Supremum of series difference question

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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?