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

[evinda]

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)

 

[Evgeny.Makarov]

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 $A$").

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.

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 $\lfloor x_0/n\rfloor=-1$.

 

[evinda]

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?

 

[Evgeny.Makarov]

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 $-1<x<0$?

 

[evinda]

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?

 

[Evgeny.Makarov]

These are all equal to $-1$..right?

Yes.

 

[evinda]

Yes.

Nice..thank you very much! :)