Thread: proving pointwise convergence for piecewise function

1. proving pointwise convergence for piecewise function

Show that the sequence ${f_n}$ defined by

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

converges pointwise on $\mathbb{R}$ 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...

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, $f_n(x)$ will be 0 for all n>N.

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

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

Thanks!

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

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