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Assume $f_n(x)=x^n$
Let $g(x)$ be continuous on $[0,1]$ with $g(1)=0$
Prove ${g(x)f_n(x)}$ is uniformly convergent on $[0,1]$

I know how to prove this using Dini's theorem about compact sets and decreasing functions. But I want to know if there's another way to prove it.

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1 Answer 1

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Let $\varepsilon >0$. We can find $\delta$ such that $|g(x)|\leq \varepsilon$ if $1-x\leq \delta$. We have $\sup_{x\in\left[0,1\right]}|g(x)f_n(x)|=\max(\sup_{x\in\left[0,1-\delta\right]}|g(x)f_n(x)|,\sup_{x\in\left[1-\delta,1\right]}|g(x)f_n(x)|)$. The first $\sup$ is $\leq (1-\delta)^n\sup_{x\in\left[0,1\right]}|g(x)|$ and the second $\leq \varepsilon$.

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