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Suppose f(x) is continuous on [0,1], and that f(1)=0. Prove that [itex]x^n f(x)[/itex] converges uniformly on [0,1] as [itex]n \rightarrow \infty[/itex]

By continuity, if [itex]|x-1|< \delta [/itex] then [itex]|f(x)|< \epsilon[/itex] for [itex]x \in [x_0 ,1][/itex] for some [itex]x_0 \in [0,1][/itex].

And there is some N such that if n>N, then [itex]|x^n|<\epsilon[/itex] since [itex]x^n \rightarrow 0[/itex] for [itex]x \in [0,1][/itex].

The endpoints work since x^nf(x) is 0 there. So I have an N that works for [itex] \{ 0 \} \cup [x_0, 1][/itex].

I'm having trouble getting the rest of the interval. I thought about covering the set and using compactness, but was wondering if there was a better way.

By continuity, if [itex]|x-1|< \delta [/itex] then [itex]|f(x)|< \epsilon[/itex] for [itex]x \in [x_0 ,1][/itex] for some [itex]x_0 \in [0,1][/itex].

And there is some N such that if n>N, then [itex]|x^n|<\epsilon[/itex] since [itex]x^n \rightarrow 0[/itex] for [itex]x \in [0,1][/itex].

The endpoints work since x^nf(x) is 0 there. So I have an N that works for [itex] \{ 0 \} \cup [x_0, 1][/itex].

I'm having trouble getting the rest of the interval. I thought about covering the set and using compactness, but was wondering if there was a better way.

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