Suppose is a continuous function (where I is an interval). Now, for every
and
, we have a
such that
for
. This tells us that f is bounded in the interval
. Does it mean that the function is bounded in its entire domain? What we have shown is that given an
, there is an interval
and two real numbers
and
such that
for all
.
Surely . But, if we could choose finitely many intervals out of the collection
, say,
such that
, then we would get
, where
and
. That, we can indeed make such a choice is a property of a closed bounded interval I in
and is given by the following theorem, the proof of which, is given below:
Theorem (Heine-Borel):
Let and let I be a family of open intervals covering
, that is, for all
, there exists
such that
. Then, we can find finitely many open intervals
such that
.
Proof:
Suppose our contention is false: Let us take the intervals and
where
. If the hypothesis is false, then it should be false for at least one of the intervals
or
. Otherwise, we could find
and
such that
and
and then
would be the finite family of intervals for which
.
So let us assume that at least for one of the intervals or
the assumption of the theorem is false. Call it
. Again let
. Now since the claim of the theorem is false for
it should be false for at least
or
by the above argument. Call it
. We have
. We can continue this process to get a sequence of intervals
for which the assertion is false. Observe further that
and that we have
.
This gives us a monotonically increasing sequence which is bounded above and a monotonically decreasing sequence
bounded below. So
and
must converge to say
and
respectively. Then,
because
. Since
covers
,
must belong to J for some
. Also, since
, there exists an
such that
for all
. Now let
. Therefore, we conclude that
for all
. But, this violates our hypothesis that we cannot choose finitely many members of
whose union will contain
for any n. QED.
Corollary.
A continuous function on a closed interval is bounded.
The proof of the corollary is already given just before the Heine-Borel theorem. So, if we have a continuous function and
and
, the above corollary says
. Next, we ask the natural question: do there exist two points
such that
and
In other words, does a continous function on a closed interval attain its bounds? The answer is yes.
Theorem:
Suppose is continuous, and
and
. Then, there are two points
such that
and
.
Note: these points and
need not be unique.
Proof by contradiction:
Suppose there is no point such that
, then we would have
or
for all
. Let us define
by
Since vanishes nowhere, y is also a continuous function. So, by the corollary above it ought to be bounded above, and below. Let
, for all
. On the other hand, by the property of a supremum we note that there exists an
such that
, which implies that
or
, which is contradiction. Therefore,
must attain the value M at some point
. The proof of the other part is very similar. QED.
The above theorem together with the corollary says that on a closed interval, a continuous function is bounded and attains its bounds. This, again by the intermediate value theorem, means that the function must attain all the values between its supremum and infimum. Thus, the image of a closed interval under a continuous map is a closed interval. However, if f is a continuous map on an open interval, then the function need not be bounded.
Example.
Let be defined by
. This is surely continuous but the limit,
, which means that given any
, we can always find x such that
, viz., choose
.
If f is a continuous function, then given , for each
fixed, we can find
such that
whenever
Here depends upon
.
Can we choose such that it works for all
? The answer in general is no.
Example.
Let be defined by
. If we fix any
, then for
,
, and hence as x becomes large, the difference between
and
also becomes large for every fixed
. So for say
, we cannot choose
such that
such that
for all
and all x. We thus have the following definition:
Definition:
Let be a continuous function where
or
or (a,b). Then, f is said to be uniformly continuous if for all
, there exists a
such that
for all
with
We have seen above that every continuous function need not uniformly continuous. When , however, every continuous function is uniformly continuous as the next result shows.
Theorem:
Let be continuous. Then, f is uniformly continuous.
Proof.
Fix . The continuity of f implies that for every
, we can choose
such that
whenever
and
Now, let
Then, clearly covers
as
. By the Heine Borel theorem, we can get finitely many intervals out of this family,
,
, …,
such that
.
Let
Then, (note that minimum of finitely many positive numbers is always positive). Next we claim that if
,
then
Since , we can find
such that
, that is,
. Now,
.
Hence, and
and therefore,
. QED.
More later,
Nalin Pithwa