Is this sequence of functions uniformly convergent on [0, 2] ??

Question

Define a sequence of functions $f_n : [0,2] \to \Bbb R$ as:

$$f_n(x) = \frac {1-x} {1+x^n}$$

Is this sequence of functions uniformly convergent on $[0,2]$?

Answer 1

Let $f_n(x)=\frac{1-x}{1+x^n}$ and $f(x)=\begin{cases}1-x&,0\le x\le 1\\\\0&,1\le x\le 2\end{cases}$

Clearly we have

$$\lim_{n\to \infty}f_n(x)=f(x)$$

Furthermore, we see that

$$|f_n(x)-f(x)|=\begin{cases}\frac{(1-x)x^n}{1+x^n}&,0\le x\le 1\\\\\frac{x-1}{1+x^n}&,1\le x\le 2\end{cases}$$

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Next, we have the following estimates for $x\in [0,1]$

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

whenever $n>1+\frac1\epsilon$.

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Similarly, we have the following estimates for $x\in[1,2]$

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

whenever $n>1+\frac1\epsilon$.

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Putting it all together, we see that for all $\epsilon>0$

$$|f_n(x)-f(x)|<\epsilon$$

whenever $n>1+\frac1\epsilon$ for all $x\in [0,2]$.

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The convergence is uniform.