Does x^n f(x) Converge Uniformly on [0,1] as n Approaches Infinity?

[sparkster]

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

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

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

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

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.

 

[ZioX]

You get uniform continuity on compact sets, my friend.

 

[sparkster]

You get uniform continuity on compact sets, my friend.

I'm asking about uniform convergence.

ETA: But the function would be uniformly continuous, giving me delta that works for all x in the interval.

Thanks.

 

[AKG]

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

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

What's the point of this?

"If |x-1|< \delta then |f(x)|< \epsilon"

means that for x \in (1-\delta ,1], |f(x)|<\epsilon. Why add the part about "for x \in [x_0, 1] for some x_0 \in [0,1]."?

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

This isn't true for x=1.

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

What is the definition of uniform convergence?