Transcript Lectures on Calculus - The University of West Georgia

Lectures on Calculus
The Inverse Function Theorem
by William M. Faucette
University of West Georgia
Adapted from
Calculus on Manifolds
by Michael Spivak
Lemma One
Lemma: Let ARn be a rectangle and let
f:ARn be continuously differentiable.
If there is a number M such that
|Djf i(x)|≤M for all x in the interior of A,
then
f (x) - f (y) £ n M x - y
2
for all x, y2A.
Lemma One
Proof: We have
n
f i (y) - f i (x) = å[ f i (y1,… , y j , x j +1,… , x n )
j=1
- f (y ,… , y , x ,… , x )].
i
1
j-1
j
n
Lemma One
Proof: Applying the Mean Value Theorem
we obtain
i
j
f (y ,… , y , x
1
j +1
,… , x ) - f (y ,… , y , x ,… , x )
n
i
j-1
1
j
n
= (y - x ) × D j f (zij )
for some zij between xj and yj.
j
j
i
Lemma One
Proof: The expression
(y - x ) × D j f (zij )
j
j
i
has absolute value less than or equal to
M× y -x .
j
j
Lemma One
Proof: Then
n
f i (y) - f i (x) £ å M × y j - x j £ nM y - x
j=1
since each |yj-xj|≤|y-x|.
Lemma One
Proof: Finally,
n
f (y) - f (x) £ å f (y) - f (x) £ n M × y - x ,
i
i
2
i=1
which concludes the proof. QED
The Inverse Function
Theorem
The Inverse Function
Theorem
Theorem: Suppose that f:RnRn is
continuously differentiable in an open set
containing a, and det f (a)≠0. Then there is
an open set V containing a and an open set
W containing f(a) such that f:VW has a
continuous inverse f -1:WV which is
differentiable and for for all y2W satisfies
[
]
-1
¢
-1
( f ) (y) = f ¢( f (y)) .
-1
The Inverse Function
Theorem
Proof: Let  be the linear transformation
Df(a). Then  is non-singular, since
det f (a)≠0. Now
D( l
-1
f )(a) = D( l
-1
)( f (a))
Df (a)
= l Df (a)
is the identity linear transformation.
-1
The Inverse Function
Theorem
Proof: If the theorem is true for -1f, it is
clearly true for f. Therefore we may
assume at the outset that  is the
identity.
The Inverse Function
Theorem
Whenever f(a+h)=f(a), we have
f (a + h) - f (a) - l(h) h
= = 1.
h
h
But
The Inverse Function
Theorem
This means that we cannot have f(x)=f(a)
for x arbitrarily close to, but unequal to,
a. Therefore, there is a closed
rectangle U containing a in its interior
such that
f (x) ¹ f (a) if x Î U and x ¹ a.
The Inverse Function
Theorem
Since f is continuously differentiable in an
open set containing a, we can also
assume that
det f ¢(x) ¹ 0 for x Î U.
1
i
i
D j f (x) - D j f (a) < 2 for all i, j, and x Î U.
2n
The Inverse Function
Theorem
Since
1
D j f (x) - D j f (a) < 2 for all i, j, and x Î U.
2n
i
i
we can apply Lemma One to g(x)=f(x)-x
to get
f (x1 ) - x1 - ( f (x 2 ) - x 2 )
1
£ x1 - x 2 .
2
The Inverse Function
Theorem
Since
x1 - x 2 - f (x1 ) - f (x 2 ) £ f (x1 ) - x1 - ( f (x 2 ) - x 2 )
1
£ x1 - x 2 ,
2
we have
x1 - x 2 £ 2 f (x1) - f (x 2 ) for x1, x 2 Î U.
The Inverse Function
Theorem
Now f(boundary U) is a compact set
which does not contain f(a). Therefore,
there is a number d>0 such that
|f(a)-f(x)|≥d for x2boundary U.
The Inverse Function
Theorem
Let W={y:|y-f(a)|<d/2}. If y2W and
x2boundary U, then
y - f (a) < y - f (x).
The Inverse Function
Theorem
We will show that for any y2W there is a
unique x in interior U such that f(x)=y.
To prove this consider the function
g:UR defined by
n
g(x) = y - f (x) = å ( y - f (x)) .
2
i
i=1
i
2
The Inverse Function
Theorem
This function is continuous and therefore
has a minimum on U. If x2boundary U,
then, by the formula on slide 20, we
have g(a)<g(x). Therefore, the
minimum of g does not occur on the
boundary of U.
The Inverse Function
Theorem
Since the minimum occurs on the interior
of U, there must exist a point x2U so
that Djg(x)=0 for all j, that is
n
å2( y
i=1
i
- f (x)) × D j f (x) = 0 for all j.
i
i
The Inverse Function
Theorem
Since the Jacobian [Djf i(x)] is nonsingular, we must have
y - f (x) = 0 for all i.
i
i
That is, y=f(x). This proves the existence
of x. Uniqueness follows from slide 18.
The Inverse Function
Theorem
If V=(interior U)f-1(W), we have shown that
the function f:VW has inverse f -1:WV.
We can rewrite
As
f
x1 - x 2 £ 2 f (x1) - f (x 2 ) for x1, x 2 Î V
-1
( y1) - f ( y 2 ) £ 2 y1 - y 2
-1
for y1, y 2 Î W .
This shows that f-1 is continuous.
The Inverse Function
Theorem
We only need to show that f-1 if differentiable.
Let =Df(x). We will show that f-1 is
differentiable at y=f(x) with derivative -1.
The Inverse Function
Theorem
Since =Df(x), we know that
Setting (x)=f(x+h)-f(x)-(h), we know that
The Inverse Function
Theorem
Hence, we have
The Inverse Function
Theorem
Therefore,
m
-1
( f (x1) - f (x)) = x1 - x + m (f (x1 - x)).
-1
The Inverse Function
Theorem
-1
-1
m ( f (x1 ) - f (x)) = x1 - x + m (f (x1 - x)).
Since every y12W is of the form f(x1) for some
x12V, this can be written
m ( y1 - y ) = f
-1
or
f
-1
-1
( y1) - f ( y) + m (f ( f ( y1) - f ( y))).
-1
-1
-1
-1
( y1) = f ( y) + m ( y1 - y ) - m (f ( f ( y1) - f ( y))).
-1
-1
-1
-1
-1
The Inverse Function
Theorem
It therefore suffices to show that
Since - is a linear transformation, it suffices
to show that
The Inverse Function
Theorem
Recall that
f
-1
( y1) - f ( y 2 ) £ 2 y1 - y 2
-1
Also, f-1 is continuous, so
for y1, y 2 Î W .
The Inverse Function
Theorem
Then
f ( f (y1) - f (y))
-1
-1
y1 - y
f ( f (y1 ) - f (y))
-1
=
-1
-1
-1
f (y1 ) - f (y)
×
f -1(y1) - f -1 (y)
y1 - y
where the first factor goes to 0 and the second
factor is bounded by 2. This completes the
proof. QED
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