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.
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Have a look at this: http://math.stackexchange.com/questions/441604/uniformly-convergence-in-compact-sets – MPW Feb 13 '14 at 02:28
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Thanks, but I was instructed it is possible to construct a much simpler proof without Dini's Theorem. – userzzz23 Feb 13 '14 at 02:34
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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 $\frac{\ln\epsilon}{\ln a}$.
André Nicolas
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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.
Batman
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