Editing Extreme value theorem (section)

==Proving the theorems==

We look at the proof for the [[upper bound]] and the maximum of <math>f</math>. By applying these results to the function <math>-f</math>, the existence of the lower bound and the result for the minimum of <math>f</math> follows. 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 extreme value theorem. The basic steps involved in the proof of the extreme value theorem are:

# Prove the boundedness theorem.
# Find a sequence so that its [[Image (mathematics)|image]] converges to the [[supremum]] of <math>f</math>.
# Show that there exists a [[subsequence]] that converges to a point in the [[domain of a function|domain]].
# Use continuity to show that the image of the subsequence converges to the supremum.

===Proof of the boundedness theorem===
{{Math theorem
|name=Boundedness Theorem
|If <math>f(x)</math> is continuous on <math>[a,b],</math> then it is bounded on <math>[a,b].</math>}}
{{Math proof|
Suppose the function <math>f</math> is not bounded above on the interval <math>[a,b]</math>. Pick a [[sequence]] <math>(x_n)_{n \in \mathbb{N}}</math> such that <math>x_n \in [a,b]</math> and <math>f(x_n)>n</math>. Because <math>[a,b]</math> is bounded, the [[Bolzano–Weierstrass theorem]] implies that there exists a convergent subsequence <math>(x_{n_k})_{k \in \mathbb{N}}</math> of <math>({x_n})</math>. Denote its limit by <math>x</math>. As <math>[a,b]</math> is closed, it contains <math>x</math>.  Because <math>f</math> is continuous at <math>x</math>, we know that <math>f(x_{{n}_{k}})</math> converges to the real number <math>f(x)</math> (as <math>f</math> is [[sequentially continuous]] at <math>x</math>). But <math>f(x_{{n}_{k}}) > n_k \geq k </math> for every <math>k</math>, which implies that <math>f(x_{{n}_{k}})</math> diverges to [[Extended real number line|<math>+ \infty </math>]], a contradiction. Therefore, <math>f</math> is bounded above on <math>[a,b]</math>.&nbsp;[[Q.E.D.|∎]]}}
{{Math proof
|title=Alternative proof
|Consider the set <math>B</math> of points <math>p</math> in <math>[a,b]</math> such that <math>f(x)</math> is bounded on <math>[a,p]</math>.  We note that <math>a</math> is one such point, for <math>f(x)</math> is bounded on <math>[a,a]</math> by the value <math>f(a)</math>.  If <math>e > a</math> is another point, then all points between <math>a</math> and <math>e</math> also belong to <math>B</math>.  In other words <math>B</math> is an interval closed at its left end by <math>a</math>.

Now <math>f</math> is continuous on the right at <math>a</math>, hence there exists <math>\delta>0</math> such that <math>|f(x) - f(a)| < 1</math> for all <math>x</math> in <math>[a,a+\delta]</math>. Thus <math>f</math> is bounded by <math>f(a) - 1</math> and <math>f(a)+1</math> on the interval <math>[a,a+\delta]</math> so that all these points belong to <math>B</math>.

So far, we know that <math>B</math> is an interval of non-zero length, closed at its left end by <math>a</math>.

Next, <math>B</math> is bounded above by <math>b</math>.  Hence the set <math>B</math> has a supremum in <math>[a,b]</math> ; let us call it <math>s</math>. From the non-zero length of <math>B</math> we can deduce that <math>s > a</math>.

Suppose <math>s<b</math>.  Now <math>f</math> is continuous at <math>s</math>, hence there exists <math>\delta>0</math> such that <math>|f(x) - f(s)| < 1</math> for all <math>x</math> in <math>[s-\delta,s+\delta]</math> so that <math>f</math> is bounded on this interval.  But it follows from the supremacy of <math>s</math> that there exists a point belonging to <math>B</math>, <math>e</math> say, which is greater than <math>s-\delta/2</math>.  Thus <math>f</math> is bounded on <math>[a,e]</math> which overlaps <math>[s-\delta,s+\delta]</math> so that <math>f</math> is bounded on <math>[a,s+\delta]</math>.  This however contradicts the supremacy of <math>s</math>.

