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Let $(a_n)$ and $(b_n)$ be two real-valued and bounded sequences.

  1. $\limsup \max \{a_n,b_n\} = \max\{\limsup a_n, \limsup b_n\}$.
  2. $\limsup\min\{a_n,b_n\} = \min\{\limsup a_n, \limsup b_n\}$.

EDIT: Thanks to Trevor Wilson, I could find a counterexample to 2. Let $$a_n= 1,0,1,0,1,0,...$$ and $$b_n=0,1,0,1,0,1,...$$

Here, min $\{a_n,b_n\}= 0,0,0,0,0,...$ Thus, lim sup min $\{a_n,b_n\} = 0 \neq 1 =$ min$\{1,1\}$

For the first, I checked on few examples and the statement look true. But I am yet to come up with proofs or counterexample.

Any hint will be much appreciated!

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    $\begingroup$ For (2), can you find a counterexample where $\min\{a_n,b_n\}$ is always zero? $\endgroup$ – Trevor Wilson Sep 29 '13 at 17:46
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For (1):

$$\max\{a_n,b_n\} \geq a_n,$$ and $$\max \{a_n,b_n \} \geq b_n.$$ Applying $\limsup$ on both of these inequalities gives $$\limsup \max \{a_n,b_n \} \geq \limsup a_n,$$ and $$\limsup \max \{a_n,b_n \} \geq \limsup b_n .$$ So, clearly $\limsup \max \{a_n,b_n \} \geq \max \{ \limsup a_n, \limsup b_n \}$.

It follows from the definition of $\limsup$ as the maximal subsequential limit that there exists some subsequence $(\max\{a_{n_k},b_{n_k} \} )_k$ such that $$\limsup \max \{ a_n,b_n \}= \lim_{k \to \infty} \max \{a_{n_k},b_{n_k} \}.$$ But each of the elements $\max\{ a_{n_k},b_{n_k} \}$ is either $a_{n_k}$ or $b_{n_k}$. Thus, the subsequence $(\max\{a_{n_k},b_{n_k} \} )_k$ must contain infinitely many elements of $(a_n)_n$ or infinitely many elements of $(b_n)_n$. WLOG, suppose that there are infinitely many elements of $(a_n)_n$, that is, there exists a subsequence $(\max\{ a_{n_{k_l}},b_{n_{k_l}} \})_l$ such that for all $l$, $\max\{ a_{n_{k_l}},b_{n_{k_l}} \}=a_{n_{k_l}}$. Moving to that subsequence doesn't affect the limit: $$\limsup \max \{a_n,b_n\}=\lim_{l \to \infty} \max\{ a_{n_{k_l}},b_{n_{k_l}} \}=\lim_{l \to \infty} a_{n_{k_l}} \leq \limsup a_n \leq \max \{\limsup a_n,\limsup b_n \}. $$

In summary we have proven $$\limsup \max \{a_n,b_n \} \geq \limsup a_n,$$ as well as $$\limsup \max \{a_n,b_n \} \leq \limsup a_n,$$ so that (1) is true.

Regarding (2), consider $a_n=b_n=(-1)^n$.

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You could simply use definition, i.e., $\limsup x_n= \lim\limits_{N\to\infty} \sup\{x_n, n\ge N\}$.
(If you have been taught different definition of $\limsup x_n$, it is equivalent to this one.)

So you get $$\limsup\max\{a_n,b_n\}= \lim_{N\to\infty} \sup_{n\ge N} \max\{a_n,b_n\}\overset{(1)}= \lim_{N\to\infty} \max\{\sup_{n\ge N} a_n,\sup_{n\ge N} b_n\}\overset{(2)}= \max\{\lim_{N\to\infty}\sup_{n\ge N} a_n,\lim_{N\to\infty}\sup_{n\ge N} b_n\}= \max\{\limsup a_n,\limsup b_n\} $$

It remains to find and argument why steps (1) and (2) are ok. (Can we exchange $\sup$ and $\max$ in (1)? Why? Can we exchange limit and maximum, provided we know that all limits in question exist?)

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