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Global Minimum and Global Maximum
Motivation:
In the previous section, the method about how to find local maximum or local minimum has been introduced. Naturally, we would like to know how to find the “global maximum” and “global minimum” . Some functions sometimes would go to infinitely large ( either positive or negative ) place. We just start our study on the behavior of the function in an interval, and then we think something else.
Usually, we use the notation [a,b] ( a, b R ) to stands for all the set A such that A={ x | a x b }
In other words, A is the set containing all the numbers between a and b that a and b are also included in the set .
And the notation (a,b) ( do not confuse with vector notation or coordinate notation ) stands for the set B such that
B = { x | a < x < b }
So, B is the set containing all the numbers between a and b , but a and b are not included in this set. Sometimes, you might hear that [a,b] is called as “closed interval” and (a,b) is called as “open interval” . Because of that, sometimes you might see a notation like -- that stands for the interval that a is not included, but b is included. Similarly, stands for the interval that a is included but b is not included. Why do we have to emphasize on this? Let’s look at the following question:
Question: f(x)=x and find the maximum and minimum in the interval (0,1) .
You might say: this question is easy; f(0)=0 and f(1)=1 ... But please notice that x=0 and x=1 are not included in our question for minimum and maximum. Our question is on “the interval (0,1)” . In this question, you know that f(0) is always larger than 0 and smaller than 1. But you just can not determine the maximum value and minimum value. Actually, maximum and minimum do not exist in this question. ( But supremum (sup) and infimum ( inf ) exist – that will be discussed in advanced math courses ) . So, the type of interval might affect the answer. In the course “Topology” or other advanced courses, the concept for “open set” or “closed set” will be carefully introduced. Here, we only introduce very basic concept.
Definition ( Open Set and Closed Set in R ) A set O in R is an open set if the following statement holds:
for any x O, there exist a set B such that B={ t : |x-t| < r for some r > 0 } and B O .
And a set E is an closed set if its complement is an open set.
Basically, the definition says: for any point in this set in R, if you can find an interval ( without including the end points ) to contain this point, and this interval is still in this set, then this set is an open set. For in the case of Rn , it turns out that we use an “disk” or “ball” to contain a point in a set to check if it is open. At this moment, we only focus on the set of R ( one-dimensional case ).
Definition ( Inverse Image ) Let f by a function such that f: A B . Let S be a subset of B, i.e. S B . Then the inverse image of S under the effect of f is denoted as f-1(S) such that
f-1(S) A and for every element a f-1(S), it has f(a) B .
Please note that f-1(S) is just a notation to indicate the set { x | f(x) B } . It does not imply the inverse function of f exists. That’s why it is called inverse image.
Theorem: Let f be a continuous function such that f: A B and A R, B R . Then for any open set U B, the inverse image f-1(U) is still an open subset of A.
Proof: U is an open set . Thus, for any t U, there exists such that
V U, where V={ y: |y-t| < }
And t U B, thus there exists at least one element c A such that
f(c)= t
Since f is continuous function, for this given , there exists a corresponding such that
| f(x) –f (c) | < when | x-c | <
Hence, for any c satisfying f(c)=t, we can find a set W={x: | x-c | < } to contain the point c. So, f-1(U) is an open set in A.
In more advanced math, the result here is just directly used as the definition of the continuity of a function.
Theorem: Let f be a continuous real function. If A R is a closed set, then f(A) is also a closed set.
Proof: Let’s assume B=f(A) is an open set. Since f is a continuous function, then the inverse mapping f-1(B) is an open set. But we know it is a closed set ( contradiction ).
Thus, f maps a closed set to a closed set.
Theorem: Let f be a continuous real function defined on a closed interval [c,d] . Then the maximum and minimum of f exist.
Proof: f is a continuous function so that f maps a closed interval to a closed interval. For a closed interval, you can find maximum and minimum from that.
Global Maximum and Minimum in an Interval Since the theorem above confirms the existence of maximum and minimum on a closed interval [c,d] , we will start with differentiable functions on a closed interval [c,d]. For a differentiable function f , the local maxima and minima can be found out via derivative functions. To find the global maximum and minimum, just compare those local maxima and minima with the values of f at end points : f(c) and f(d) .
The reason to do that is : for all the local maxima and minima, we know the behavior of the function on their neighborhoods. But for the end points, the function might be monotonically increasing or decreasing around the neighborhoods of the two end points.
To find the global maximum and minimum on an open interval (c,d), we find out those local maxima and minima at first. Then check the behavior of the function by approaching the points c and d .
Example: f(x)=x+ . Find the maximum and minimum of f(x) on the interval (0,2] .
Sol: f’(x) = 1 + f”(x)=
Thus, f(x) has local minimum at x=1 , f(1)=0 .
f(x) as x 0. And f(2)= 2.5
So, f(x) has minimum at x=1 with value f(1)=0 . There is no maximum .
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