Class notes Nov. 6
Continuity
Below are some complete notes on what we did in class.
Example 1. Theorem 4.3.9.(Abbott): Let f : A  R, g : B  R, f ( A)  B , so that the
composition g f ( x)  g ( f ( x)) is well-defined on A. If f is continuous at c  A , and g is
continuous at f (c)  B , then g f is continuous at c.
Let   0 . Since g is continuous at f (c)  B , there exists 1  0 such that for any
y  B :| y  f (c) | 1 , then | g ( y )  g ( f (c)) |  .
Now, since f is continuous at c , for 1  0 , there exists   0 such that for any
x  A :| x  c |  , we have | f ( x)  f (c) | 1 . But since f ( x)  B, and | f ( x)  f (c) | 1 ,
as shown above, it follows that | g ( f ( x))  g ( f (c)) |  , proving that the composite
function g f is continuous at c.
Example 2. Contraction Mapping Theorem (ex. 4.3.9. Abbott)
Let f : R  R and assume that there is a constant 0  c  1:| f ( x)  f ( y ) | c | x  y | for
all x, y  R .
a) We first prove that f is continuous on R. Let a  R be arbitrary, and let   0 . We
need to find   0 such that for any x :| x  a |  ,| f ( x)  f (a) |  . Using the given
property of the function we may write: | f ( x)  f (a) | c | x  a | c   if | x  a |  .
Therefore, in the above we may take    , and the continuity at a is proved.
Since a was arbitrary, it follows that f is continuous on R.
Notice that we did not use the fact that 0  c  1
b) Construct the sequence y1 , y2  f ( y1 ),..., yn1  f ( yn ),... .We show that the sequence
( yn ) is a Cauchy sequence.
Let   0 . We need to find the rank n0  1 , such that for any n, m  n0 : | y n  ym |  . (1)
Notice first that for any n  2 :
| yn  yn 1 || f ( yn 1 )  f ( yn  2 ) | c | yn 1  yn  2 | ...  c n  2 | y2  y1 | (2)
The above result can be easily proved by induction.
Now, if n, m  n0 , we may assume that n  m , and for computational purposes we shall
assume that n  m  p, p  0 . Then we write:
| y n  ym || y m  p  ym || y m  p  ym  p 1 |  | y m  p 1  ym  p  2 | ... | y m 1  ym | (3)
Using (2) for n  m  p, m  p  1,..., m  1 , (3) becomes:
| y n  ym | c m p  2 | y 2  y1 | c m p 3 | y 2  y1 | ...  c m1 | y 2  y1 |
p 1
1 c p
(4)
1 c
k 0
Now we can try to find n0  1 such that | y n  ym |  for any n, m  n0 .
| y 2  y1 |  c m k 1 | y 2  y1 | c m1
1 c p
1
| y2  y1 | c n0 1
  , so choose n0  1
1 c
1 c
  (1  c) 
 (1  c)
to satisfy: c n0 1 
, or (n0  1) ln(c)  ln 
 . Since 0  c  1, ln(c)  0 , we
| y2  y1 |
 | y2  y1 | 
From (4) we have | yn  ym || y2  y1 | c m1
  (1  c) 
obtain n0  1  ln 
 / ln(c) (5), and therefore we may choose n0  1 to be the
 | y2  y1 | 
smallest positive integer satisfying (5).
Then we proved that the sequence ( yn ) is Cauchy.
One remark is in place: Notice that in solving for n0 we divided by | y2  y1 | , and one
may argue that this quantity may be zero. If it is, then it means that y1  y2  f ( y1 ) , so
we found the fixed point of the function f.
Since the sequence ( yn ) is Cauchy, it is convergent. Let y  lim yn  lim f ( yn 1 ) . We
n 
n 
proved in (a) that the function is continuous, so lim f ( yn 1 )  f ( y ) and we get y  f ( y ) .
n 
c) We just proved above that the limit point of the sequence is a fixed point of f, i.e.
y  f ( y ) . Now we have to show that the fixed point is unique. Suppose there are two
such points, x, y : x  f ( x), y  f ( y ) . Using the contraction property given in the
hypothesis we have: | y  x || f ( y )  f ( x) | c | y  x || y  x | since c  1 . But the last
inequality is true only if x  y , and the uniqueness is proved.
d) Since we proved that the fixed point is unique, and the construction in b) always leads
to a sequence that converges to a fixed point, it follows that regardless of the initial value
y1 , the sequence will converge to the fixed point of the function.
Here are some applications of the Contraction Mapping Theorem:
Example 3. Consider the function f ( x)  sin x .
How do we know if f ( x)  sin x satisfies the contraction property? Well, it almost
satisfies it since for arbitrary x, y  R , using the Mean Value Theorem, there exists
sin x  sin y
d
x  c  y such that
 ( sin)(c)  cos c , and therefore we have
x y
dx
| sin x  sin y || cos c || x  y || x  y |
But we cannot be certain that the constant is less than one since | cos x | 1 . However, if
we apply the iterative process to f ( x)  sin x we obtain the fixed point.
Let us have a look at the graph first. Since we try to find the value for which sin x  x we
graph both functions in the same system of coordinates:
We see that the solution seems to be x  0 . With a program in Maple (see file) the
solution found to an approximation of 0.0001, when y1  2 is x  0.08425963516 .
Notice that sin(0.08425963516)=0.08424094766 , so the approximation is within the
desired margin of error.
Example 4. Let us look at the function f ( x)  e x . This function does not satisfy the
contraction property on the whole real line, but only for positive real numbers. Indeed,
using again the mean value theorem, for any x, y  0 :| e x  e y || e c || x  y || x  y |
since ec  1, for c  0 . This however will not hurt since if the start with any value
y1  R, e  y1  0 so starting with the second iteration all values will be positive.
Let us then approximate the solution to e x  x .
If we start with y1  0 and want a margin of error of 0.0001 we get after 16 iterations the
(approximate) solution x=0.5671860501. Check that e0.571860501  0.5671190685 , whereas
the solution found with Maple to the equation e x  x is 0.5671432904.
Below you will see the graphs of the two functions: y  e x , y  x .
Finally, some hints for the homework problems:
Exercise 4.2.8: Denote h( x)  f ( x)  g ( x) . So h( x)  0 for all x  A . We want to show
that lim f ( x)  lim g ( x) , which is equivalent to lim h( x)  0 . Suppose that the opposite
x c
x c
x c
happens, that is, lim h( x)  L  0 . Show then that there must be a neighborhood of c on
x c
which h( x)  0, for all x V (c) .
Exercise 4.3.7. Use the definition of a closed set that it must contain its accumulation
points. If K '   , then it follows that K is closed and you are done. If K '   so there
exists a  K ' then look at the characterization of an accumulation point and use the
continuity of the function.
Exercises:
sin x
 1 . Do not use an argument such as L’Hospital’s Rule.
x 0
x
1. Show that lim