5
Continuous functions
5.1
Some examples
Consider the functions  () = 2 and  () =
||
 .
Figure 8: The graph of  () = 2 (a continuous function).
Figure 9: The graph of  () =
||

(a function discontinuous at  = 0).
By inspecting the graphs of  and , you can notice the following difference: the graph of  can be drawn with
a single line stroke (i.e. without lifting the pencil off the paper), but the graph of  cannot be drawn like this (once
you arrive at the point  = 0, you have to lift the pencil off the paper to continue graphing).
In this section we will describe this type of behaviour of functions, known as continuity.
5.2
The continuity of a real-valued function of a real variable
Intuitively, a function is continuous at a point if its graph does not have jumps at that point. The formal definition
is the following.
Definition 5.1 (Definition of continuity at a point) Let  :  ⊂ R → R and 0 ∈ .
We say that  is continuous at  = 0 if for any   0 there exists  ()  0 such that
| () −  (0 )|  
for all  ∈  with | − 0 |   ()
(17)
or equivalent5
lim  () =  (0 )
→0
(18)
(the limit lim→0  () exists and equals  (0 )).
If  is not continuous at  = 0 , we say that  is discontinuous at  = 0 .
If  is continuous at every point 0 ∈  we say that  is continuous on .
Remark 5.2 Note that unlike the definition of the limit lim→0  (), the definition of continuity of  at  = 0
requires  to be defined at 0 (i.e. 0 ∈ ).
5 In fact, for the two definitions to be equivalent we need to assume that  ∈  is not an isolated point of . If  ∈  is an isolated
0
0
point of , then the  −   definition shows that  is continuous at 0 .
48
3
Example 5.3 Consider the function  :  = (−1 1) ∪ (1 3] ∪ {5} → R,  () = −1
.
We first note that 0 = 5 ∈  is an isolated point of , so by the definition  is continuous at 0 = 5 (every
function is continuous at an isolated point).
The function  is continuous at 0 = 2, because
lim  () =
3
= 3 =  (2) 
2−1
lim  () =
3
3
= =  (3) 
3−1
2
→2
Similarly, at 0 = 3, we have
%3
hence  is continuous at  = 3.
Note that at  = −1 the function  is not continuous (the function is not defined for  = −1).
Also, the function  is not continuous at  = 1 since  is not defined at this point. Another reason for
which the function  is not continuous at  = 1 is that the lim→1  () does not exist (lim%1  () = −∞ and
lim&1  () = +∞).
The above discussion can be summarized by saying that  is continuous on  = (−1 1) ∪ (1 3] ∪ {5}.
Remark 5.4 As it can be seen from the previous example, if 0 ∈  is an isolated point of , then any function
 :  ⊂ R → R is continuous at 0 (i.e. any function is continuous at an isolated point).
Just as in the case of sided limits, we can define the sided continuity at a point. For example, the notion of
continuity from the left is defined as follows (continuity from the right is defined similarly):
Definition 5.5 Let  :  ⊂ R → R and 0 ∈ . We say that  is continuous from the left at  = 0 if for any
  0 there exists  ()  0 such that
| () −  (0 )|  
for all  ∈  with 0 −  ()   ≤ 0 
or equivalent
lim  () =  (0 )
%0
(lim%0  () exists and equals  (0 )).
Remark 5.6 From the definitions above it is easy to see that  is continuous at  = 0 if and only if it is continuous
both from the left and from the right at  = 0 , that is:
lim  () = lim  () =  (0 ) 
%0
&0
(19)
Example 5.7 It is easy to see that the function  : R → R defined by
½
−1
≤0
 () =
1
0
is continuous from the left at 0 = 0, but it is not continuous from the right at 0 = 0 (lim%0  () =  (0) = −1 6=
1 = lim&0  ()).
The following shows that the continuity is preserved under the usual operations with functions:
Theorem 5.8 Let   :  ⊂ R → R be continuous at  = 0 ∈ .
a) The functions  ± ,  ·  and   (provided  () 6= 0  ∈ ) are also continuous at  = 0 ;
b) If  () ⊂  and  :  ⊂ R → R is continuous at  (0 ), then the composition  ◦  :  ⊂ R → R is also
continuous at  = 0 .
Proof. The statement follows from the definition with limits of continuity and from the corresponding properties
of limits.
Theorem 5.9 (Weierstrass’s boundedness theorem) Let   ∈ R be real numbers with   , and let  :
[; ] → R be a continuous function. Then  is bounded on [ ] and attains its bounds. That is, there exist points
   ∈ [ ] such that for all  ∈ [ ] we have
 ( ) ≤  () ≤  ( ) 
49
(20)
Proof. Let  = inf ∈[]  () and  = sup∈[]  (), and therefore
 ≤  () ≤ 
 ∈ [ ] 
In order to show that  is bounded we have left to show that  and  are finite, more precisely we have to
show that  6= −∞ and  6= +∞.
If we assume that  = −∞, from the definition of  = inf ∈[]  () it follows that there exists a sequence
( )≥1 ⊂ [ ] such that lim→∞  ( ) =  = −∞.
Since ( )≥1 is a bounded sequence ( ∈ [ ],  ≥ 1), it contains a convergent subsequence ( )≥1 , that is
lim→∞  =  ∈ [ ]. Since the function  is continuous on [ ], it is continuous in particular at  = , so
 () = lim  ( ) = lim  ( ) = −∞
→∞
→∞
which is a contradiction.
Therefore  cannot equal −∞, and similarly it can be shown that  cannot equal +∞, so we have shown that
 is bounded on [ ].
Next, we will show that  must attains its bounds on [ ]. Assume that there is no  ∈ [ ] such that
 ( ) =  = inf ∈[]  (), and therefore  ()   for all  ∈ [ ].
The function  : [ ] → R defined by
 () =
1

