I want to find out if the sequence $f_n = x^n - x^{2n}$ converges uniformly for $x \in [0,1]$.
Actually I know that it converges poinwise but It doesn't converge uniformly and I just don't see why.
If I calculate the limit $f(x)=\lim_{n\to \infty} (x^n - x^{2n})$ for $x=0$ and $x=1$ it is zero (note sure about for $x \in (0,1)$)
And if the sequence converges uniformly, we have:
$\lim_{n\to \infty} \sup_{x \in [0,1]}|f_n(x)-f(x)| = 0$ $\;$ $\;$ $\;$$\;$ (*)
I think the problem would be in evaluating the limit of $f_n(x)$ for $x \in (0,1)$ becauce in $x=1$ and $x=0$ it converges and that would be the pointwise convergence. But I want to know how (*) is going to look like.
Note that $$ f_n(x)=x^n(1-x^n) $$ So this function converges point-wise to $f(x) = 0$ for $x \in [0,1]$.
Let $\epsilon > 0$ given. Then we must show that there is $N$ such that $$ |f_n(x) - f(x)|=|f_n(x)| \le \epsilon $$ for all $n \ge N$ and all $x\in[0,1]$. Write $z=x^n$. Then $$ f_n(z) = z(1-z) $$ which attains its maximum at $z=1/2$. This occurs when $x=(1/2)^{1/n}$ Hence $$ f_n(x) \le (1/2)^{1/n}(1-(1/2)^{1/n}) < 1-(1/2)^{1/n}. $$ The last expression converges to zero. Hence for sufficiently large $N$, $$ f_n(x) \le \epsilon $$ for $n \ge N$ and for all $x \in [0,1]$.