We must therefore have <math>s=b</math>.  Now <math>f</math> is continuous on the left at <math>s</math>, hence there exists <math>\delta>0</math> such that <math>|f(x) - f(s)| < 1</math> for all <math>x</math> in <math>[s-\delta,s]</math> so that <math>f</math> is bounded on this interval.  But it follows from the supremacy of <math>s</math> that there exists a point belonging to <math>B</math>, <math>e</math> say, which is greater than <math>s-\delta/2</math>.  Thus <math>f</math> is bounded on <math>[a,e]</math> which overlaps <math>[s-\delta,s]</math> so that <math>f</math> is bounded on <math>[a,s]</math>.&nbsp;&nbsp; 
[[Q.E.D.|∎]]}}

===Proofs of the extreme value theorem===

{{Math proof
|title=Proof of the Extreme Value Theorem
|proof=By the boundedness theorem, ''f'' is bounded from above, hence, by the  [[Dedekind-complete]]ness of the real numbers, the least upper bound (supremum) ''M'' of ''f'' exists. It is necessary to find a point ''d'' in [''a'', ''b''] such that ''M'' = ''f''(''d''). Let ''n'' be a natural number. As ''M'' is the ''least'' upper bound, ''M'' – 1/''n'' is not an upper bound for ''f''. Therefore, there exists ''d<sub>n</sub>'' in [''a'', ''b''] so that ''M'' – 1/''n'' < ''f''(''d<sub>n</sub>''). This defines a sequence {''d<sub>n</sub>''}. Since ''M'' is an upper bound for ''f'', we have ''M'' – 1/''n'' < ''f''(''d<sub>n</sub>'') ≤ ''M'' for all ''n''.  Therefore, the sequence {''f''(''d<sub>n</sub>'')} converges to ''M''.

The [[Bolzano–Weierstrass theorem]] tells us that there exists a subsequence {<math>d_{n_k}</math>}, which converges to some ''d'' and, as [''a'', ''b''] is closed, ''d'' is in [''a'', ''b'']. Since ''f'' is continuous at ''d'', the sequence {''f''(<math>d_{n_k}</math>)} converges to ''f''(''d''). But {''f''(''d<sub>n<sub>k</sub></sub>'')} is a subsequence of {''f''(''d<sub>n</sub>'')} that converges to ''M'', so ''M'' = ''f''(''d''). Therefore, ''f'' attains its supremum ''M'' at ''d''.&nbsp;
[[Q.E.D.|∎]]}}

{{Math proof
|title=Alternative Proof of the Extreme Value Theorem
|proof=The set {{math|1= {''y'' &isin; '''R''' : ''y'' = ''f''(''x'') for some ''x'' &isin; [''a'',''b'']}<nowiki/>}} is a bounded set. Hence, its [[least upper bound]] exists by [[least upper bound property]] of the real numbers. Let {{math|1=''M'' = sup(''f''(''x''))}}&nbsp;on&nbsp;{{closed-closed|''a'', ''b''}}.  If there is no point ''x'' on [''a'',&nbsp;''b''] so that ''f''(''x'')&nbsp;=&nbsp;''M'', then
{{math|''f''(''x'') < ''M''}} on [''a'',&nbsp;''b''].  Therefore, {{math|1/(''M'' &minus; ''f''(''x''))}} is continuous on [''a'', ''b''].