 () − 
 ∈ [ ] 
is continuous on [ ] (the function  is continuous on [ ] and  6=  on [ ]) and also  ()  0 for  ∈ [ ]
(since  ()   for  ∈ [ ]).
By the previous part of the proof it follows that  is bounded on [ ], hence  0 = sup∈[]  ()  0. We
1
≤  0 or equivalent (recall that by assumption  ()   for all  ∈ [ ])
have  () =  ()−
 () ≥  +
1
 
0
which shows that
inf  () ≥  +
∈[]
1
 
0
in contradiction with our assumption that inf ∈[]  () = . The contradiction obtained shows that there exists
 ∈ [ ] such that  ( ) = .
Similarly it can be shown that there exists  ∈ [ ] such that  ( ) =  , concluding the proof.
The next result shows that a continuous function defined on a closed and bounded interval assumes every
intermediate value. This means for example that if the function takes the values −1 and 3, then it must take any
value between −1 and 3; it cannot “jump” from −1 to 3 without taking all the intermediate values between them.
The precise statement is the following.
Theorem 5.10 (Intermediate Value Theorem) Let   ∈ R be real numbers with   , and let  : [ ] → R
be a continuous function on [ ]. Then for every  between  () and  () there exists 0 ∈ [ ] such that  (0 ) = .
Proof. ¡Without
loss of generality we may assume that  () ≤  ≤  ().
¢
If  +
the proof.
=
,
then we can choose 0 = +
2 ¢
2 and conclude
¡ + ¢
¡ +
+
If  2  , consider 1 = 2 and 1 = , and if  2  , consider 1 =  and 1 = +
2 .
¡
¡ 1 +1 ¢
¢
1 +1
1
1
Next, if  1 +
=
and
conclude
the
proof.
If

=
,
we
can
choose


,
consider
2 = 1 +
0
2
2
2
2
¡ 1 +1 ¢
1 +1
and 2 = 1 , and if 
 , consider 2 = 1 and 2 = 2 .
2
¡
¢

