Prove that if, for all $n$, $a_n>0$ and $b_n \geq 0$, then:
$\limsup(a_n b_n) \leq \limsup(a_n)\limsup(b_n)$
Provided that the product on the right is not of the form $0 \times \infty$
$proof:$
Case 1: $\limsup(a_n)=A < \infty$, $\limsup(b_n)=B < \infty$
Say $\limsup(a_nb_n)=C$. Then there exists a subsequence of $(a_nb_n)$, say $(a_{n_k}b_{n_k})$ s.t. $\lim_{k \rightarrow \infty}(a_{n_k}b_{n_k})=C$
Thus we have the following:
$\limsup(a_nb_n) = \lim_{k \rightarrow \infty}(a_{n_k}b_{n_k})=\lim_{k \rightarrow \infty}(a_{n_k})\lim_{k \rightarrow \infty}(b_{n_k}) \leq AB$
Case 2: $\limsup(a_n)=\infty$ and $\limsup(b_n)=\infty$
In this case, $\limsup(b_n)\limsup(a_n)=\infty$ and so the inequality holds trivially.
My instructor said that in regards to case 1, $\lim_{k \rightarrow \infty}(a_{n_k})$ and $\lim_{k \rightarrow \infty}(b_{n_k})$ may not exist. I see now what he meant. Can somebody help me make this proof correct? Thanks!
Best Answer
There are a few ways to do it, but since you like subsequences, you can start with a subsequence of $a_n b_n$ that converges to $\limsup_{n\to\infty} a_nb_n$, say, $a_{n_m}b_{n_m}$. Next, take the sequence $a_{n_m}$, and find a (sub)subsequence $a_{n_{m_k}}$ of $a_{n_m}$ that converges to some $A \le \limsup_{n\to\infty} a_n$. Finally, take a (subsub)subsequence $b_{n_{m_{k_l}}}$ of $b_{n_{m_k}}$ that converges to $B \le \limsup_{n\to\infty} b_n$.
Now, given that sub$^n$sequences of convergent sequences converge to the same limit, we have \begin{align*} a_{n_{m_{k_l}}} &\to A \\ b_{n_{m_{k_l}}} &\to B \\ a_{n_{m_{k_l}}}b_{n_{m_{k_l}}} &\to \lim_{m\to\infty} a_{n_m} b_{n_m} = \limsup_{n\to\infty} a_n b_n. \end{align*} By algebra of limits, and the uniqueness of limits, $$\limsup_{n\to\infty} a_n b_n = AB.$$ Note that the positivity of $a_n$ and $b_n$ imply that $A \ge 0$ and $B \ge 0$, hence $$\limsup_{n\to\infty} a_n b_n = AB \le \left(\limsup_{n\to\infty} a_n\right) \cdot \left(\limsup_{n\to\infty} b_n\right).$$