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

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 $[0,2]$?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Trevor Gunn, Saad, TheSimpliFire, Claude Leibovici, Matthew Leingang
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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.

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