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 $x \in [0,1)$

  • 3
    $\begingroup$ 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

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.


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 $[0,1].$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.