Prove that $(f_n)_n$ is uniformly convergent.

Question

Let $g$: $[0,1]\to\mathbb{R}$ be continuous and $g({1})=0$. Define $f_n(x)= x^{n}{g(x)}$.

Prove that $(f_n)_n$ is uniformly convergent.

Answer 1

First, if $g(x)=0$ for all $x\in [0,1]$, then there is nothing to prove. Now, because the function $g$ is continuous, then it sends compact set to compact set, so $g([0,1])$ is compact, hence its bounded; from this (and from the fact that $g(1)=0$) it follows that for every $x\in [0,1]$, $lim_{n} f_n(x)=0$. Hence, $f_n$ converge pointwise to the zero function. Now, product of continuous functions is continuous, hence $|x^n g(x)|$ is continuous; by the extreme value theorem $|x^n g(x)|$ must attain a maximum at some point $a\in (0,1)$ (We can assume $a$ different from $0$ and $1$, because otherwise g would be trivial), then $sup_{ x\in [0,1]} |x^n g(x)|= |a^ng(a)|$. If you take the limit, then you have the uniform convergence.

Answer 2

Let $\varepsilon > 0$. Since $g$ is continuous with $g(1) = 0$, there exists a $\delta$, $0 < \delta < 1$ such that for all $x$, $1 - \delta < x \le 1$ implies $|g(x)| < \varepsilon$. Let $N$ be a positive integer greater than $\log(\varepsilon)/\log(1 - \delta)$. If $n\ge N$, then

1.$|f_n(x)| \le (1 - \delta)^n < \varepsilon$ if $x\in [0, 1 - \delta]$.

2.$|f_n(x)| \le |g(x)| < \varepsilon$ if $x\in (1 - \delta, 1)$.

3.$|f_n(x)| = 0 < \varepsilon$ if $x = 1$.

Hence, $|f_n(x)| < \varepsilon$ for all $n\ge N$ and $x\in [0,1]$. Consequently, $f_n$ converges uniformly to $0$ on $[0,1]$.