1. ## functions

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 $f(x) := 0$?

2. 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

3. 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 $x \in [1/2, 1 ]$.

4. 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...

5. 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 $f^{n}=0$