Supremum of series difference question

[lom]

$$f_n(x)=1,1\leq x\leq n\\$$
$$f_n(x)=0,1< n< \infty$$
f_n converges to f which is 1
at the beggining f_n is 0 but when n goes to infinity its 1

so why sup(f_n(x)-f(x))=1 ?

f is allways 1

but f_n is 0 and going to one

in one case its 1-1
in the other its 0-1

the supremum is 0

so the supremumum of their difference is 0 not 1



?

 

[LCKurtz]

$$f_n(x)=1,1\leq x\leq n\\$$
$$f_n(x)=0,1< n< \infty$$

You obviously have typos there. Is the second one supposed to read:

$$f_n(x)=0,n < x < \infty$$?

f_n converges to f which is 1
at the beggining f_n is 0 but when n goes to infinity its 1

That's "beginning". What do you mean by "at the beginning f_n = 0"? If you state things more precisely, it might help you understand the problem better.

so why sup(f_n(x)-f(x))=1 ?

Given any n, can you find an x where |f_n(x) - f(x)| = 1?