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I came across this statement while I was reading some topic: we have a sequence $\{ a_n \}$ of real numbers, such that $|f(x)|\geq \frac{1}{a_n}$ for all $n\in \mathbb Z$, then

" if $\inf a_{n}=a>0$, then $\sup |f(x)| \geq \frac{1}{a} $"

How can the above conclusion be true?

As I know since $\inf a_{n}=a$ then $a_n \geq a$ for all $n$, and $\frac{1}{a_n} \leq \frac{1}{a} $!!

Thanks

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    What is the contradiction ? $\frac{1}{a_n} \le \frac{1}{a} \le \sup_x |f(x)|$ which does not contradict $\vert f(x) \vert \ge \frac{1}{a}$.2011-11-01
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    Basically, you are arguing the following way: I want to prove that $9 \geq 4$, and I know that $9 \geq 3$ so how can that statement be true since $3 < 4$!!2011-11-01

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