# supremum difference question..

• Nov 15th 2009, 03:47 AM
transgalactic
supremum difference question..
$\displaystyle f_n(x)=1,1\leq x\leq n\\$
$\displaystyle f_n(x)=0,1< n< \infty$
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

so the supremumum of their difference is 0 not 1

?
• Nov 15th 2009, 03:52 AM
Sampras
Quote:

Originally Posted by transgalactic
$\displaystyle f_n(x)=1,1\leq x\leq n\\$
$\displaystyle f_n(x)=0,1< n< \infty$
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

so the supremumum of their difference is 0 not 1

?

you are taking the supremum of the difference. Not just $\displaystyle f_{n}(x)$. So you are looking for the least upper bound of the differences.
• Nov 15th 2009, 05:05 AM
transgalactic
exacty
in one case its 1-1
in the other its 0-1

the supremum is 0