Prove or disprove.

$\forall n \in \mathbb N$, $\forall \epsilon >0$, $\exists \delta >0$ such that $\forall x,y \in [0,1]$

$|x-y| < \delta$ implies that $|f_n(x)-f_n(y)| < \epsilon$.

Let $f_n(x)=x^n$.

I am pretty sure this is false because the limit function is not continuous and this problem is basically asking about the continuity of each approximating function in the sequence of functions.

How would I formally prove this??

  • $\begingroup$ We need to know more about the sequence of functions. This is certainly not true for every sequence of functions. For example, it is false for $f_n(x) = x^n$. $\endgroup$ – Paul Hurst Jun 9 '14 at 2:09
  • $\begingroup$ Since we choose a delta for each n, this is true. Just show that $x^n$ is uniformly continuous for fixed n. Actually a real valued continuous function on a compact domain is uniformly continuous. $\endgroup$ – ThePortakal Jun 9 '14 at 2:16

By Mean Value Theorem: $f_n(x) - f_n(y) = f'_n(c)(x-y)$ for some $c \in (x,y)$. Thus:

$|f_n(x) - f_n(y)| = |f'_n(c)||x-y| = nc^{n-1}|x-y| < n|x-y|$. Thus for $\epsilon > 0$ given, choose $\delta = \dfrac{\epsilon}{n}$, then:

if $|x-y| < \delta$, then: $|f_n(x) - f_n(y)| < n|x-y| < n\cdot \dfrac{\epsilon}{n} = \epsilon$


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