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Let $f: [0,1]\rightarrow\mathbb{R}$ be continuous. For $n\geq 1$, define $f_n(x)=f(x)x^n$ for all $x\in[0,1]$. Show that if $f(1)=0$, then the sequence $\{f_n\}$ is uniformly convergent.

Certainly it converges to $0$ point wise. I tried to show that $\forall\epsilon>$, there is some $n$ big enough such that $|f_n(x)|=|f(x)x^n|<\epsilon$.

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    $\begingroup$ To show that $f_n \to 0$ uniformly, you should be looking at $|f_n(x) - 0|$, not $|f_n(x) - f(x)|$, no? $\endgroup$ – Xander Henderson Dec 29 '17 at 21:56
  • $\begingroup$ @XanderHenderson Right thank you! I meant the limit... $\endgroup$ – 2ndaccount Dec 29 '17 at 21:58
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Find some $\delta\in(0,1/2)$ such that $|f(x)|<\epsilon$ for all $x\in[1-\delta,1]$. Then by choosing $N$ large enough such that $(1-\delta)^{n}<\epsilon$ for all $n\geq N$, then \begin{align*} |f_{n}(x)|=|f(x)x^{n}|\leq\max_{x\in[0,1-\delta]}|f(x)x^{n}|+\max_{x\in[1-\delta,1]}|f(x)x^{n}|\leq M(1-\delta)^{n}+\epsilon<(M+1)\epsilon, \end{align*} here $M=\max_{x\in[0,1]}|f(x)|$.

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Take $\varepsilon>0$. Let $M=\max|f|$. Since $f$ is continuous at $1$, there is a $\delta>0$ such that $1-\delta<x\leqslant1\implies\bigl|f(x)\bigr|<\varepsilon$. If $n$ is big enough, then $|x^n|<\frac\varepsilon M$ when $x\in[0,\delta]$. Therefore, if $n$ is big enough, then$$(\forall x\in[0,1]):\bigl|f(x)x^n\bigr|<\varepsilon$$because $[0,1]=[0,\delta]\cup(\delta,1]$.

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