Suppose $x = (x_1, x_2, ...)$ is a bounded increasing sequence and $x \in l^\infty$. Thus its limit is the least-upper bound of the sequence, i.e. $sup(x_i), i \in \Bbb N $.
Is it always true for such a sequence that $|sup(x_i)|\le sup(|x_i|) $?
Properties of infimum and supremum of monotone bounded sequences
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sequences-and-series
convergence
supremum-and-infimum
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1Since $|\sup A|\le\sup|A|$ for any set $A$ (here $|A|=\{|x|\mid x\in A\}$), for your sequence it's also true. – 2017-01-26
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0So if I now say that my sequence x is bounded decreasing, would it be right to say that |inf(x)| $\le$ sup|x|? – 2017-01-26
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1It would be correct. – 2017-01-26
1 Answers
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First Suppose that $\sup a_i \geq 0$. Let $|a_i|$ be the sequence of absolute values. Then we must have $$ a_i \leq |a_i|.$$ Hence $$ 0\leq \sup a_i \leq \sup |a_i|$$ and if you take absolute value in both sides you have the result.
Second Suppose $\sup a_i \leq 0$. Then $$ \sup a_i \leq |a_i|, \text{for every}~i $$ and so $$ \sup a_i \leq \sup |a_i| $$ and by taking the absolute value in both sides you have the result.