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Thread: If lim sup a_n = lim inf a_n = a, then a_n->a

  1. #1
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    If lim sup a_n = lim inf a_n = a, then a_n->a

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    Or maybe perhaps someone can provide a different/simpler proof for theorem 1?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    Theorem 1: It can easily be seen that if $\displaystyle S$ is the set of all sub sequential limits of $\displaystyle \left\{a_n\right\}$ then $\displaystyle \limsup\text{ }a_n=\sup\text{ }S$ and $\displaystyle \liminf\text{ }a_n=\inf\text{ }S$. But, since $\displaystyle \lim\text{ }a_n$ exists and equals $\displaystyle a$ we know that every subsequential limit converges to $\displaystyle a$ so that $\displaystyle \limsup\text{ }a_n=\sup\text{ }S=\sup\text{ }\{a\}=a=\inf\text{ }\{a\}=\inf\text{ }S=\liminf\text{ }a_n$

    Theorem 2: Using the same idea as above we see that since $\displaystyle a=\inf\text{ }S\leqslant s\leqslant a=\sup\text{ }S$ for every $\displaystyle s\in S$ we see that $\displaystyle S=\{a\}$ and so every subsequential limit converges to $\displaystyle a$. The conclusion follows.
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    Smile

    Quote Originally Posted by Drexel28 View Post
    Theorem 1: It can easily be seen that if $\displaystyle S$ is the set of all sub sequential limits of $\displaystyle \left\{a_n\right\}$ then $\displaystyle \limsup\text{ }a_n=\sup\text{ }S$ and $\displaystyle \liminf\text{ }a_n=\inf\text{ }S$. But, since $\displaystyle \lim\text{ }a_n$ exists and equals $\displaystyle a$ we know that every subsequential limit converges to $\displaystyle a$ so that $\displaystyle \limsup\text{ }a_n=\sup\text{ }S=\sup\text{ }\{a\}=a=\inf\text{ }\{a\}=\inf\text{ }S=\liminf\text{ }a_n$

    Theorem 2: Using the same idea as above we see that since $\displaystyle a=\inf\text{ }S\leqslant s\leqslant a=\sup\text{ }S$ for every $\displaystyle s\in S$ we see that $\displaystyle S=\{a\}$ and so every subsequential limit converges to $\displaystyle a$. The conclusion follows.
    Somehow my textbook presents lim sup in a different way. It defines the lim sup as simply the limit of the seqeunce of supremums of the tails, and nothing related to subsequences and subsequential limits. (so I admit there is some gap between your proof and my background)

    Is there an alternative way to prove theorem 1 using more basic definitions/principles? (e.g. epsilon-limit proofs, upper bounds, least upper bounds, etc.)

    Thank you!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    Somehow my textbook presents lim sup in a different way. It defines the lim sup as simply the limit of the seqeunce of supremums of the tails, and nothing related to subsequences and subsequential limits. (so I admit there is some gap between your proof and my background)

    Is there an alternative way to prove theorem 1 using more basic definitions/principles? (e.g. epsilon-limit proofs, upper bounds, least upper bounds, etc.)

    Thank you!
    I mean, it's just direct application. If $\displaystyle \forall\varepsilon>0$ there exists some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\leqslant n\implies \left|a_n-a\right|<\varepsilon$ what can we say about $\displaystyle \left|\sup_{N\leqslant n}a_n-a\right|$?
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    I mean, it's just direct application. If $\displaystyle \forall\varepsilon>0$ there exists some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\leqslant n\implies \left|a_n-a\right|<\varepsilon$ what can we say about $\displaystyle \left|\sup_{N\leqslant n}a_n-a\right|$?
    |a_n -a| < ε for all n≥N
    => a-ε < a_n < a+ε for all n≥N
    => in particular,
    a-ε < a_N < a+ε........................(1)

    sup a_n ≥ a_n for all n≥N (by definition of an upper bound)
    n≥N
    => in particular,
    sup a_n ≥ a_N............................(2)
    n≥N

    (1) and (2) show that we will always have
    sup a_n > a-ε
    n≥N

    But how to do it for the "upper" bound a+ε?
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