$\limsup(a\cdot a_n)=a\cdot \limsup(a_n)$

Question

I know it's quite obvious that $\limsup(a\cdot a_n)=a\cdot \limsup(a_n)$ for $a$ a real number >0, but I don't know how to prove it.

My second question is whether the following proof works for: $$\limsup(a + b) \leq \limsup(a) + \limsup(b)$$ for a and b as sequences.

http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2001;task=show_msg;msg=0119.0001.0001 Thanks!

Answer 1

Assuming $ a>0$.

\begin{equation} \limsup x_n = \lim_n (\sup \{x_m : m\ge n\}) \end{equation} Thus, assuming $a>0$ and $\sup a_n \ge 0$ we have. \begin{eqnarray} \limsup( a \cdot a_n \_n) &=& \lim_n (\sup \{a \cdot a_m : m\ge n\}) \\ &=& \lim_n [a \cdot(\sup \{ a_m : m\ge n\})]\\ &=& a \cdot \lim_n (\sup \{a_m : m\ge n\})\\ &=& a \cdot \limsup a_n. \end{eqnarray} Also, \begin{eqnarray} \limsup (a_n + b_n ) &=& \lim_n (\sup \{a_m + b_m : m\ge n\}) \\ &\le& \lim_n [(\sup \{a_m : m\ge n\}) + (\sup \{a_m : m\ge n\})]\\ &=& \lim_n (\sup \{a_m : m\ge n\} + \lim_n (\sup \{b_m : m\ge n\} \\ &=& \limsup a_n + \limsup b_n . \end{eqnarray}