However, to every positive number ''&epsilon;'', there is always some ''x'' in [''a'',&nbsp;''b''] such that {{math|''M'' &minus; ''f''(''x'') < ''&epsilon;''}} because ''M'' is the least upper bound. Hence, {{math|1/(''M'' &minus; ''f''(''x'')) > 1/''&epsilon;''}}, which means that {{math|1/(''M'' &minus; ''f''(''x''))}} is not bounded.  Since every continuous function on [''a'', ''b''] is bounded, this contradicts the conclusion that {{math|1/(''M'' &minus; ''f''(''x''))}} was continuous on [''a'',&nbsp;''b''].  Therefore, there must be a point ''x'' in [''a'',&nbsp;''b''] such that ''f''(''x'')&nbsp;=&nbsp;''M''.
[[Q.E.D.|∎]]}}

===Proof using the hyperreals===
{{Math proof
|name=Proof of Extreme Value Theorem
|proof=In the setting of [[non-standard calculus]], let ''N''&thinsp; be an infinite [[hyperinteger]].  The interval [0,&nbsp;1] has a natural hyperreal extension.  Consider its partition into ''N'' subintervals of equal [[infinitesimal]] length 1/''N'', with partition points ''x<sub>i</sub>''&nbsp;= ''i''&nbsp;/''N'' as ''i'' "runs" from 0 to ''N''.  The function ''&fnof;''&thinsp; is also naturally extended to a function ''&fnof;''* defined on the hyperreals between 0 and 1.  Note that in the standard setting (when ''N''&thinsp; is finite), a point with the maximal value of ''&fnof;'' can always be chosen among the ''N''+1 points ''x<sub>i</sub>'', by induction.  Hence, by the [[transfer principle]], there is a hyperinteger ''i''<sub>0</sub> such that 0&nbsp;≤ ''i''<sub>0</sub>&nbsp;≤ ''N'' and <math>f^*(x_{i_0})\geq f^*(x_i)</math>&thinsp; for all ''i''&nbsp;=&nbsp;0,&nbsp;...,&nbsp;''N''.  Consider the real point
<math display="block">c = \mathbf{st}(x_{i_0})</math>
where '''st''' is the [[standard part function]].  An arbitrary real point ''x'' lies in a suitable sub-interval of the partition, namely <math>x\in [x_i,x_{i+1}]</math>, so that&thinsp; '''st'''(''x<sub>i</sub>'')&nbsp;= ''x''.  Applying '''st''' to the inequality <math>f^*(x_{i_0})\geq f^*(x_i)</math>, we obtain <math>\mathbf{st}(f^*(x_{i_0}))\geq \mathbf{st}(f^*(x_i))</math>.  By continuity of ''&fnof;''&thinsp; we have
:<math>\mathbf{st}(f^*(x_{i_0}))= f(\mathbf{st}(x_{i_0}))=f(c)</math>.
Hence ''&fnof;''(''c'')&nbsp;≥ ''&fnof;''(''x''), for all real ''x'', proving ''c'' to be a maximum of ''&fnof;''.<ref>{{cite book |last=Keisler |first=H. Jerome |title=Elementary Calculus : An Infinitesimal Approach |publisher=Prindle, Weber & Schmidt |location=Boston |year=1986 |isbn=0-87150-911-3 |url=https://www.math.wisc.edu/~keisler/chapter_3e.pdf#page=60 |page=164 }}</ref>
[[Q.E.D.|∎]]
}}

===Proof from first principles===

'''Statement'''  &nbsp;&nbsp;&nbsp;&nbsp; If <math>f(x)</math> is continuous on <math>[a,b]</math> then it attains its supremum on <math>[a,b]</math>

{{Math proof
|name=Proof of Extreme Value Theorem
|proof=By the Boundedness Theorem, <math>f(x)</math> is bounded above on <math>[a,b]</math> and by the completeness property of the real numbers has a supremum in <math>[a, b]</math>.  Let us call it <math>M</math>, or <math>M[a,b]</math>.  It is clear that the restriction of <math>f</math> to the subinterval <math>[a,x]</math> where <math>x\le b</math> has a supremum <math>M[a, x]</math> which is less than or equal to <math>M</math>, and that <math>M[a,x]</math> increases from <math>f(a)</math> to <math>M</math> as <math>x</math> increases from <math>a</math> to <math>b</math>.