Inductively, either at some point   +
=  (concluding the proof), or we can construct a sequence [   ]
2
of nested intervals (i.e. [+1  +1 ] ⊂ [   ] for each  ≥ 1), such that  ( ) ≤  () ≤  ( ) and  −  = −
2 ,
for all  ≥ 1.
Note that ( )≥1 is an increasing and bounded sequence, so it is a convergent sequence. Let 0 = lim→∞ 
be its limit.
Since  =  + −
2 by construction, it follows that ( )≥1 is also a convergent sequence, and
lim  = lim  + lim
→∞
→∞
→∞
50
−
= 0 + 0 = 0 .
2
We have shown that lim→∞  = lim→∞  = 0 , and since  is continuous at  = 0 (it is continuous at any
point of [ ]), it follows that
 (0 ) = lim  ( ) = lim  .
→∞
→∞
From the construction of the intervals [   ] we have that  ( ) ≤  ≤  ( ) for any  ≥ 1, and therefore by
passing to the limit with  → ∞ we obtain
 (0 ) = lim  ( ) ≤  ≤ lim  ( ) =  (0 ) 
→∞
→∞
and therefore  (0 ) =  for some 0 ∈ [ ], concluding the proof.
Example 5.11 Consider the function  () = 3 − 3 − 1. Note that  (−1) = 1  0 and  (0) = −1  0. Since 
is continuous on the closed bounded interval [−1 0], by the Intermediate value theorem it follows that there exists
 ∈ [−1 0] such that  () = 0.
In other words, we have shown that the equation 3 − 3 − 1 = 0 has a solution in the interval [−1 0].
Example 5.12 (Bisection method) Consider again the function  () = 3 − 3 − 1.
Proceeding as in the proof of the Intermediate value theorem, you can find the approximation of the solution of
the equation 3 − 3 − 1 = 0, as follows (the method is known as the bisection method).
Since  (−1) = 1  0 and  (0) = −1  0, you know that the equation  () = 0 has a solution in the interval
[−1 0] 
¡
¢
Since  ¡ −1+0
2
¢=  (−05)  0 and  (0)  0, you know that the equation has a solution in the interval [−05 0].
Since  −05+0
=  (−025)  0 and  (−05)  0, you know that the equation has a solution in the interval
2
[−05 −025].
¡
¢
Since  −05−025
=  (−0375)  0 and  (−025)  0, you know that the equation has a solution in the
2
interval [−0375
−025].
¡
¢
Since  −0375−025
=  (−03125)  0 and  (−0375)  0, you know that the equation has a solution in the
2
interval [−0375
−03125].
¡
¢
Since  −0375−03125
=  (−034375)  0 and  (−0375)  0, you know that the equation has a solution in
2
the interval¡[−0375 −034375].
¢
Since  −0375−034375
=  (−0359375)  0 and  (−034375)  0, you know that the equation has a solution
2
in the interval [−0359375 −034375].
So one of the solutions of the equation 3 − 3 − 1 = 0 is approximately 03. Few more iterations will give you
a better approximation of the solution (more decimals).
5.2.1
Uniform continuity
In the Definition 5.1 of continuity at a point  = 0 , the number  =  () depends in general on the choice of the
point 0 . If however  does not depend on the choice of 0 , the continuity of  is called “uniform”, in the sense
that  is the “same” (or uniform) for all points 0 . The formal definition is as follows:
Definition 5.13 We say that the function  :  → R is uniformly continuous on  if for any   0 there exists
  0 such that
| () −  ()|  
for any   ∈  with | − |  .
We have the following:
Proposition 5.14 If  :  → R is uniformly continuous on  then it is continuous on  (i.e. it is continuous at
any point  = 0 ∈ ).
Proof. Follows immediately from the definition of uniform continuity by considering  = 0 ∈ .
The converse of the above theorem is not generally true, as can be checked by considering the function  :
(0 1) → R with  () = 1 which is continuous on (0 1) but it is not uniformly continuous on (0 1).
However, under the additional hypothesis that the domain  of  is a closed interval, the converse of the above
proposition is true:
Theorem 5.15 If  : [ ] → R is continuous on the closed interval [ ], then it is uniformly continuous on [ ].
51
Proof. Assuming  is not uniformly continuous on [ ], there exists   0 such that for any    0 there exist
points   ∈ [ ] with | − |    such that | () −  ()| ≥ .
In particular, choosing   = 1 for  = 1 2   , we can construct two sequences ( )≥1 and ( )≥1 of points
in [ ], such that
1
 = 1 2   
| −  | 
and
| ( ) −  ( )| ≥ 

Since ( )≥1 ⊂ [ ] is a bounded sequence, it contains a convergent subsequence, say lim→∞  =  ∈ [ ].
Since | −  |  1 for all  ≥ 1, it follows that ( )≥1 is also a convergent subsequence, and lim→∞  = .
Passing to the limit in the inequality | ( ) −  ( )| ≥  with  → ∞, and using the continuity of  we
obtain
¯
¯
¯
¯
¯
 ≤ ¯ lim  ( ) − lim  ( )¯¯ = | () −  ()| = 0
→∞
→∞
contradicting   0.
The contradiction obtained shows that  must be uniformly continuous on [ ], concluding the proof.
The following property is useful for showing a certain function is uniformly continuous:
Proposition 5.16 If  :  ⊂ R → R and there exists a constant  ∈ R such that
| () −  ()| ≤  | − | 
  ∈ 
then  is uniformly continuous on .
Proof. Given   0, we can choose   =


