Q:

Uniform convergance for $f_n(x)=x^n-x^{2n}$

the function $f_n(x)=x^n-x^{2n}$ converge to $f(x)=0$ in $(-1,1]$. Intuativly the function does not converge uniformally in (-1,1]. How can I prove it?
I tried using the definition $\lim \limits_{n\to\infty}\sup \limits_{ x\in (-1,1]}|f_n(x)-f(x)|$ function is continial fractional on $[-1,1]$ and $x=0,(\frac 1 2 )^{\frac 1 n}$ are the roots of the derivative. I found that the second derivative is negative in the second point. then $\sup=1/4$ and the function does not converge uniformally?

A:

Choose an arbitrarily large odd value of $n$.  There exists some $0<x<1$ such that $x^n>\dfrac 12$.
Then $$\begin{array}{rl}f_n(-x) &= (-x)^n - (-x)^{2n}
\\ &= -\left(x^n + x^2n\right) \\ &\leq -\frac 34 \end{array}$$
So $f_n$ does not converge uniformly on $(-1,1]$.

