provine value of sup

[orir]

f is a continous function in (0,1) which maintains \(\displaystyle \lim_{x\rightarrow0^{+}}f(x)=-1 \), and \(\displaystyle \lim_{x\rightarrow1^{-}}f(x)=1 \)

let's define:\(\displaystyle A={x\in(0,1) | f(x)=0\} \); s=supA. i need to prove that\(\displaystyle f(s)=0\) .

well, i know that in oter words if s is part of the group Ai need to prove that it's its maximum, and if \(\displaystyle s\notin A \) i don't have any idea how to prove that. (...my proffesor wants from us a formal writen proof [i saw him proving existence of sup using the definition of the limit (epsilon-delta) and by that contradicted the existence of another upper-bound number which smaller than the one we needed to prove]

 

[pka]

f is a continous function in (0,1) which maintains \(\displaystyle \lim_{x\rightarrow0^{+}}f(x)=-1 \), and \(\displaystyle \lim_{x\rightarrow1^{-}}f(x)=1 \)
let's define:\(\displaystyle A=\{x\in(0,1) | f(x)=0\} \); s=supA. i need to prove that\(\displaystyle f(s)=0\) .

You need to have proven the lemma: if \(\displaystyle f\) is a continuous function and \(\displaystyle f(s)\ne 0\) then there exists \(\displaystyle \delta>0\) such that if \(\displaystyle t\in (s-\delta,s+\delta)\) then \(\displaystyle f(s)~\&~f(t)\) have the same sign.

If you have that lemma, then it implies that \(\displaystyle s\) is not a the \(\displaystyle \sup(A)\), because \(\displaystyle (s-\delta,s+\delta)\) can contain no point of\(\displaystyle A\).

If you need to prove the lemma, then note that \(\displaystyle |f|\) is also continuous. Use \(\displaystyle \delta=\dfrac{|f(s)|}{2}\).

 

[orir]

If you have that lemma, then it implies that \(\displaystyle s\) is not the \(\displaystyle \sup(A)\), because \(\displaystyle (s-\delta,s+\delta)\) can contain no point of\(\displaystyle A\).

why is that so? i didn't understand the end of the sentence...

If you need to prove the lemma, then note that \(\displaystyle |f|\) is also continuous. Use \(\displaystyle \delta=\dfrac{|f(s)|}{2}\).

how can i prove this lemma?

 

[pka]

why is that so? i didn't understand the end of the sentence...
how can i prove this lemma?

Please correct this impression. It seems that you are not really sure what this is all about?

(**) If \(\displaystyle s=\sup(A)~\& ~ \beta>0 \) then \(\displaystyle \exists t\in A\cap(s-\beta,s]\).
Do you understand why that is true?

If \(\displaystyle s\in A \) then \(\displaystyle f(s)=0 \). Why?

If \(\displaystyle s\notin A \) then \(\displaystyle f(s)\ne 0 \). Why?

Let’s say that \(\displaystyle f(s)>0 \). From continuity we know that \(\displaystyle \left( {\exists \beta > 0} \right)\left[ {\left| {x - s} \right| < \beta \Rightarrow \left| {f(x) - f(s)} \right| < \frac{{f(s)}}{2}} \right] \)

Using the properties of absolute values we get that
\(\displaystyle 0 < \frac{{f(s)}}{2} < f(x) \) is true for all \(\displaystyle x\in (s-\beta,s] \).

Does that contradict (**)?

You can do the case where \(\displaystyle f(s)<0 \)