absolute maximum in [a,b)

**alphabeta89** · December 13th 2011, 05:43 PM

Suppose that the function $f$ is continuous on $[a,b)$ and the limit $L=\lim_{x\rightarrow{b^-}}{f(x)}$ exists.
(i) Prove that if there is an $x_0\in{[a,b)}$ such that $f(x_0)>L$ , then $f$ has an absolute maximum in $[a,b)$ .
How to approach this question?

(ii) If there is an $x_1\in{[a,b)}$ such that $f(x_1)=L$ , does $f$ necessarily have an absolute maximum in $[a,b)$ ?

**Jose27** · December 13th 2011, 09:19 PM

I'm assuming L is a finite real number, if that's the case then

(i) By the definition of limit there is an $s>0$ such that if $0<|x-b|<s$ then $f(x)-L\leq |f(x)-L|< f(x_0)-L$ so that $f(x)<f(x_0)$ on $(b-s,b)$ . From here it should be obvious.

(ii) If we define $f(b)=L$ then $f$ is continous on $[a,b]$ , by the extreme value theorem we have an absolute maximum $M$ , then $M\geq L$ . Consider the cases $M=L$ , $M>L$ .

**CaptainBlack** · December 13th 2011, 10:50 PM

Originally Posted by Jose27

I'm assuming L is a finite real number, if that's the case then

(ii) If we define $f(b)=L$ then $f$ is continous on $[a,b]$ , by the extreme value theorem we have an absolute maximum $M$ , then $M\geq L$ . Consider the cases $M=L$ , $M>L$ .

This method can also be adapted to do part (i) thus saving work by using the same idea for both.

CB

**alphabeta89** · December 14th 2011, 12:03 AM

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