Uniform convergence problem

Question

I would like some help with the following problem:

Suppose $f:[0,1]\to\mathbb{R}$ is continuous. Define $f_n:[0,1]\to\mathbb{R}$ by $$ f_n(t)=t^n\cdot f(t). $$ Show that if $f(1)=0$, then $(f_n)$ converges uniformly to $0$.

Thanks in advance.

Answer 1

Choose any positive real number $\epsilon > 0$. The goal is to show that there is an $N>0$ so that $\sup |f_n| \le \epsilon$ for all $n > N$.

Since $f$ is continuous at $1$, there exists $0<K<1$ so that on $(K,1]$, $|f(t)| < \epsilon$.

Note that $|f_n(t)| \le t^n |f(t)| \le |f(t)|$. It follows that

$|f_n(t)|<\epsilon$ for $t \in (K,1]$.

On $[0,K]$, $|f|$ attains a maximum $M$. It follows that

$|f_n(t)| \le t^n M \le K^n M$ for $t\in [0,K]$.

Putting the two inequalities together, we have

$|f_n(t)| \le \max(K^n M, \epsilon)$ on $[0,1]$.

From here, you should be able to complete the proof.