Proving uniform convergence of sequence of functions

Question

Let $f$ be continuous on $[0,1]$ with $f(1)=0.$ How can I show that the sequence $\{x^nf(x)\}$ converges uniformly on $ [0,1]$?

Can I deduce that $f(x)$ is bounded? Then I can get $lim_{n \to \infty} x^nf(x)=0.$

Please help...

Answer 1

See you can show that $f$ is bounded on $[0,1].$

First observe that $[0,1]$ is closed and bounded in $\Bbb R$ hence it is compact by Heine Borel theorem.

Now, we know that continuous image of a compact set is compact. Hence image of $f$ on $[0,1]$ is compact and hence again closed and bounded as you want.

Let $ g_n(x) $ = $x^nf(x)$. then since $f$ is bounded on $[0,1]$, therefore $g_n(x)$ are also bounded and $\lim_{n\to \infty}x^nf(x)$=0 on $[0,1].$ (becuase, at $x=1$ , $f(x)=0$).