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Let $f_n(x)=x^n$. The sequence $\{f_n(x)\}$ converge pointwise but no uniformly on $[0,1]$. Let $g$ be continuous on $[0,1]$ with $g(1)=0$. Prove that the sequence $\{g(x)x^n\}$ converge uniformly on $[0,1]$.

My attempt: Since $g$ is continuous on $[0,1]$ it is uniformly continuous on $[0,1]$. So $$\forall \; \epsilon >0, \exists \; \delta(\epsilon): |x-y|<\delta \, \Rightarrow \, |g(x)-g(y)|<\epsilon$$

It is clear that $\lim_{n\rightarrow\infty}g(x)x^n=0$ for all $x \in [0,1]$. Then, i need to prove that, given $\epsilon > 0 $, there is a positive integer $N$ such that as $n \geq N$, then $|g(x)x^n|<\epsilon$. Since $g$ is continuous at $1$, then $$|x-1|<\delta \, \Rightarrow |g(x)-g(1)|=|g(x)|<\epsilon$$ So, we have $$|g(x)x^n|<|g(x)|<\epsilon,$$

but i'm stuck in the case $x \in [0,1)$

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    $\begingroup$ If $x\in[0,1)$ then $f_n$ converges uniformly on $[0,x]$ and $g$ is bounded so $g\cdot f_n$ also converges uniformly. $\endgroup$ – Mario Carneiro Mar 4 '15 at 23:34
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The trick is to think of $[0,1]$ as $[0,1-\delta] \cup [1-\delta,1]$, where $\delta=\delta(\varepsilon)$ is a small number. First we choose $\delta$ so that $g f_n \leq \varepsilon$ on $[1-\delta,1]$. Then we choose $N$ large enough to get $g f_n \leq \varepsilon$ on $[0,1-\delta]$ provided $n \geq N$. The former requires that $g$ is continuous and satisfies $g(1)=0$, while the latter requires that $g$ is bounded.

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The uniform limit of $\{{g(x)x^n}\}$ on $[0,1]$ is the zero function.

To prove that the uniform limit is 0, i.e. to prove that $g$ converges uniformly, we want to show:

$$given \ \epsilon > 0, \exists N \in \Bbb N \ s.t. \lvert g(x)x^n - 0 \rvert = \lvert g(x)x^n \rvert < \epsilon.$$

Since $g$ is continuous on $[0,1]$, a closed and bounded interval, then $g$ attains its min and its max (by the Extreme Value Theorem). Specifically, it will be bounded on $[0,1]$. In other words, $\exists M\in\Bbb R, M < \infty$, such that $\lvert g(x) \rvert <M $ for all$ \ x \in [0,1].$ So we have:

$$\lvert g(x) \rvert \lvert x^n \rvert \le \lvert M \rvert \lvert x^n \rvert$$

Since on $[0,1), \ x^n \rightarrow 0 \ $as$ \ n \rightarrow \infty $ and $M < \infty$, we can make the quantity on the right hand side as arbitrarily small, i.e. as arbitrarily close to $0$, as we want. Since $x^n \rightarrow 0, \ $by definition there exists an $N \in \Bbb N \ s.t. x^n < \frac {\epsilon} {M}$.

Then for all $n \ge N,$ $$\lvert g(x)x^n \rvert = \lvert g(x) \rvert \lvert x^n \rvert \le \lvert M \rvert \lvert x^n \rvert < M \frac {\epsilon} {M} = \epsilon. $$

For the point $x=1, \ g(1)=0,$ so$ \ g(1)(1)^n = 0 < \epsilon$ for any $N$.

$\therefore \{{g(x)x^n}\}$ converges uniformly on $[0,1].$

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