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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 $[0,2]$?

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closed as off-topic by Trevor Gunn, Saad, TheSimpliFire, Claude Leibovici, Matthew Leingang Mar 22 '18 at 11:33

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  • 1
    $\begingroup$ I think you mean to ask whether the sequence of functions $\{f_n\}$ is uniformly convergent. $\endgroup$ – 16278263789 Mar 22 '18 at 2:27
  • $\begingroup$ yes.............. $\endgroup$ – purecj Mar 22 '18 at 2:34
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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

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


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 $n>1+\frac1\epsilon$.


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 $n>1+\frac1\epsilon$.


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

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

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


The convergence is uniform.

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  • $\begingroup$ Thanks a lot !! $\endgroup$ – purecj Mar 22 '18 at 6:01
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Mar 22 '18 at 13:16

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