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Thread: Is this proof correct?

  1. #1
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    Is this proof correct?

    prove that if f(x) is monotone increasing on an open interval (a,b) and x_0 is any point in (a,b), then lim_{x \to x_0+} f(x) exists and satisfies lim_{x to x_0+} f(x) >= f(x_0). (from the right side)

    Is this proof correct? if not could someone please correct it/ explain what I'm doing wrong?

    Fix $\displaystyle x_0$ in (a,b). First consider x in ($\displaystyle x_0$,b] then f(x)>= f($\displaystyle x_0$) so inf f(x) ($\displaystyle x_0$<x<b) >=f($\displaystyle x_0$)

    Claim: lim f(x)=inf f(x) ($\displaystyle x_0$<=x<b) . Let S=inf f(x). Let $\displaystyle \epsilon$>0. Then there exists x$\displaystyle \epsilon$ in ($\displaystyle x_0$,b) such that f(x$\displaystyle \epsilon$)<S-$\displaystyle \epsilon$.

    But if x is in (x$\displaystyle \epsilon$,$\displaystyle x_0$) then f(x$\displaystyle \epsilon$)>=f(x) so S<=f(x)<S-$\displaystyle \epsilon$

    Hence |f(x)-s|<$\displaystyle \epsilon$ for all x in (x$\displaystyle \epsilon$,$\displaystyle x_0$).

    therefore, lim f(x) (x->$\displaystyle x_0$+) exists and equals S since S>=f($\displaystyle x_0$) we get lim (x->$\displaystyle x_0$+) >= f($\displaystyle x_0$)
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  2. #2
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    Frankly, I find your post very hard to read. Let me give you some alternative ways to see this.

    If $\displaystyle a<c<b$ then let $\displaystyle f(c+)=\inf\{f(x):x\in (c,b]\}$ and $\displaystyle f(c-)=\{\sup\{f(x):x\in [a,c)\}$.
    It is not hard the see that for an increasing function each of these exists.

    Clearly $\displaystyle \lim _{x \to c^ + } f(x) = f(c + ) \ge f(c)$ and likewise Clearly $\displaystyle \lim _{x \to c^ - } f(x) = f(c - ) \le f(c)$.

    I hope this helps you with notation.
    I will be glad to help with questions you may have.
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  3. #3
    Member mabruka's Avatar
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    I think the property yo uare using in your wrong has a little mistake:

    Claim: lim f(x)=inf f(x) (<=x<b) . Let S=inf f(x). Let >0. Then there exists x in (,b) such that f(x)<S-.
    I think it should say:


    For ever \epsilon there exists $\displaystyle x_\epsilon \in (x_0,b)$ such that $\displaystyle f(x_\epsilon) <\inf\{f(x):x_0\leq x \leq b \}+\epsilon $


    what do you think?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    Whatever you said..
    I think you mean to prove that if $\displaystyle fa,b)\mapsto\mathbb{R}$ is increasing then $\displaystyle \lim_{x\to x_0^-}f(x)=\sup_{x\in (a,x_0)}f(x)$. To see this let $\displaystyle \varepsilon>0$ be given. By virtue of the definition of supremum we know that there exists some $\displaystyle f(\xi)\in f((a,x_0))$ such that $\displaystyle \sup_{x\in(a,x_0)}f(x)-\varepsilon<f(\xi)\implies \sup_{x\in(a,x_0)}f(x)-f(\xi)<\varepsilon$. Notice though by definition we have that $\displaystyle \sup_{x\in(a,x_0)}f(x)-f(\xi)=\left|\sup_{x\in(a,x_0)}f(x)-f(\xi)\right|$. But, by $\displaystyle f$'s monotonicity we have that for any $\displaystyle \xi<\xi'<\xi$ that $\displaystyle f(\xi)<f(\xi')<\sup_{x\in(a,x_0)}f(x)\implies \sup_{x\in(a,x_0)}f(x)-f(\xi')<\sup_{x\in(a,x_0)}f(x)-f(\xi)<\varepsilon$. In particular if $\displaystyle x\in(\xi,x_0)\implies \left|\sup_{x\in(a,x_0)}f(x)-f(x)\right|=\sup_{x\in(a,x_0)}f(x)-f(x)<\sup_{x\in(a,x_0)}f(x)-f(\xi)<\varepsilon$. It follows that $\displaystyle \lim_{x\to x_0^-}f(x)=\sup_{x\in(a,x_0)}f(x)$.
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