How to show that $f_{n}(x)=\frac{1}{1+x^{n}}$ converges uniformly to $1$ on $[0,a]$ with $0<a<1$?

Question

The problem is as follows. I have a secuence $f_{n}(x)=\frac{1}{1+x^{n}}$ and I have to show why $f_{n}$ converges uniformly on $[0,a]$, $0<a<1$ and why does not converge uniformly on $[0,1]$.

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I know that $f_{n}$ is a secuence of continuous functions on $[0,1]$, but its limit function $$f(x)=\cases{1 & if $x<1$ \\ 1/2 & if $x=1$ }$$

has a discontinuity at $1$, hence $f_{n}$ does not converges uniformly on $[0,1]$. Now, I only have to show that $f_{n}$ converges uniformly on $[0,a]$ but here is where I am stuck. Without success, I have tried to do it directly from definition. Do you have any suggestion that can help me?

Answer 1

Note that for $x \in [0,a]$ you have $1 \ge \frac{1}{1+x^n} \ge \frac{1}{1+a^n}$. Now show that $\frac{1}{1+a^n}$ tends to $1$ as $n$ goes to $\infty$.

Answer 2

Hint. One has that $$ f_{n}'(x)=-\frac{nx^{n-1}}{(1+x^{n})^2}\leq0,\quad x \in [0,a], $$ thus $f_n$ is decreasing over $[0,a]$ and $$\sup_{x \in [0,a]}f_{n}(x)=f_n(0)=1. $$ Now we have $$ \sup_{x \in [0,a]}\left|f_{n}(x)-1\right|=\sup_{x \in [0,a]}\left|-\frac{x^n}{1+x^n}\right|\leq \frac{a^n}{1+a^n}\leq a^n, $$ then use $0<a<1$.

Answer 3

I thought it would be useful to present a prof that directly uses the definition of uniform convergence. To that end, we proceed.

Given $\epsilon>0$ we have for $x\in [0,a]$ with $0<a<1$

$$\begin{align} \left|\frac{1}{1+x^n}-1\right|&=\frac{x^n}{1+x^n}\\\\ &\le x^n\\\\ &\le a^n\\\\ &<\epsilon \end{align}$$

whenever $n>\log(\epsilon)/\log(a)$