Show for a fixed $0<a<1$ that the sequence of functions on $[-a,a]$ defined by $f_n = x^n$ converges uniformly.

Question

I can show very easily that this same sequence of functions does not converge uniformly on (-1,1) and [0,1] but I am having trouble seeing how to prove that the sequence of functions does show uniform convergence on the suggested interval.

Answer 1

We go back to the definition of uniform convergence.

The limit is clearly $0$ for all $x$ in our interval. We want to show that for any $x$ in $[-a,a]$, and for any $\epsilon \gt 0$, there is an $N$ independent of $x_0$ such that if $n\ge N$, then $|x^n|\lt \epsilon$.

We have $|x^n|\le a^n$. So if $a^N\lt \epsilon$ and $n\ge N$, the desired inequality will hold. Let $N$ be the smallest positive integer greater than $\frac{\ln\epsilon}{\ln a}$.

Answer 2

Obviously, $f_n(x) \to 0$ for all $x \in [-a,a]$ for $a \in (0,1)$. Now just use the definition.

Note that $\sup_{x \in [-a,a]} |f_n(x) - f(x)| = \sup_{x \in [-a,a]} |f_n(x) - 0 | = |f_n(a)| = a^n$, and $f_n \to f$ uniformly if and only if $\sup_{x \in [-a,a]} |f_n(x) - f(x)| \to 0$. Since $a^n \to 0$ for $a \in (0,1)$, $f_n \to 0$ uniformly.