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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} $!!


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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 $\vert f(x) \vert \ge \frac{1}{a}$. – Sasha Nov 1 '11 at 18:50
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 $3 < 4$!! – N. S. Nov 1 '11 at 18:56

Prove these two lemmas.

Lemma 1: Let $\{b_n\}$ be a sequence so that $|f(x)|\geq b_n$ for all $n$. Then $|f(x)|\geq \sup {b_n}$.

Lemma 2: If $\{a_n\}$ is a sequence such that $\inf{a_n}>0$, then: $\sup{\frac{1}{a_n}} = \frac{1}{\inf{a_n}}$.

Then, letting $b_n = \frac{1}{a_n}$, you get $|f(x)|\geq b_n$ for all $n$, hence:

$$|f(x)| \geq \sup{b_n} = \sup {\frac{1}{a_n}} = \frac{1}{\inf a_n} = \frac{1}{a}$$

Lemma 1 is essentially the definition of $\sup$.

Lemma 2 is a very little bit harder. In particular, you've already shown that $\frac{1}{a}=\frac{1}{\inf {a_n}}\geq \sup {\frac{1}{a_n}}$, so you only need to show that $\frac{1}{a}\leq \sup{\frac{1}{a_n}}$.)

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Let $a_n$ be a sequence of non-zero real numbers and in addition suppose that $\mathrm{inf}\;a_n=a\not=0$.

Then, yes, $a_n \geq a$ and so $1/a_n \leq 1/a$. If $1/a_n \leq c$ then $a_n \geq 1/c$. Since $a$ is the infimum, it is the largest number smaller than all of the $a_n$. So $1/c \leq a$ (if $1/c$ were bigger, then $a$ could not be the inf). Therefore, $c \geq 1/a$. Thus $1/a$ is the smallest number larger than all $1/a_n$. Therefore, $\mathrm{sup}\;1/a_n=1/a$.

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