,
{(x,y)?W : f(x,y) = 0} = (F
-1
?F)

W nf
-1
{0}


= F
-1

W n(R
m
×{0})


= {

x,g(x)


: x?X} .
One readily checks that g is continuously di?erentiable with
?g
i
?x
j
(x) =
?(F
-1
)
m+i
?x
j
(x,0)
for i = 1,...,n, j = 1,...,m, and x ? W. The formula for (d
x
g)
x
follows from
di?erentiating the identity
f

x,g(x)


= 0 on W ,
and using the chain rule. Smoothness of g follows from smoothness of f by
repeatedly di?erentiating this identity.
4