Another Uniform Convergence problem

Question

I have another uniform convergence question:

Suppose $f:[0,1]\to{\mathbb R}$ is continuous. Define $f_n:[0,1]→ℝ$ by $f_n(t)=t^nf(t)$. Prove if $f(1)\neq 0$, then $f_n$ is not uniformly convergent.

Answer 1

Choose $\epsilon=\frac{f(1)}{4}$, say. Since $f$ is continuous, there is a $\delta$ so that if $1-\delta<t<1$, then $f(x)\geq \frac{f(1)}{2} >0$. Choose $x_n>\max\{(\frac{1}{2})^{\frac{1}{n}}, 1-\delta\}$. Then $f_n(x)=(x_n)^n f(x_n)\geq \frac{1}{2} \frac{f(1)}{2}=\epsilon$.

Answer 2

Hint

If $t\in [0,1)$ what's $\displaystyle\lim_{n\to\infty} t^n$ and then $\displaystyle\lim_{n\to\infty} f_n(t)?$

Now if $f(1)\ne0$ and since the limit function isn't continuous (why?) what result we can apply?

Answer 3

First note that $[0,1]$ is compact and $f$ is continuous, so the range of f is bounded. This helps you find the limit function. The rest is straightforward.