Absolute convergence for $\sum_1^n(f(x))^n$ where $f\in C([0,1])$ and $\|f\|_\infty=1$

Question

I am studying for a final exam I have tomorrow and am having trouble solving this practice problem.

Let $f \in C [0, 1]$, such that $\|f\|_\infty =1$, where $\| \cdot \|_\infty $ denotes the sup-norm. Show that the $\sum_1^n (f(x))^n$ $n\in N$ cannot converge absolutely for all $x\in [0,1]$ (in other words, for at least one $x\in [0,1]$, it does not converge absolutely).

I am not 100% sure what the correct answer would be but this is what I have so far:

Assume the statement is false. let $f(x)=x$ since $f(x)=x$ is continuous on $[0,1]$ with $\|f(x)=x\|_ \infty = 1$. Then $\lim \sum_1^n (f(x))^n$ diverges when $x=1$ and converges when $x<1$ by comparison with geometric sereis. But this is a contradition thus proving the above statement

Would this be correct? Any and all help with this problem is much appreciated

Answer 1

Suppose there exists $x_0$ such that $|f(x_0)|=1$. Done since $\sum_n 1$ and $\sum_n(-1)^n$ does not converge. Since $f$ is continue, $|f|$ is continue, so there exists $x_0$ such that $|f(x_0)|=sup_{x\in [0,1]}|f(x)|=1$ since the domain of $|f|$ is compact. done.