If <math>f(a)=M</math> then we are done.  Suppose therefore that <math>f(a)<M</math> and let <math>d=M-f(a)</math>.   Consider the set <math>L</math> of points <math>x</math> in <math>[a,b]</math> such that <math>M[a,x]< M</math>.

Clearly  <math>a\in L</math> ; moreover if <math>e>a</math> is another point in <math>L</math> then all points between <math>a</math> and <math>e</math> also belong to <math>L</math> because <math>M[a,x]</math> is monotonic increasing.  Hence <math>L</math> is a non-empty interval, closed at its left end by <math>a</math>.

Now <math>f</math> is continuous on the right at <math>a</math>, hence there exists <math>\delta>0</math> such that <math>|f(x)-f(a)| < d/2</math> for all <math>x</math> in <math>[a,a+\delta]</math>. Thus <math>f</math> is less than <math>M-d/2</math> on the interval <math>[a,a+\delta]</math> so that all these points belong to <math>L</math>.

Next, <math>L</math> is bounded above by <math>b</math> and has therefore a supremum in <math>[a,b]</math>:  let us call it <math>s</math>.   We see from the above that <math>s > a</math>.  We will show that <math>s</math> is the point we are seeking i.e. the point where <math>f</math> attains its supremum, or in other words <math>f(s)=M</math>.

Suppose the contrary viz. <math>f(s)<M</math>.  Let <math>d=M-f(s)</math> and consider the following two cases:

# <u><math>s<b</math></u>.&nbsp;&nbsp; As <math>f</math> is continuous  at <math>s</math>, there exists <math>\delta>0</math> such that <math>|f(x)-f(s)| < d/2</math> for all <math>x</math> in <math>[s-\delta,s+\delta]</math>.    This means that <math>f</math> is less than <math>M-d/2</math> on the interval <math>[s-\delta,s+\delta]</math>.   But it follows from the supremacy of <math>s</math> that there exists a point, <math>e</math> say, belonging to <math>L</math> which is greater than <math>s-\delta</math>.  By the definition of <math>L</math>, <math>M[a,e]< M</math>.  Let  <math>d_1=M-M[a,e]</math> then for all <math>x</math> in <math>[a,e]</math>, <math>f(x)\le M-d_1</math>.  Taking <math>d_2</math> to be the minimum of <math>d/2</math> and <math>d_1</math>, we have <math>f(x)\le M-d_2</math> for all <math>x</math> in <math>[a,s+\delta]</math>. {{pb}} Hence <math>M[a,s+\delta]<M</math> so that <math>s+\delta \in L</math>.  This however contradicts the supremacy of <math>s</math> and completes the proof.
# <u><math>s=b</math></u>.&nbsp;&nbsp;  As <math>f</math> is continuous on the left at <math>s</math>, there exists <math>\delta>0</math> such that <math>|f(x)-f(s)| < d/2</math> for all <math>x</math> in <math>[s-\delta,s]</math>.    This means that <math>f</math> is less than <math>M-d/2</math> on the interval <math>[s-\delta,s]</math>.   But it follows from the supremacy of <math>s</math> that there exists a point, <math>e</math> say, belonging to <math>L</math> which is greater than <math>s-\delta</math>.  By the definition of <math>L</math>, <math>M[a,e]< M</math>.  Let  <math>d_1=M-M[a,e]</math> then for all <math>x</math> in <math>[a,e]</math>, <math>f(x)\le M-d_1</math>.  Taking <math>d_2</math> to be the minimum of <math>d/2</math> and <math>d_1</math>, we have <math>f(x)\le M-d_2</math> for all <math>x</math> in <math>[a,b]</math>.  This contradicts the supremacy of <math>M</math> and completes the proof. [[Q.E.D.|∎]]
}}