Prove uniform convergence of $x^{\frac{1}{n}}+(1-x)^{n}$ Is it true or not that the this succession converges uniformly on $(C[0, 1],\Vert . \Vert_{\infty})$: $$f_{n}=x^{\frac{1}{n}}+ (1-x)^{n}$$I have found an elementary solution, but I would like to understand fully this problem with a serious and complete proof.
 A: The function $$f_{n}=x^{\frac{1}{n}}+ (1-x)^{n}$$ converges pointwise to $$f(x)=1$$ For uniform convergence we should study $$ \lim_{n\to \infty} \sup \mid x^{\frac{1}{n}}+(1-x)^{n}-1\mid. $$If $x=\frac{1}{n}$we can observe that: $$\lim_{n\to \infty}\mid\frac{1}{n}^{\frac{1}{n}}+(1-\frac{1}{n})^{n}-1\mid=\frac{1}{e}\leq \lim_{n\to \infty} \sup \mid x^{\frac{1}{n}}+(1-x)^{n}-1\mid.$$This proves that $f_{n}$ is not uniformly convergent. I happend to find this solution, but understood only little about the problem.
A: Well, you're looking at a compact set $K=[0,1]$ and at the space of continuous functions $\mathscr C [0,1]$ over it. If your sequence $\{f_n\}$ converged uniformly, then it would be equicontinuous over $[0,1]$. But you have shown that $$1+e^{-1}=\lim_{n\to \infty} f_n(n^{-1})\neq \lim_{n\to\infty}{f_n \left(\lim_{n\to\infty} n^{-1}\right)}=1$$
which violates equicontinuity.
Recall that a familiy $\{f_n\}$ of functions is said to be equicontinious over a set $K$ if for each $\epsilon >0$ there exists a $\delta >0$ such that, for any choice of $x,y\in K$ with $|x-y|<\delta$ and $f_k$ we have $$|f_k(x)-f_k(y)|<\epsilon$$
Can you see how the above implies $$\lim_{n\to\infty} f_n(\alpha_n)=\lim_{n\to\infty} f_n \left(\lim_{n\to\infty} \alpha_n \right)\text{ ? }$$
A: What your solution shows is the following: uniform convergence means that given $\varepsilon>0$, there exists $k>0$ such that $n>k$ implies $|f_n(x)-1|<\varepsilon$, for all $x$. But if you take $\varepsilon=e^{-1}$, then $|f_n(1/n)-1|=e^{-1}\geq\varepsilon$ even if $n\geq k$ for the corresponding $k$.
