pointwise convergence and a function f_n

Question

I'm trying to do a task in my Fourier class. I have to check for pointwise convergenge and i can't really understand some terms. I tried reading in the book and around the internet, i might just confuse myself.

Anyways i have a function.

$ f_n(t)= \begin{cases} 1, & 0 \leq t \leq\frac{1}{n}\\ 0, & \text{otherwise} \end{cases} $

and my book says there will be pointwise convergence if there exist a positive integer $N$ such that:

$n \geq N $ then $| f_n(t) - f(t)| \lt \epsilon $

But i can't understand this condition: what is $N$?

$f(t)$ doesn't make sense to me.. what should $n$ be then?

What limit should $ \epsilon$ be?

many problems here.. I would love a few tips.. Thank you

score 0 · Answer 1 · answered Oct 08 '17 at 15:10

0

Hint

Let $x $ be a real.

If $x\le 0$ then for $n>0$,

$$f_n (x)=0$$ and $$\lim_{n\to+\infty}f_n (x)=0$$

If $x>0$ then for enough large $n $

or for $n>\frac {1}{x} $ we will have $x>\frac {1}{n} $ and $$f_n (x)=0$$ thus the limit is zero.

answered Oct 08 '17 at 15:10

hamam_Abdallah

62,951

Thank you, but what about N and epsilon? I am not sure why this says anything point wise convergence – nammerkage Oct 09 '17 at 18:27

score 0 · Accepted Answer · answered Oct 08 '17 at 15:10

A sequence $(f_n)_{n \in \mathbb N}$ converges pointwise to a limit $f$, if for every $t \in \mathbb R$ and every $\epsilon > 0$ there exists a $N \in \mathbb N$, such that $\vert f_n(t) - f(t) \vert < \epsilon$ for all $n \geq N$. That is what the books means. In your specific case you can use this in the following way:

Let $t < 0$ then you have $f_n(t) = 0$ for all $n \in \mathbb N$. Thus you can choose $N = 1$ for every $\epsilon > 0$ and you will get $\vert f_n(t) \vert = 0< \epsilon$ for all $n \geq N$. Therefore $f(t) = 0$ for all $t < 0$.

Let $t = 0$ then you have $f_n(t) = 1$ for all $n \in \mathbb N$. Thus you can choose $N = 1$ for every $\epsilon > 0$ and you will get $\vert f_n(t) - 1\vert = 0< \epsilon$ for all $n \geq N$. Therefore $f(t) = 1$ for $t = 0$.

Those were the easy cases. I leave the case $t > 0$ for you to try yourself: Just let $t > 0$ and $\epsilon > 0$ and try to find $f(t)$ and $N \in \mathbb N$, such that you get $\vert f_n(t) - f(t) \vert < \epsilon$ for all $n \geq N$. The Archimedean principle might help you :)

Thank you very much. I’m trying to do the task as you did, but I can’t get it to work. Can we talk about your first check? If I let t<0, I can’t see why that would mean f_n(t) = 0. Wouldn’t that depend on what 1/n is in the function? Also what are we getting for checking each of these steps? Would love your help. — nammerkage, Oct 09 '17 at 18:36
$f_n(t) = 0$ for $t < 0$ follows from the definition of $f_n$ that you gave. In the case $t > 0$ the value of $f_n(t)$ depends on $1/n$ thats why you have to be a little bit more clever there. By checking these steps you get the limit funciton $f$. Currently we know $f(t) = 0$ for $t < 0$ and $f(t) = 1$ for $t = 0$ :) — Yaddle, Oct 09 '17 at 19:11

pointwise convergence and a function f_n

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