Editing Extreme value theorem (section)

===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.|∎]]
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