functions

**manjohn12** · April 22nd 2009, 05:08 PM

Let $f^{(n)}(x) := 2n$ when $x \in [1/2n, 1/n]$ and $f^{(n)}(x) := 0$ for all other values of $x$ . Why does $f^{(n)}$ converge pointwise to $f(x) := 0$ ?

**TheEmptySet** · April 22nd 2009, 06:17 PM

Originally Posted by manjohn12

Let $f^{(n)}(x) := 2n$ when $x \in [1/2n, 1/n]$ and $f^{(n)}(x) := 0$ for all other values of $x$ . Why does $f^{(n)}$ converge pointwise to $f(x) := 0$ ?

Pointwise convergence depends on both n and x.

So for any fixed x if it is out side of $[0,1]$

$f^{n}=0$ for all n

so we are okay.

Now for any $x \in [0,1]$

There exits an N such that for all n> N $f^{n}=0$

The Key point is that it depends both on x and n. I hope this helps

**manjohn12** · April 22nd 2009, 07:09 PM

Originally Posted by TheEmptySet

Pointwise convergence depends on both n and x.

So for any fixed x if it is out side of $[0,1]$

$f^{n}=0$ for all n

so we are okay.

Now for any $x \in [0,1]$

There exits an N such that for all n> N $f^{n}=0$

The Key point is that it depends both on x and n. I hope this helps

I think you mean $[1/2, 1]$ . Also why is this true: There exits an N such that for all n> N $f^{n}=0$ ? Because $f^{(100)}(x) := 200$ for all $x \in [1/2, 1 ]$ .

**Jhevon** · April 22nd 2009, 07:21 PM

Originally Posted by manjohn12

I think you mean $[1/2, 1]$ . Also why is this true: There exits an N such that for all n> N $f^{n}=0$ ? Because $f^{(100)}(x) := 200$ for all $x \in [1/2, 1 ]$ .

pay attention to your interval. $[1/2,1]$ is the BIGGEST it will be. for your example, $x = 100$ , we have the interval $[1/200, 1/100]$ . this is why TES considers the interval $(0,1]$ , because $[1/(2n), 1/n] \to \{ 0 \}$ (note, 0 is never actually in the interval)

so as $n$ gets large, the interval for which $f^{(n)} \ne 0$ gets smaller and smaller...

**TheEmptySet** · April 22nd 2009, 07:21 PM

Originally Posted by manjohn12

I think you mean $[1/2, 1]$ . Also why is this true: There exits an N such that for all n> N $f^{n}=0$ ?

I meant $(0,1]$

What we have is a sequnce of intervals.

When n=1 $[\frac{1}{2},1]$
n=2 $[\frac{1}{4},\frac{1}{2}]$
n=3 $[\frac{1}{6},\frac{1}{3}]$
n=4 $[\frac{1}{8},\frac{1}{4}]$
...

So eventally every x in $(0,1]$

Is not in $[\frac{1}{2n},\frac{1}{n}]$

so $f^{n}=0$

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