Proof of the Extreme Value Theorem

In summary: The part that might need a bit more explanation is why f(c) is in f([a,b]). But you sort of gave that explanation by saying that f(d) <= f(x) <= f(c) for all x in [a,b].In summary, the Extreme Value Theorem states that if a function f(x) is continuous on a closed and bounded interval [a,b], then it attains its maximum and minimum values at some points c and d within the interval. This is proven by showing that f([a,b]) is also bounded, closed, and that f(c) is in f([a,b]).
  • #1
jgens
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Homework Statement



Essentially, prove the Extreme Value Theorem.

Homework Equations



n/a

The Attempt at a Solution



Proof: Let a function f(x) be continuous on the closed interval [a,b]. Moreover, define a set A such that A={x ϵ [a,b]}. Since f(x) satisfies the condition for the Boundedness Theorem, sup{f(A)}= M. If f(x) does not attain its supremum in [a,b] then it must asymptotically converge to some valve infinitesimally near M in the interval [a,b]. We may define this value such that f(c_n ) > M-1/n, where as n→∞, f(c_n )→f(c). We are then left with the resultant inequality: M-1/n < f(c_n ) <M, and by the Squeeze Theorem, lim(x→c)f(x) = M. Therefore, under the initial premise of continuity, f(x) attains its supremum M at some value x=c within the closed interval [a,b].

Given that f(x) satisfies the Boundedness Theorem, we may also claim inf{f(A)} = m. If f(x) does not attain its infimum in [a,b] then it must asymptotically converge to some value infinitesimally near m in the interval [a,b]. We may define this value such that f(d_n) < m+1/n where as n→∞, f(d_n )→f(d). We are then left with the resultant inequality: m+1/n > f(d_n ) > m, and by the Squeeze Theorem, lim(x→d)f(x) = m. Therefore, under the initial premise of continuity, f(x) attains its infimum m at some value x=d within the closed interval [a,b].

Therefore, f(d)≤f(x)≤f(c) for all x ϵ [a,b]. Q.E.D.

Any suggestions would be appreciated since I'm fairly certain the above proof is incorrect - I'm not certain how I would fix any errors though.

Thanks.
 
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  • #2
Your proof looks pretty good to me. Essentially, because f is continuous on the closed and bounded interval [a,b], f([a,b]) is also bounded and so has sup and inf (the "boundedness" theorem). f([a,b]) is also closed, which is what you show with your limit argument.
 

Related to Proof of the Extreme Value Theorem

What is the Extreme Value Theorem?

The Extreme Value Theorem states that if a function is continuous on a closed interval, then it must have a maximum and minimum value on that interval.

What is the importance of the Extreme Value Theorem?

The Extreme Value Theorem is important because it allows us to determine the maximum and minimum values of a function on a given interval, which is useful in many real-world applications. It also helps us prove the existence of solutions to optimization problems.

How is the Extreme Value Theorem used in calculus?

In calculus, the Extreme Value Theorem is used to find the absolute maximum and minimum values of a continuous function on a closed interval. This is done by finding the critical points of the function and evaluating them at the endpoints of the interval.

What are the conditions for the Extreme Value Theorem to hold?

The Extreme Value Theorem will hold if the function is continuous on a closed interval and the interval is finite. If the interval is infinite, then the function must have a horizontal asymptote on both ends.

Can the Extreme Value Theorem be applied to all functions?

No, the Extreme Value Theorem only applies to continuous functions. If a function is not continuous on a given interval, then it may not have a maximum or minimum value on that interval.

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