such that for any   ∈  with | − |    we have
| () −  ()| ≤  | − |  

= 

and therefore  is uniformly continuous on .
Example 5.17 Using the above proposition we see that the function  : [0 3] → R given by  () = 3 is uniformly
continuous on [0 3], since there exists  = 27 such that
¯
¯ ¯
¡
¢¯
| () −  ()| = ¯3 −  3 ¯ = ¯( − ) 2 +  +  2 ¯
³
´
2
2
≤
|| + || + || | − |
¢
¡
≤ 32 + 3 · 3 + 32 | − |
= 27 | − | 
for any   ∈ [0 3].
Also note that since  is uniformly continuous on [0 3], it is also continuous on [0 3] by Theorem 5.15.
5.3
Continuity of a real-valued function of vector variable
Consider now a function  =  (1       ) of  variables (or a function of a vector variable). We define the
continuity of the function  at a point 0 similar to the case  = 1 (Definition 5.1), as follows.

¡Definition
¢ 5.18 Let  :  ⊂ R → R and 0 ∈  be a point of . We say that  is continuous at 0 =
0
0
1       , if it satisfies for any   0 there exists  ()  0 such that
| () −  ()0 |  
for all  = (1       ) ∈  with k − 0 k   () 
or equivalent
lim  () =  (0 )
→0
(lim→0  () exists and equals  (0 )).
52
Example 5.19 Consider the function  : R2 − {(0 0)} → R defined by
(
1 22
if (1  2 ) 6= (0 0)
2 +2 

 (1  2 ) =
1
2
0
if (1  2 ) = (0 0)
¡
¢
The function  is continuous at the point 0 = 01  02 = (0 0) by the definition with  of continuity, because
given   0 there exists  () = 2  0, such that
¯
¯
¯
¯
¯1 22 ¯
¯ 1 22
¯
| (1  2 ) −  (0 0)| = ¯¯ 2
− 0¯¯ ≤ p 2 2
1 + 22
2 |1 2 |
1
=
|2 |
2
1
≤
||||
2
1

2 = 
2
p
for any  = (1  2 ) ∈ R2 − {(0 0)} with 0  || − 0|| = 21 + 22   () = 2.
Since for (1  2 ) = (0 0) we have trivially | (0 0) −  (0 0)| = 0  , it follows that for any   0 there exists
 () = 2 such that | () −  (0 0)|   for any  = (1  2 ) ∈ R with || − 0||   (), which proves the claim.
Similar to the case when  = 1, the properties of sum/difference, product/quotient and composition of continuous
functions (Theorem 5.8) also hold in this case.
5.4
Continuity of a vector-valued function of vector variable
¡
¢
Consider the function  = (1       ) :  ⊂ R → R and let 0 = 01      0 ∈ .
¡
¢
Definition 5.20 We say that  is continuous at 0 = 01      0 ∈ , if for each  = 1      the function
 :  ⊂ R → R is continuous at 0 ∈ .
We say that the function  is continuous on the set  if  is continuous at any point 0 ∈ .
We say that the function  is uniformly continuous on  if for any   0 there exists   0 such that
|| () −  ()||  
for any   ∈  with || − ||  
(or equivalently, that for each  = 1     , the function  :  ⊂ R → R is uniformly continuous on ).
From the above definition, we see that the study of the continuity of a function  = (1       ) :  ⊂ R → R
reduces to the study of the continuity for each of the  coordinate functions 1       :  ⊂ R → R.
Example 5.21 The function  : R2 → R3 ,  (1  2 ) = (1 + 1  2 + 2  3 + 1 2 ) is continuous at the point 0 =
(0 0) ∈ R2 , since each of the functions 1 (1  2 ) = 1 + 1 , 2 (1  2 ) = 2 + 2 and
¡ 3 (
¢ 1  2 ) = 3 + 1 2 are
continuous at this point. Replacing the point 0 = (0 0) by an arbitrary point 0 = 01  02 ∈ R2 , we see that the
function  is continuous at any point 0 ∈ R2 (since 1 , 2 and 3 are continuous at any point 0 ∈ R2 ).
6
Exercises
1. Study the continuity of the following functions:
½ 2
 
 ∈ [0 1]
(a)  : [0 2] → R,  () =
;
 + 1  ∈ (1 2]
½
sin 1   6= 0
(b)  : R → R,  () =
;
0
=0
½
 sin 1   6= 0
(c)  : R → R,  () =
;
0
=0
53
2. Consider the function  () =
(0 1), (0 1], [1 ∞), (1 ∞)?
√

