Suppose $\displaystyle f : [0,1]\rightarrow R$ is continuous.

Show that the sequence $\displaystyle (x^{n}f(x))$ is uniformly convergent on [0,1] iff f(1)=0

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- Oct 18th 2008, 10:50 AMszpengchaoshow this sequence of functions is uniformly convergent
Suppose $\displaystyle f : [0,1]\rightarrow R$ is continuous.

Show that the sequence $\displaystyle (x^{n}f(x))$ is uniformly convergent on [0,1] iff f(1)=0 - Oct 18th 2008, 03:50 PMThePerfectHacker
There is $\displaystyle M$ so that $\displaystyle |f|\leq M$. This means $\displaystyle |x^n f(x)| \leq Mx^n \to 0$, uniformly, for $\displaystyle x\in [0,1)$. Thus, to have uniform convergence we require that $\displaystyle (1)^n f(1) \to 0$ to i.e. $\displaystyle f(1)=0$.