Supremum of series difference question

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The discussion centers on the supremum of the difference between the functions f_n(x) and f(x) as n approaches infinity. The function f_n(x) is defined as 1 for 1 ≤ x ≤ n and 0 for n < x < ∞, while f(x) is consistently 1. Participants clarify that as n increases, f_n converges to f, leading to the conclusion that the supremum of their difference, sup(f_n(x) - f(x)), is indeed 0, not 1. This conclusion arises from the fact that for any finite n, the maximum difference occurs at the boundary where f_n transitions from 0 to 1.

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lom
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[tex]f_n(x)=1,1\leq x\leq n\\[/tex]
[tex]f_n(x)=0,1< n< \infty[/tex]
f_n converges to f which is 1
at the beginning 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



?
 
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lom said:
[tex]f_n(x)=1,1\leq x\leq n\\[/tex]
[tex]f_n(x)=0,1< n< \infty[/tex]

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

[tex]f_n(x)=0,n < x < \infty[/tex]?


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

That's "beginning". What do you mean by "at the beginning fn = 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 |fn(x) - f(x)| = 1?
 

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