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That is, there exist numbers ''c'', and ''d'' within the interval ''b'' such that for ''every'' value of ''x'' in ''b'' ,

:f(c) \le f(x) \le f(d).

A weaker version of this theorem is the boundedness theorem which states that a function ''f''(''x'') continuous in the closed interval is Bounded on the interval. That is, there exist numbers ''l'' and ''L'' such that for ''every'' value of ''x'' in [''a'', ''b'' ,

:l \le f(x) \le L.

The extreme value theorem enriches the boundedness theorem by saying that not only is the function bounded, but it also attains its least upper bound as its maximum and its greatest lower bound as its minimum.

The extreme value theorem is used to prove Rolle's Theorem .


PROVING THE THEOREMS


We look at the proof for the maximum, the minimum is very similar. Also note that everything in the proof is done within the context of the Real Numbers .

We first prove the boundedness theorem, which is a step in the proof of the extrem value theorem. The basic steps involved in the proof of the extrem value theorem are:

1. Prove the boundedness theorem.

2. Find a sequence so that its Image converges to the Supremum of ''f''.

3. Show that there exists a Subsequence that converges to a point in the Domain .

4. Use continuity to show that the image of the sequence converges to the supremum.


Proof of the boundedness theorem


Suppose ''f'' is not bounded. Then, by the Archimedean Property of the real numbers, for every ''m'', there exists an ''x'' in ''b'' such that ''f''(''x'') > ''m''. In particular, for every ''k'' in N, there exists an x_k such that ''f''(x_k) > ''k''. This defines a Sequence of x_ks. Because ''b'' is bounded, by the Bolzano-Weierstrass Theorem , there exists a convergent subsequence {x_{n_k}} of {x_k}. As ''b'' is closed, {x_{n_k}} converges to some ''x'' in ''b'' . Because ''f''(''x'') is continuous over ''b'' , we know that ''f''(x_{n_k}) converges to ''f''(''x''). But, ''f''(x_{n_k}) > n_k > k for every k, which implies ''f''(x_{n_k}) diverges to Infinity . Contradiction. Therefore, ''f''(''x'') is bounded above.


Proof of the extreme value theorem


We will now show that ''f''(''x'') has a maximum in ''b'' . As, by the boundedness theorem, ''f'' is bounded from above, there exists ''c'' the least upper bound (supremum) of ''f''(''x''). It is necessary to find a x_0 in ''b'' such that c=f({x_0}). Let ''n'' be a natural number. As ''c'' is the ''least'' upper bound, c-1/n is not an upper bound for ''f''(''x''). Therefore, there exists x_n in ''b'' so that c-1/n < f(x_n). This defines a sequence {x_n}. Since ''c'' is an upper bound for ''f''(''x''), c-1/n < f(x_n) ≤ ''c'' for all ''n''. Therefore, {''f''(x_n)} converges to ''c''.

The Bolzano-Weierstrass Theorem tells us that {x_{n_k}} exists in {x_n} such that {x_{n_k}} converges to some x_0 and, as ''b'' is closed, x_0 is in ''b'' . Since ''f''(''x'') is continuous over ''b'' , {''f''(x_{n_k})} converges to ''f''(x_0). But, {''f''(x_{n_k})} is a subsequence of {''f''(x_n)} that converges to ''c'', so c = f(x_0). Then x_0 is a maximum of ''f''(''x'').


EXAMPLES


The following examples show why the function domain needs to be closed and bounded.

''Bounded''. ''f''(''x'') = ''x'' defined over [0,\infty) is not bounded from above.

''Closed''. ''f''(''x'') = ''x'' defined over [0,1) never attains its least upper bound 1.


TOPOLOGICAL FORMULATION


In General Topology , the extreme value theorem is follows from the general fact that Compactness is preserved under Continuity , and the fact that a subset of the real line is compact if and only if it is both closed and bounded.


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