Let $F:=(f_n)_{n\in\mathbb{N}}$ where $$\forall x\in[-1,1]\,\,\,\,\,\,\,\,\,f_n(x):=\mid x\mid^{1+\frac{1}{n}}$$ I have to prove that $F$ is not compact in $(C^1([-1,1]),\lVert\,\cdot\,\rVert_{\infty})$.
My idea is to prove that $F$ is not sequentially compact w.r.t. the topology induced by the infinity norm.
In particular, if I prove that $f_n\to f$ uniformly on $[-1,1]$, where $f(x):=\lvert x\rvert$, we are done. Infact, every subsequence will also converge in norm to $f\notin C^1([-1,1])$.
Is this reasoning correct? If yes, how can I prove $f_n\to f$ uniformly on $[-1,1]$?
It's enough to consider $x\geq 0$. Then $$ \left|x\right|-\left|x\right|^{1+\frac{1}{n}}= x-x^{1+\frac{1}{n}}= x\left(1-x^\frac{1}{n}\right)= $$ $$ =n^n\left(\frac{x^{\frac{1}{n}}}{n}\right)^n\left(1-x^\frac{1}{n}\right) $$ By AM-GM inequality, $$ \ldots\leq n^n\left(\frac{n\,\frac{x^{1/n}}{n} + 1 - x^{1/n}}{n+1}\right)^{n+1}= \frac{n^n}{(n+1)^{n+1}}=\frac{1}{n+1}\left(\frac{n}{n+1}\right)^n\leq \frac{1}{n+1}\to0 $$