# absolute maximum in [a,b)

• Dec 13th 2011, 05:43 PM
alphabeta89
absolute maximum in [a,b)
Suppose that the function $\displaystyle f$ is continuous on $\displaystyle [a,b)$ and the limit $\displaystyle L=\lim_{x\rightarrow{b^-}}{f(x)}$ exists.
(i) Prove that if there is an $\displaystyle x_0\in{[a,b)}$ such that $\displaystyle f(x_0)>L$, then $\displaystyle f$ has an absolute maximum in $\displaystyle [a,b)$.
How to approach this question?

(ii) If there is an $\displaystyle x_1\in{[a,b)}$ such that $\displaystyle f(x_1)=L$, does $\displaystyle f$ necessarily have an absolute maximum in $\displaystyle [a,b)$?
• Dec 13th 2011, 09:19 PM
Jose27
Re: absolute maximum in [a,b)
I'm assuming L is a finite real number, if that's the case then

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

(ii) If we define $\displaystyle f(b)=L$ then $\displaystyle f$ is continous on $\displaystyle [a,b]$, by the extreme value theorem we have an absolute maximum $\displaystyle M$, then $\displaystyle M\geq L$. Consider the cases $\displaystyle M=L$, $\displaystyle M>L$.
• Dec 13th 2011, 10:50 PM
CaptainBlack
Re: absolute maximum in [a,b)
Quote:

Originally Posted by Jose27
I'm assuming L is a finite real number, if that's the case then

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

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

CB
• Dec 14th 2011, 12:03 AM
alphabeta89
Re: absolute maximum in [a,b)
(Rofl)