−1 .
3. Determine if the function
On which of the following intervals is the function  continuous [0 1),
( ³
1+
 () =
0
||

´
+  sin 1   6= 0
=0
is continuous from the left / right at  = 0. Is  continuous at  = 0?
4. Use the properties of continuous functions (Theorem 5.8) to determine the intervals on which the following
functions are continuous:
√
√
1
√ 1
(a)  () = 9 − 2 ,  () = 3 9 − 2 ,  () = √9−
2 ,  () = 3 9−2 ;
2
1
(b)  () = − sin + ,  () = ln (1 + sin ),  () = sin −1
,  () =
1
cos  ;
5. Determine the value of the constant  such that the following functions are continuous:
½ √
2 − 2 + 2   ∈ [1 2]
;
(a)  : [1 3] → R,  () =
 + 3
 = (2 3]
½ sin((−1))
 1
−1
(b)  : R → R,  () =
;
2 +  − 1  ≥ 1
½ 1−cos 
 2   6= 0
;
(c)  : R → R,  () =
2
=0
2 
½
 − 
≤0
1
(d)  : R → R,  () =
;
(1 + )     0
½ 3+2 −27
9 −9   6= 1 ;
(e)  : R → R,  () =

=1
½ sin(−2)
2 −5+6    2 ;
(f)  : R → R,  () =
≥2
 
6. Find the constants   ∈ R so that the given functions are continuous on R.
⎧
≤0
⎨ −2 + 
 sin 1 + 1 0   ≤ 1 ;
(a)  : R → R,  () =
⎩ 3
1
 + 2
 
⎧
⎨ −3 + 1   0
 + 
0≤≤1 ;
(b)  : R → R,  () =
⎩ √
 + 2   1
⎧
⎨ −3 +    0
 + 
0≤≤1 
(c)  : R → R,  () =
⎩ √
 + 4   1
7. Study the uniform continuity of the following functions:
(a)  : [−2 2] → R,  () = 2 ;
(b)  : R → R,  () = 2 ;
(c)  : (0 1] → R,  () = cos 1 ;
Hint: a) use the fact that  is continuous on a closed bounded interval; b) show that for any  ≥ 1 there are
points    ∈ R with | −  | = 1 for which | ( ) −  ( )| ≥ 1; b) show that for any  ≥ 1 there are
points    ∈ (0 1] with | −  | = 1 for which | ( ) −  ( )| = 2;
8. Determine if the following functions are continuous at (0 0)
54
2
(a)  : R → R,  ( ) =
2
(b)  : R → R,  ( ) =
(c)  : R2 → R,  ( ) =
2
(d)  : R → R,  ( ) =
2
(e)  : R → R,  ( ) =
(f)  : R2 → R,  ( ) =
(
(
½
(
(
(
3 −3
− 
0
2  2
2 + 2 
0

2 +2 
0
 6= 
=
( ) 6= (0 0)
( ) = (0 0)
( ) 6= (0 0)
( ) = (0 0)
sin(4 + 4 )
2 + 2 
0
( ) 6= (0 0)
( ) = (0 0)
1−cos(2 + 2 )

2 +2
0
√
1−cos 2 +2
tan(2 + 2 ) 
1
2
( ) 6= (0 0)
( ) = (0 0)
( ) 6= (0 0)
( ) = (0 0)
9. Determine if the given functions are continuous at the indicated point:
(a)  : R3 → R2 ,  (  ) = ( + sin   + 2 + ), at (0 1 2) ;
³
´
p
(b)  : R3 → R3 ,  (  ) =  +  2 + 1  + sin  2  +  , at (0 0 0) ;
10. Consider the function  () = 3 + 52 − 4 − 1. Show that the equation  () = 0 has at least one solution
in the interval [0 1].
11. Show that the equation 3 + 2 +  +  = 0 with   0 has at least one solution. Does the conclusion hold
if   0? Does the conclusion hold if  = 0?
12. Show that the equation  = 2 −  has at least one solution.
13. Consider the function  () = 2 − 2 sin  − 1.
(a) Evaluate  (0).
(b) Find a real number   0 such that  ()  0. Use the Intermediate value theorem to show that there
exists a real number   0 such that  () = 0.
(c) Find a real number   0 such that  ()  0. Use the Intermediate value theorem to show that that
there exists a real number   0 such that  () = 0.
14. Suppose that the function  is continuous on the interval [0 1], and that 0 ≤  () ≤ 1 for every  ∈ [0 1].
Show that there exists  ∈ [0 1] such that  () = .
55