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

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!

For some basic information about writing math at this site see e.g. here, here, here and here. — Julian Kuelshammer, Dec 02 '12 at 15:36
A hint (assuming $a>0$): Do it for sup, then for the limits (which should be known already), then put the two together. — Harald Hanche-Olsen, Dec 02 '12 at 15:38
I am struggling to understand why an edit was reviewed and approved that made the question grammatically worse. Decapitalizing "I", replacing a period with ellipsis, and allowing the phrase "as an real number" are obviously wrong. — rschwieb, Dec 02 '12 at 15:42
@rschwieb I agree and also don't understand why the link was deleted which at least at my PC seems to work. — Julian Kuelshammer, Dec 02 '12 at 15:44
Dear Promtea, show us what you have done and the community here will help you. I downvoted because the question is unclear (what are $a,b$? sequences, real numbers???) and shows no effort — Nameless, Dec 02 '12 at 15:45
@amWhy (and other people involved in the editing process): You can (and should) use \limsup rather than \lim\sup or \mbox{limsup}. — Asaf Karagila, Dec 02 '12 at 15:50
It is my first time posting on math.stackexchange, so I'm sorry I'm struggling with different functionalities. — Promtea, Dec 02 '12 at 15:51
For the inequality, you can look at these questions: How to prove these inequalities in real analysis?, Properties of $\liminf$ and $\limsup$ of sum of sequences or Subadditivity of the limit superior. — Martin Sleziak, Dec 04 '12 at 15:19

score 1 · Answer 1 · answered Nov 17 '15 at 03:12

$$\{{a_m}+{b_m}:m\geqslant n\}\subseteq \{{a_m}+{b_k}:m,k\geqslant n\}$$

since we are pairing elements from two sets together in the first set while drawing each elements at random from two sets in the second set. By taking the supremum we have:

$$\sup\{{a_m}+{b_m}:m\geqslant n\}\leqslant\sup\{{a_m}+{b_k}:m,k\geqslant n\}\\=\sup\{\{{a_m}:m\geqslant n\}+\sup\{\{{b_m}:m\geqslant n\}$$

Using $\textbf{lemma}$ : $\sup (A+B)=\sup A+ \sup B$ , where $(A+B)=\{a+b:a\in A,b\in B\}$ Taking limit of above iequality gets:

$$\lim_{n\to\infty}\sup\{{a_m}+{b_m}:m\geqslant n\}\leqslant \lim_{n\to\infty}\sup\{{a_m}+{b_k}:m,k\geqslant n\}\\=\lim_{n\to\infty}\sup\{\{{a_m}:m\geqslant n\}+\lim_{n\to\infty} \sup\{\{{b_m}:m\geqslant n\}$$ $$Q.E.D$$

Proof of $\textbf{lemma}$:

$$\forall c\in A+B,\exists a\in A,b\in B,s.t.c=a+b\leqslant \sup A +\sup B$$

So $A+B$ is bounded by $\sup A +\sup B$

$$\forall \varepsilon \gt 0,\exists a \in A,b \in B ,s.t.a \gt \sup A-\varepsilon ,b \gt \sup B -\varepsilon ,a+b\gt \sup A +\sup B -2\varepsilon$$

So any number less than $\sup A +\sup B $ is not an upper bound. Thus $\sup A +\sup B $ is the least upper bound.

In fact, you don't need the lemma: since $a_m \le \sup A$ and $b_k \le \sup B$, we must have $\sup (A+B) \le \sup A + \sup B$, whence the desired result follows. — , Nov 17 '15 at 03:37

user29999 · Answer 2 · 2012-12-02T16:13:02.370

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}

edited Dec 02 '12 at 16:13

answered Dec 02 '12 at 15:50

user29999

5,211

Can you give more detail on why it is correct that $\sup{a\times a_m:m\ge n}=a\sup{a_m:m\ge n}$ – Mathematics Dec 02 '12 at 16:00
2

Ok, I must assume that $ \sup a_n \ge 0$, In fact, if this is done and $S= \sup a_n$, we have $ a\cdot a_m \le a \cdot S $ and for any $ \varepsilon > 0$ given, we can choose $ a_m $ such that $ | s - a_m | < a / \varepsilon $. So, $ | a S - a a_m | = a |a - _m | < \varepsilon. $ – user29999 Dec 02 '12 at 16:23

score 0 · Answer 3 · answered May 31 '19 at 14:04

0

You may find something useful here: Limit superior and limit inferior - properties, and here: Calculus - Proofs of Some Basic Limit Rules.
Also note that , as it might be helpful.

answered May 31 '19 at 14:04

Anas Khaled

775

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

3 Answers3