Show that the sequence of functions $f_n(x) = x^n$ defined on $[0, 1]$ converges pointwise but not
uniformly. Now suppose $g \in C([0, 1])$ such that $g(1) = 0$. Show that the sequence $f_ng$ converges uniformly.
When n goes to infinity, $f_n(x)$ goes to $f(x)$, i.e. it goes to infinity, too. So, I am writing, this sequence converges pointwise. Then, should I do something uniform convergence. Is there a way to show clearly pointwise but not uniformly? Also the second part the same as my first sentence? Is this enough to write?
Best Answer
$(f_ng)$ converges uniformly but $(f_n)$ does not. Since $\sup f_n(x)=1$ it follows that $(f_n)$ does not converge uniformly. Let $\epsilon >0$ and choose $\delta >0$ such that $|g(x)| <\epsilon$ for $|x-1|<\delta$. Then $|f_n(x)g(x)| \leq |g(x)| <\epsilon$ for $1-\delta <x\leq 1$ and $|f_n(x)g(x)| \leq (1-\delta)^{n}$ for $0 \leq x \leq 1-\delta$. Since $(1-\delta)^{n} \to 0$ we get uniform convergence of $(f_ng)$.