Pointwise, and uniform convergence

Asked Mar 28 '14 at 05:37

Active Mar 28 '14 at 06:00

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Suppose $f$ is continuous on $[0,1]$ and $f_n(x) = f(x)^n$. Explain why this sequence does not converge uniformly when $f(x)=1$?

When $f(x)=1$, $f_n(x)$ is $1$ for all $n$. But I don't see why this means there is no uniform convergence. Can't we say $|f_n(x)-f(x)|=|1-1|<\epsilon$ which is uniformly convergent?

edited Mar 28 '14 at 06:00

mookid

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asked Mar 28 '14 at 05:37

user138602

The constant sequence $(f_n)_n=(1,1,1, \dots)$ indeed converges uniformly to the constant function $1$. – user1337 Mar 28 '14 at 05:45
But what is the difference between the constant sequence and f(x)? – user138602 Mar 28 '14 at 05:51
It is zero, which is lesser than any positive $\epsilon$. – user1337 Mar 28 '14 at 05:52
yes, that is my understanding too, but the problem is asking to show that f(x) does not converge... – user138602 Mar 28 '14 at 05:53
... then the problem is asking to prove something incorrect. Unless of course, by "$f(x)=1$", the problem meant $f(x)=x$, i.e., "$f$" is the identity function on $[0,1]$. – Amitesh Datta Mar 28 '14 at 05:54
Does it converge uniformly if f(x)<1? Perhaps there is a typo – user138602 Mar 28 '14 at 05:59
Yes, if $f(x)<1$ then ${f_n}\to 0$ uniformly – Jlamprong Mar 28 '14 at 06:20

Pointwise, and uniform convergence

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