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

If $\displaystyle x_n\geq 0, y_n\geq 0$ and $\displaystyle \lim \limits_{n \to \infty }x_n$ exists, we have $\displaystyle \limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\limsup\li mits_{n\to\infty}y_n)$. But if $\displaystyle \lim\limits_{n\to\infty}x_n<0$, do we have analog equation(I guess $\displaystyle \limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\liminf\li mits_{n\to\infty }y_n)$)? 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!

2. Originally Posted by zzzhhh
If $\displaystyle x_n\geq 0, y_n\geq 0$ and $\displaystyle \lim \limits_{n \to \infty }x_n$ exists, we have $\displaystyle \limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\limsup\li mits_{n\to\infty}y_n)$. But if $\displaystyle \lim\limits_{n\to\infty}x_n<0$, do we have analog equation(I guess $\displaystyle \limsup\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\liminf\li mits_{n\to\infty }y_n)$)? 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 $\displaystyle x_n\leqslant 0,\lim\text{ }x_n=\xi$ and $\displaystyle y_n\geqslant 0$ that $\displaystyle \limsup\text{ }x_ny_n=\xi\liminf\text{ }y_n$. To do this we note that in general $\displaystyle \limsup\text{ }a_n=\sup\text{ }S$ where $\displaystyle S$ is the set of all convergent subsequential limits of $\displaystyle a_n$ and $\displaystyle \liminf\text{ }a_n=\inf\text{ }S$.

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

We may assume (since otherwise this is easy) that $\displaystyle \xi<0$. So, let $\displaystyle \alpha\in S_1$ then $\displaystyle \alpha=\lim\text{ }z_{\varphi(n)}=\lim\text{ }x_{\varphi(n)}y_{\varphi(n)}$ for some injection $\displaystyle \varphi:\mathbb{N}\to\mathbb{N}$. Thus, we note that $\displaystyle \frac{\alpha}{\xi}=\frac{\lim\text{ }x_{\varphi(n)}y_{\varphi(n)}}{\lim\text{ }x_{\varphi(n)}}$ (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 $\displaystyle \frac{\alpha}{\xi}=\lim\text{ }y_{\varphi(n)}\implies \alpha=\xi\lim\text{ }y_{\varphi(n)}$ and thus $\displaystyle \alpha$ is $\displaystyle \xi$ times a subsequential limit of $\displaystyle y_n$ and thus $\displaystyle \alpha\in \xi S_2$.

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

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

NOTE: It's late and I haven't checked over the above too carefully. I have been known to make light night stupid errors.

3. 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 $\displaystyle x_n$ and $\displaystyle y_n$, only $\displaystyle \lim\limits_{n\to\infty}x_n<0$ and finite is sufficient. We can also obtain $\displaystyle \liminf\limits_{n\to\infty}(x_n\cdot y_n)=(\lim\limits_{n\to\infty}x_n)\cdot(\limsup\li mits_{n\to\infty }y_n)$ similarly.
Thank you very very much, Drexel28! I would like to give you double thanks. Have a good dream:-)