Every function on [0,1] attains its maximum

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Homework Help Overview

The discussion revolves around the statement "Every function on [0,1] attains its maximum." Participants are exploring the implications of this claim within the context of real analysis, particularly focusing on the properties of functions defined on the closed interval [0,1].

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are questioning the meaning of "attains its maximum" and whether it implies the existence of a maximum value within the function's image. There is discussion about the definitions of maximum and supremum, as well as the implications of continuity and boundedness of functions on the interval.

Discussion Status

The discussion is active, with various interpretations being explored. Some participants suggest that the statement may be false unless additional conditions, such as continuity, are specified. Others are clarifying definitions and providing examples of functions that do not meet the criteria.

Contextual Notes

There is an underlying assumption that the functions in question may not necessarily be continuous or bounded, which is central to the debate. The phrasing of the original question is also noted as potentially misleading.

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Homework Statement


True or False: Every function on [0,1] attains its maximum

I was working on a practice midterm, and this question stumped me

The Attempt at a Solution


I know that by "every function on [0,1]" means that dom f = [0,1] but what exactly does it mean by "attains its maximum" simply that it has a limit, as in, it doesn't approach +∞?

If this is what it means, it would be false, correct? Because you could have a function that approaches ±∞ as it approaches 1 from below.

For instance, take f(x)=[itex]\frac{1}{1-x}[/itex]
 
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PirateFan308 said:
but what exactly does it mean by "attains its maximum"
The question presumes the function has a maximum value, and asks if that maximum value is in the function's image.

It's a bit oddly phrased, since a maximum is, by definition, in the function's image. It's either a 'trick' question, or they meant "has a maximum" or "attains its supremum". Or, you learned a slightly non-standard definition of those terms in class.

For instance, take f(x)=[itex]\frac{1}{1-x}[/itex]
That's not a function on [0,1].
 
The main point is that [0, 1] is a closed interval.

It's certainly not true that "Every function on (0, 1) attains its maximum."
 
I think the question really means every bounded function on [0,1] attains its least upper bound. But I may be wrong.
 
Funny nobody has mentioned the word 'continuous' yet. All continuous functions on [0,1] have an upper bound and attain it. The given problem didn't say 'continuous'. So it's false.
 
I would take that to mean there is a [itex]c[/itex] in [itex][0, 1][/itex] such that [itex]\forall x[/itex] in [itex][0, 1], f(x)\leq f(c)[/itex].

Certainly this is true for all continuous functions (Use the fact that if K is compact, f(K) is also compact for continuous f, and [0, 1] is compact. It can be shown that compact sets have a maximum and minimum value).

For discontinuous functions though, take
[itex]\begin{align*}f(x)&=\frac{1}{x} \text{ when }x≠0\\ &=0 \text{ when } x=0\end{align*}[/itex]

Then [itex]f(x)[/itex] does not attain it's maximum on [itex][0, 1][/itex]
 
Thanks guys! This helped a lot!
 

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