Conditions for convergence

Question

Let f be a continuous function on [0,1], and $f_n(x)=f(x)^n$. Under what conditions on f will the sequence converge point wise? Uniformly?

I think it will converge uniformly if $-1\leq x \leq 1$ but I am not sure about point wise.

Answer 1

Study the convergence of the sequence $(y^n)$ for any real number $y$. Once you've done that you can answer the first question which is the point wise convergence($-1<f\leq 1$). Next you can show that it converges uniformly if and only if $-1<f<1$ or $f=1$.

Answer 2

If $|f(x)|>1$, then $|f_{n+1}(x)|>|f_n(x)|$ and so you cannot have point-wise convergence. There is also no reason to have uniform convergence when $|f(x)|\leq 1$. Let $f(x)=x$. Then $f_n(x)=x^{n!}$. $f_n(x)$ converges point-wise to a discontinuous function, $F(x)$ such that $F(1)=1$ and $F(x)=0$ on $[0,1)$. But each $f_i$ is continuous, and the uniform limit of continuous functions is continuous. Thus we can't have uniform convergence to $F(x)$.