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Math Help - A question about property of liminf and limsup

  1. #1
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    A question about property of liminf and limsup

    If x_n\geq 0, y_n\geq 0 and \lim \limits_{n \to \infty }x_n exists, we have \limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\limsup\li  mits_{n\to\infty}y_n). But if \lim\limits_{n\to\infty}x_n<0, do we have analog equation(I guess \limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\liminf\li  mits_{n\to\infty }y_n))? and what change should be made to conditions to achieve the analog equation? Formal source of reference such as textbooks or webpages is recommended. Thanks!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zzzhhh View Post
    If x_n\geq 0, y_n\geq 0 and \lim \limits_{n \to \infty }x_n exists, we have \limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\limsup\li  mits_{n\to\infty}y_n). But if \lim\limits_{n\to\infty}x_n<0, do we have analog equation(I guess \limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\liminf\li  mits_{n\to\infty }y_n))? and what change should be made to conditions to achieve the analog equation? Formal source of reference such as textbooks or webpages is recommended. Thanks!
    I would agree with you assertion, namely:

    If x_n\leqslant 0,\lim\text{ }x_n=\xi and y_n\geqslant 0 that \limsup\text{ }x_ny_n=\xi\liminf\text{ }y_n. To do this we note that in general \limsup\text{ }a_n=\sup\text{ }S where S is the set of all convergent subsequential limits of a_n and \liminf\text{ }a_n=\inf\text{ }S.

    So, let S_1 denote the set of all subsequential limits of z_n=x_ny_n and \xi S_2 the multiplication by \xi of each subsequential limit of y_n. We claim that S_1=\xi S_2.

    We may assume (since otherwise this is easy) that \xi<0. So, let \alpha\in S_1 then \alpha=\lim\text{ }z_{\varphi(n)}=\lim\text{ }x_{\varphi(n)}y_{\varphi(n)} for some injection \varphi:\mathbb{N}\to\mathbb{N}. Thus, we note that \frac{\alpha}{\xi}=\frac{\lim\text{ }x_{\varphi(n)}y_{\varphi(n)}}{\lim\text{ }x_{\varphi(n)}} (notice that if a limit converges to a values so does every subsequential limit) and thus by the assumption that both of the limits converge we see that \frac{\alpha}{\xi}=\lim\text{ }y_{\varphi(n)}\implies \alpha=\xi\lim\text{ }y_{\varphi(n)} and thus \alpha is \xi times a subsequential limit of y_n and thus \alpha\in \xi S_2.

    Conversely, if \alpha\in \xi S_2 then \alpha=\xi\lim\text{ }y_{\varphi(n)}=\lim\text{ }x_{\varphi(n)}\lim\text{ }y_{\varphi(n)} and once again since we assumed both of these limits exists we see that \alpha=\lim\text{ }x_{\varphi(n)}y_{\varphi(n)} and thus \alpha\in S_1

    Thus, \limsup\text{ }x_ny_n=\sup\text{ }S_1=\sup\text{ }\xi S_2=\xi\inf\text{ }S_2=\xi\liminf\text{ }y_n



    NOTE: It's late and I haven't checked over the above too carefully. I have been known to make light night stupid errors.
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  3. #3
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    Great! I checked it and found no problem. It's a rigorous and beautiful proof. From the proof, the conclusion holds even without any assumption regarding the sign of real terms x_n and y_n, only \lim\limits_{n\to\infty}x_n<0 and finite is sufficient. We can also obtain \liminf\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\limsup\li  mits_{n\to\infty }y_n) similarly.
    Thank you very very much, Drexel28! I would like to give you double thanks. Have a good dream:-)
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