1. ## functions

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

2. Originally Posted by manjohn12
Let $\displaystyle f^{(n)}(x) := 2n$ when $\displaystyle x \in [1/2n, 1/n]$ and $\displaystyle f^{(n)}(x) := 0$ for all other values of $\displaystyle x$. Why does $\displaystyle f^{(n)}$ converge pointwise to $\displaystyle f(x) := 0$?

Pointwise convergence depends on both n and x.

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

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

so we are okay.

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

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

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 $\displaystyle [0,1]$

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

so we are okay.

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

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

The Key point is that it depends both on x and n. I hope this helps
I think you mean $\displaystyle [1/2, 1]$. Also why is this true: There exits an N such that for all n> N $\displaystyle f^{n}=0$? Because $\displaystyle f^{(100)}(x) := 200$ for all $\displaystyle x \in [1/2, 1 ]$.

4. Originally Posted by manjohn12
I think you mean $\displaystyle [1/2, 1]$. Also why is this true: There exits an N such that for all n> N $\displaystyle f^{n}=0$? Because $\displaystyle f^{(100)}(x) := 200$ for all $\displaystyle x \in [1/2, 1 ]$.
pay attention to your interval. $\displaystyle [1/2,1]$ is the BIGGEST it will be. for your example, $\displaystyle x = 100$, we have the interval $\displaystyle [1/200, 1/100]$. this is why TES considers the interval $\displaystyle (0,1]$, because $\displaystyle [1/(2n), 1/n] \to \{ 0 \}$ (note, 0 is never actually in the interval)

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

5. Originally Posted by manjohn12
I think you mean $\displaystyle [1/2, 1]$. Also why is this true: There exits an N such that for all n> N $\displaystyle f^{n}=0$?
I meant $\displaystyle (0,1]$

What we have is a sequnce of intervals.

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

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

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

so $\displaystyle f^{n}=0$