# Thread: proving pointwise convergence for piecewise function

1. ## proving pointwise convergence for piecewise function

Show that the sequence $\displaystyle {f_n}$ defined by

$\displaystyle f_n(x) = \left\{ \begin{array}{c l} n & x\geqslant n \\ 0 & x<n \end{array} \right.$

converges pointwise on $\displaystyle \mathbb{R}$ to the zero function $\displaystyle f(x)=0$

I know that I can pick any N such that when x < n and n > N i can get |f_n(x) - f(x)| = |f_n(x)| < epsilon. But I'm not sure how to show the convergence for x >orequalto n...

2. Originally Posted by sfspitfire23
Show that the sequence {f_n} defined by
f_n(x) = n if x >orequalto n
= 0 if x < n

converges pointwise on R to the zero function f(x)=0

I know that I can pick any N such that when x < n and n > N i can get |f_n(x) - f(x)| = |f_n(x)| < epsilon. But I'm not sure how to show the convergence for x >orequalto n...
Pointwise convergence means that you can pick the x before the N. So given a fixed x all you have to do is to take N > x. Then, for that x, $\displaystyle f_n(x)$ will be 0 for all n>N.

3. So, Let $\displaystyle \epsilon >0$ and fix $\displaystyle x\in \mathbb{R}$ . Then, take $\displaystyle N>x$. Then we have $\displaystyle f_n(x)=0, f(x)=0$. So, for n>N we have $\displaystyle |0-0|=0<\epsilon$ .

How is this? Also, this convergence is NOT uniform, right?

Thanks!

4. Originally Posted by sfspitfire23
So, Let $\displaystyle \epsilon >0$ and fix $\displaystyle x\in \mathbb{R}$ . Then, take $\displaystyle N>x$. Then we have $\displaystyle f_n(x)=0, f(x)=0$. So, for n>N we have $\displaystyle |0-0|=0<\epsilon$ .

How is this? Also, this convergence is NOT uniform, right?

Thanks!
Correct and correct!

5. ## wrong?

Doesn't this violate the fact that n>N if you say N>x?

Remember x≥n so

Right?