Uniform Convergence of (x^n)f(x)

**eigensheep** · October 18th 2011, 06:43 AM

I'm having difficulty showing that if f(1)=0 and f:[0,1] to R is continuous then (x^n)f(x) is uniformly convergent.
I get that I'll probably need the fact f is continuous on a closed bounded interval so is bounded but this seems to result in requiring x^n to be uniformly convergent on (0,1) which I don't think it is (since it's pointwise convergent to 0, but take 0<e<1 and then given N take x bigger than the (N+1)th [positive] root of e. Then there is an n>N [namely N+1] for which |x^n| > e)

Am I wrong in thinking x^n is not uniformly convergent?

Can anyone give me a hint?

Thanks.

**FernandoRevilla** · October 18th 2011, 07:03 AM

Originally Posted by eigensheep

if f(1)=0 and f:[0,1] to R is continuous then (x^n)f(x) is uniformly convergent.

We have $g(x)=\lim_{n\to +\infty}x^nf(x)=0$ for all $x$ in the compact $[0,1]$ i.e. the limit is a continuous function. Besides $g_n(x)=x^nf(x)$ is a decreasing monotonous sequence. According to the Dini's Theorem, $g_n\to 0$ uniformly in $[0,1]$ .

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