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We will prove the implicit function theorem.
Let f (x, y) be a continuous function of two variables so that it is
continuous, and differentiable with respect to y when x is kept
fixed; in particular, we have expression
f (x, y) = f_{0}(x, y_{0}) + f_{1}(x, y_{0})(y  y_{0}) + o_{x}(y  y_{0})
where the subscript in the o_{x} denotes the dependence of the
condition on x. The function f_{1}(x, y) is denoted by
f /y and is called the partial derivative of f with respect
to y (we have already seen this for the case of polynomials). We
further assume that f_{1} is continuous.
Figure 7:
The implicit function

Now suppose that f (a, b) = 0 and
c = (f /y)(a, b) 0. We want to find the implicit function h(x) defined by f = 0
(see 7). We do this by showing that for each x near a
there is a unique y so that f (x, y) = 0. Equivalently, we need to
show that the function
g(x, y) = y  f (x, y)/c has a unique fixed point
for any chosen x near a.
Contractions give rise to functions with a unique fixed point.
Exercise 49
Let 0 < c < 1 be a constant. Let g(x) be a function so that g
send the interval [a, b] to itself and
 g(x)  g(y) < c x  y
for all x and y in [a, b]. Show that
g(x_{0}) = x_{0} for exactly
one point x_{0} in [a, b].
Consider the function
g(x, y) = y  f (x, y)/c as above; we have
g(a, b) = b and
(g/y) = 0. For each fixed x we
would like g(x, y) to be a contraction on some interval around
b. As a first step:
Exercise 50
Using the continuity of
f_{1} show that there is an interval
[
a 
r,
a +
r] around
a and an interval [
b 
s,
b +
s] around
b so
that
(
g/
y)(
x,
y)
1/2 for
x and
y in
these respective intervals. (Hint: Write
g/
y in
terms of
f_{1} to show that it is continuous). In particular, by the
mean value theorem show that

g(
x,
y) 
g(
x,
y')
1/2(
y 
y') on these intervals.
Now by the continuity of f (and thus of g) we can choose a smaller
r so that
 g(x, b)  g(a, b) s/2 for x in the interval
[a  r, a + r]. Since g(a, b) = b, it follows that
g(x, y) = b + (g(x, y)  g(x, b)) + (g(x, b)  g(a, b)) lies in the interval
[b  s, b + s]. Applying the above exercise it follows that for every x
there is a unique point y so that g(x, y) = y or equivalently
f (x, y) = 0. We denote this point y as h(x). This function h is
the required implicit function.
We have the identity,
Applying the mean value theorem to g(x', y) we obtain

h(
x) 
h(
x')

g(
x,
h(
x)) 
g(
x',
h(
x) +

h(
x) 
h(
x')
Exercise 51
Use the above inequality and the continuity of g(x, y_{0})
for every fixed y_{0} to conclude the h(x) is continuous.
Now if we assume in addition that f is differentiable to order k
then it follows that h is also differentiable to order k by an
entirely similar reasoning to the one in the above exercise.
Exercise 52
If
f (
x,
y) has the form (near the point (
a,
b))
with
f_{0, 1} 0, then show that the implicit function
g(
x)
has the form (near
x =
a),
g(
x) =
b 
(
x 
a) +
(
x 
a)
^{2} +
o((
x 
a)
^{2})
Next: Integration
Up: New Functions from old
Previous: Inverse functions
Kapil H. Paranjape
20010120