I am doing a larger problem, but part of it is the following:
Let $f:[0, 1] \to \mathbb{R}$ be continuous, and define $f_n(x) = x^n f(x)$. Prove that if $f(1) = 0$, then the sequence $(f_n)_{n=0}^{\infty}$ converges uniformly to zero.
Using Dini's Theorem, this problem has a one line solution. However, I wanted to do it from the definitions, and I was unable to. It's easy to prove uniform convergence on $[0, a]$ for any $0 \le a<1$, but I am unable to do it for $[0, 1]$. I thought about splitting the interval and showing uniform convergence on $[0, a]$ and $[a, 1]$ for some $a$, but I encounter the same difficulty on $[a, 1]$ that I do on $[0, 1]$. I could see how to do it if $f$ were Lipschitz (since then we could bound $|f(1-x)| \le mx)$, but I can't see how to do it with just continuity/uniform continuity.
Additionally, I am doing this problem in preparation for an Analysis exam, so if there is any intuition you can offer on how to come up with the solution, it would be greatly appreciated.
Thank you very much for all your help.
Best Answer
Given a $\epsilon \gt 0$, since $f$ is continuous, there is a number $0 \lt K \lt 1$ such that for all $x \in [K,1]$, $\lvert f(x)\rvert\lt \epsilon$. It is also true that $f$ is bounded on $[0,1]$ (since it is continuous on $[0,1]$), and let $U = \max\{\vert\sup_{[0,1]} f\vert,\vert\inf_{[0,1]}f\vert\}$. Choose $K^NU\lt \epsilon$, or $N=\log _K \frac {\epsilon} U$, and for all $n \gt N$, for all $x \in [0,1]$, $\lvert f_n(x) \rvert \lt \epsilon$. Therefore the sequence converges uniformly to zero.