Decreasing nonnegative sequence and nonincreasing functions

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Discussion Overview

The discussion revolves around the properties of a nonnegative nonincreasing sequence converging to a limit and its relationship with a nondecreasing function. Participants explore the implications of these properties, particularly focusing on the behavior of the function values at the sequence terms and the limit.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant states that since the sequence $\{p_n\}$ is nonincreasing and converges to $p$, it follows that $p_n \ge p$ for all $n$, leading to the conclusion that $f(p_n) \ge f(p)$ due to the nondecreasing nature of $f$.
  • Another participant emphasizes that the inequalities involving $f(p_n)$ and $f(p)$ are not strict, indicating that $f(p_n)$ could equal $f(p)$.
  • A participant questions the validity of the statement $p_{n+1} \le p_n - f(p)$, asserting that this need not be true and expressing confusion about its origin.
  • The same participant acknowledges a misunderstanding regarding the inequality and indicates a correction of their earlier claim.

Areas of Agreement / Disagreement

Participants generally agree on the properties of the nonincreasing sequence and the nondecreasing function, but there is disagreement regarding the specific inequality involving $p_{n+1}$ and $f(p)$. The discussion remains unresolved on this particular point.

Contextual Notes

The discussion does not clarify the assumptions or definitions that might affect the interpretation of the inequalities discussed, particularly regarding the behavior of the sequence and the function.

ozkan12
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Let $\{p_n\}$ be a nonnegative nonincreasing sequence and converges to $p \ge 0$. Let $f : [0,\infty)\to[0,\infty)$ be a nondecreasing function. So, since f is a nondecreasing function, $f(p_n)>f(p)>0$. How did this happen?
 
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Hi ozkan12,

Since $\{p_n\}$ is nonincreasing and converging to $p$, $p = \inf\{p_n : n\in \Bbb N\}$, so $p_n \ge p$ for all $n\in \Bbb N$. Consequently, as $f$ is nondecreasing, $f(p_n) \ge f(p)$. So we have $f(p_n) \ge f(p) \ge 0$.

Note that the inequalities above are not strict.
 
Euge said:
Hi ozkan12,

Since $\{p_n\}$ is nonincreasing and converging to $p$, $p = \inf\{p_n : n\in \Bbb N\}$, so $p_n \ge p$ for all $n\in \Bbb N$. Consequently, as $f$ is nondecreasing, $f(p_n) \ge f(p)$. So we have $f(p_n) \ge f(p) \ge 0$.

Note that the inequalities above are not strict.

ok thanks a lot :) but I will ask one question : so, pn+1<=pn-f(p)...how this happened ?
 
ozkan12 said:
ok thanks a lot :) but I will ask one question : so, pn+1<=pn-f(p)...how this happened ?

I'm sorry, I don't understand -- where did you get $p_{n+1} \le p_n - f(p)$? This need not be true.
 
Euge said:
I'm sorry, I don't understand -- where did you get $p_{n+1} \le p_n - f(p)$? This need not be true.

ok :) I see my false :) I repair that :)
 

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