MHB Does the function $f_n(x)$ converge pointwise to zero on $(0,+\infty)$?

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evinda
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Hi! :)

Let $f_n(x)=\frac{n}{x}[\frac{x}{n}], \forall x \neq 0$ ($[\frac{x}{n}]$ is the integer part of $\frac{x}{n}$).
  • Prove that $f_n \to 0$ pointwise at $(0,+\infty)$
  • Prove that,if $A \subseteq \mathbb{R}- \{0\}$ contains at least one negative number, $f_n$ does not converge pointwise at any function in $A$.

  • Let $x>0$. $\exists n\in \mathbb{N}$ such that $n>x \Rightarrow [\frac{x}{n}]=0$(if $n$ is great enough).So, $f_n(x)=\frac{n}{x}[\frac{x}{n}] \to 0$ pointwise.
  • Let $x_0<0$. $\exists n \in \mathbb{N}$ such that $n>x_0$...But,why is it then like that: $[\frac{x_0}{n}]=-1$?Could you explain it to me?? (Blush)
 
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evinda said:
Prove that,if $A \subseteq \mathbb{R}- \{0\}$ contains at least one negative number, $f_n$ does not converge pointwise at any function in $A$.
"does not converge to any function" (and, I believe, "on $A$").

evinda said:
Let $x>0$. $\exists n\in \mathbb{N}$ such that $n>x \Rightarrow [\frac{x}{n}]=0$(if $n$ is great enough).So, $f_n(x)=\frac{n}{x}[\frac{x}{n}] \to 0$ pointwise.
Yes.

evinda said:
Let $x_0<0$. $\exists n \in \mathbb{N}$ such that $n>x_0$...But,why is it then like that: $[\frac{x_0}{n}]=-1$?Could you explain it to me??
There exists an $n>|x_0|=-x_0$, i.e., $x_0>-n$ Then $0>x_0/n>-n/n=-1$, so $\lfloor x_0/n\rfloor=-1$.
 
Evgeny.Makarov said:
There exists an $n>|x_0|=-x_0$, i.e., $x_0>-n$ Then $0>x_0/n>-n/n=-1$, so $\lfloor x_0/n\rfloor=-1$.

I haven't really understood why $\lfloor x_0/n\rfloor=-1$ .. (Blush) Could you explain it further to me?
 
By definition,
\[
\lfloor x \rfloor=\max\, \{m\in\mathbb{Z}\mid m\le x\}
\]
What is $\lfloor-0.5\rfloor$? $\lfloor-0.9\rfloor$? $\lfloor-0.1\rfloor$? $\lfloor x\rfloor$ where $-1<x<0$?
 
Evgeny.Makarov said:
By definition,
\[
\lfloor x \rfloor=\max\, \{m\in\mathbb{Z}\mid m\le x\}
\]
What is $\lfloor-0.5\rfloor$? $\lfloor-0.9\rfloor$? $\lfloor-0.1\rfloor$? $\lfloor x\rfloor$ where $-1<x<0$?

These are all equal to $-1$..right?
 
evinda said:
These are all equal to $-1$..right?
Yes.
 
Evgeny.Makarov said:
Yes.

Nice..thank you very much! :)
 

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