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"Let $f$ be a continuous function in $[0,1]$ such that $f(1)=0$. Prove that this functions sequence $\{g_n\}=\{x^nf\}$ converges uniformly in $[0,1]$"

I found that the punctual limit of the sequence is the function $f\equiv 0$. So, that's the only candidate to uniformly limit.

Because the $f$ can be derivable or not, I can't apply derivative calculation to bound the functions $g_n$ between smooth maximum and minimun.

I know the key is that $f(1)=0$ because if $f(1)\neq0$, the punctual limit wouldn't be zero in $x=1$ but I don't know how to apply that.

Perhaps I must use the Weierstrass theorem to use both maximum and minimum points for "smashing" the sequence $\{f_n\}$ into the uniformly limit $f\equiv 0$ but I think they're not very useful, maybe I'm wrong.

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We assume $f(x)$ is not identically $0$ else the uniform convergence is trivial.

Since $f$ is continuous on $[0,1]$, then given $\epsilon>0$, there exists a number $\delta>0$ such that $|f(x)|<\epsilon$ whenever $1-\delta<x\le1$.

For that $\epsilon>0$ and corresponding $\delta>0$ fixed, we see that if $x\in [0,1-\delta]$, then

$$\begin{align} |x^nf(x)|&\le\left(\sup_{x\in[0,1-\delta]}|f(x)|\right)\,(1-\delta)^n\\\\ &\le \left(\sup_{x\in[0,1]}|f(x)|\right)\,(1-\delta)^n\\\\ &<\epsilon \end{align}$$

whenever $n>\frac{\log(\epsilon)-\log\left(\sup_{x\in[0,1]}|f(x)|\right)}{\log(1-\delta)}$.

Thus, given $\epsilon>0$, $|x^nf(x)|<\epsilon$ whenever $n>\frac{\log(\epsilon)-\log\left(\sup_{x\in[0,1-\delta]}|f(x)|\right)}{\log(1-\delta)}$ for all $x\in [0,1]$.

And we are done!

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