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I don't understand the proof of Rudin TH.6.15

I don't understand the proof of Rudin TH.6.15

by Shamiur Rahman Ramim -
Number of replies: 4

As title suggests I am having a hard time understanding the proof for TH.6.15. I looked up an explanation online, but I still don't seem to get it. 

Attached is a photo of a part of an answer i found that i don't quite get: 

How are we sure that $M_2 - f(s) < \varepsilon$? We know that $M_2 = sup_{[s, x_2]}(f(x))$, and I'm thinking that this (even though it is bounded) could make $M_2 - f(s)$ greater than $\varepsilon$. 

I know my thinking is off, but I don't know where. Thanks in advance!

In reply to Shamiur Rahman Ramim

Re: I don't understand the proof of Rudin TH.6.15

by Sofia Tirabassi -

Hello!

Very nice question!

F is continuous and [s, x_2]  is cpt, so sup and inf ar actually max and min. So we can write M2=f(t_1) and m2= f(t_2) con ti in [t,x2]

So M2-m_2=f(t1)-f(t2)\leq |f(t1)-f(s)|+|f(s)-f(t2)|

Does this help?

Mvh

Sofia 


In reply to Sofia Tirabassi

Sv: Re: I don't understand the proof of Rudin TH.6.15

by Shamiur Rahman Ramim -
But f is only said to be continuous at s and not necessarily the whole interval [s, x_2], right?
In reply to Shamiur Rahman Ramim

Re: I don't understand the proof of Rudin TH.6.15

by Sofia Tirabassi -
Here is the correct answer.
Let $g(x)=sup_{[s,x]}f(t)$. I claim that $\operatorname{lim}_{x\rightarrow s+} g(x)=f(s)$.
In fact, fix $\epsilon>0$, then there is a $\delta>0$ such that $f(t)<f(s)+\epsilon/2$ for all t in [s,s+\delta]
then f(s)\leq g(t)\leq g(s+\delta)\leq f(s)+epilon/2 for all t in [s,s+delta]
We conclude that g(t)-f(s)\leq \epsilon/2 <\epsilon fo all t in [s,s+\delta]
This means that g(t)\rightarrow f(s).

The same statement can be proven with the inf, so both the sup and the inf of f over [s,t] tends to  f(s) when t goes to s. THis means that we can make the difference sup-inf as small as possilbe when t goes to s

Sorry for the misguidance and the mess. 
Is it clear now?
MVH
Sofia