From "Extreme value theorem" we say that both maximum and minimum is acheived by the continuous function in its closed interval domain , the proof goes by showing function is bounded and then proved that it needs to be achieved by showing contradiction of letting it not be the case , lets say that upper bound is M so from that proof we can surely say M is acheived , but suppose the maximum and minimum values are not symmetric about y axis , in that case only one is guaranteed to ve achieved from that proof isnt ? What about the other extreme value ? Like for example take this : M (maxima here will be acheived from Boundness theorem) , what about m (M>|m|) (minima how will we show that it will also be achieved ?) Proof which i am referring to Extreme Value Theorem proof help 
$\begingroup$
$\endgroup$
2
-
$\begingroup$ I don't understand what you are asking. The positions of the max and min have no general relation to x= 0. $\endgroup$– George IveyApr 23 at 15:26
-
$\begingroup$ x= 0 ? I didnt say anything related to that $\endgroup$– Orion_PaxApr 23 at 15:41
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
(This is a mirror of this proof.)
Since $f$ is continuous on $[a, b]$, the range of $f$ has a lower bound by the boundedness theorem and the greatest lower bound (call it $m$) by the completeness of real numbers.
Assume that $f(x)>m$ for all $x\in [a, b]$. Then function $g$ defined on $[a, b]$ by $$g(x) = \frac{1}{f(x)-m}$$ is continuous and positive. Hence there is a positive $K$ such that $g(x)\leq K$ for all $x\in [a, b]$ by the boundedness theorem.
Therefore, $f(x)\geq m+\frac{1}{K}$ for all $x\in [a, b]$. But then $m+\frac{1}{2K}$ which is greater than $m$ is a lower bound of the range of $f$, contradiction the fact that $m$ is the greatest lower bound of the range of $f$. Therefore, there is a $c\in [a, b]$ such that $f(c)=m$.
