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?