If f has no absolute max on [0,1], then f is discontinuous. (true or false?)
I think this is false, y=5 has no absolute maximum on the interval [0,1] and it is still continuous.
Is there a theorem I can refer to that explains this in more detail?
If f has no absolute max on [0,1], then f is discontinuous. (true or false?)
I think this is false, y=5 has no absolute maximum on the interval [0,1] and it is still continuous.
Is there a theorem I can refer to that explains this in more detail?
Last edited by stapel; 11-14-2017 at 07:27 PM. Reason: Copying exercise from subject line into post.
But it does have an absolute maximum: 5!
Perhaps you are thinking that there is no one value of x for which the function attains that maximum; but that is not what they are talking about. Remember that there is a difference between a maximum itself (its value, y) and a maximum point (an ordered pair) and the location of the maximum (the value or values of x).
What theorems have you learned that might be relevant to the existence of an absolute maximum for a continuous function?
When you're trying to disprove something, you want to show it's not true in every case. To do that, you need find one counterexample where it doesn't work. You've done just that, so you're good. Problem solved. It's false.
For an more in-depth exploration of the ideas at play here, you'd be best served by returning to the definitions of the terms involved. What is the definition of a global maximum? What is the definition of continuity? Specifically, what criteria does a function have to meet to be continuous? Do these criteria refer in any way to a global maximum? If you're finding your class notes and/or textbook to be insufficient on this topic, you might also try Googling for more information. By searching for "definition of continuity," I found this page from SOS Math, and by searching "definition of global maximum," I found this page from Math is Fun.
does that mean that "if f has no absolute max on [0,1] then f is discontinuous" is true?
I know that the extreme value theorem states that if f is continuous on a closed interval [a,b] then f attains both an absolute max and an absolute min.
if there is no absolute max, does that mean the function must therefore be discontinuous?
Yes, the given statement is true, and the EVT is the reason.
If your example of y=5 did not have a maximum, then it would be a counterexample of the theorem you cited, since it would be a continuous function that did not have an absolute max on that interval.
But the statement you are asked about is the contrapositive of the extreme value theorem applied to [0,1]:
EVT: If f is continuous, then f attains an absolute max and min.
contrapositive: If f does not attain an absolute max and min, then f is not continuous.
Not necessarily. As I mentioned earlier, you need only one counterexample to prove something false, but proving something true is much more difficult. No matter how many examples you can find that satisfy the criteria, that's never enough on its own. Although, I do need to give a small correction to my previous post, because I learned something new just today. The book I learned from defined a global maximum as:
This allows for one and only one global maximum, such that the constant function y = 5 would not have a global maximum. But I found out that this definition is apparently not the typical definition. Most books instead allow for multiple global maxima by using the criteria [tex]f(x) \le f(c)[/tex]. Let this serve as a lesson at my expense that terminology and definitions matter a great deal.f(x) has an global maximum at the point x= c, if for every x in the domain of f(x), [tex]f(x) < f(c)[/tex]
Again, this really depends on exactly how we define the terms involved. Take, for example, the function x^{2}. It is continuous on the closed interval [2, 5], so this formulation of the extreme value theorem says it ought to have a global maximum somewhere in this interval. But merely the fact that the function is continuous on this interval doesn't guarantee the existence of said maximum. The function "blows up" to infinity at either extreme, so it clearly has no global maxima. However, if we restrict the domain to be [2, 5], then it does have a global maximum.
So really, it all comes down to how do you read and interpret the problem. Whether or not the statement is true depends on whether we're looking at the function as a whole or specifically restricting the domain to [0, 1]. And only the person who wrote the problem can tell you that for sure. Perhaps your instructor has some guidance on how they'd interpret it?
Let's try to summarize. You need only one counter-example to disprove something, but examples out of an infinite number of possibilities do not constitute a proof. I can show an infinite number of even integers, but that does not prove that every integer is even.
My problem here is that though I know the meanings of "local extremum" and "global extremum," I am not sure what definition your text has assigned to "absolute extremum."
When you are stuck, going back to definitions is often helpful, particularly in proofs.
The others who are answering you seem to be forgetting the statement of the problem, "if f has no absolute max on [0,1] then f is discontinuous". They are missing the fact that it is asking specifically about the absolute maximum on the closed interval [0,1], not the global maximum for the function on its entire domain. The Extreme Value Theorem is very clear, and when stated properly includes all the definitions that are needed (https://en.wikipedia.org/wiki/Extreme_value_theorem):
"Absolute maximum" here refers to a value greater than or equal to any other value of the function in this interval - not a global maximum, and not unique.
So you are exactly right: this theorem implies that if the function does not attain a maximum in the interval, then it is not continuous on that interval.
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