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 $(-1,1]$.
A: $f$ doesn't converge uniformly in $(-1,1]$ since
$$\lim_{x\to-1}\lim_{n\to\infty}f_n(x)=0\neq\lim_{n\to\infty}\lim_{x\to-1}f_n(x)=\lim_{n\to\infty}(-1)^n-1$$
A: An idea: say for $\,n>2\,$
$$f_n(x)=x^n-x^{2n}\implies f'_n(x)=nx^{n-1}-2nx^{2n-1}=nx^{n-1}\left(1-2x^{n}\right)=0\iff$$
$$x=0\,,\,\frac1{\sqrt[n]2}$$
$$f_n''(x)=n(n-1)x^{n-2}-2n(2n-1)x^{2n-2}\implies\begin{cases}f''(0)=0\\{}\\f''\left(\frac1{\sqrt[n]2}\right)=\frac{n\sqrt[n]4}4\left(-2n\right)<0\end{cases}\;\implies$$
$$\implies\text{at}\;\;\left(x=\frac1{\sqrt[n]2}\;,\;y=f_n\left(\frac1{\sqrt[n]2}\right)=-\frac14\right)\;\text{we have a maximum for}\;\;f_n(x)\;,\;\forall\,n>2 .$$
And from here it follows at once that since $\,\displaystyle{f(x):=0=\lim_{n\to\infty}f_n(x)}\;$ , then
$$ \lim_{n\to\infty}\sup_{n}|f_n(x)-f(x)|=\lim_{n\to\infty}\frac14=\frac14$$
and thus the convergence isn't uniform.
