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I am trying to determine if the sequence $f_n$:= $\frac{x^{2n}}{1+x^{2n}}$ is uniformly convergent on $D_1:=[-q,q],0<q<1$, and $D_2:= (-\infty,r] \cup [r,\infty),r>1$.

I have determined that $\limsup \limits_{n \to \infty} f_n = 1$, and therefore $\limsup \limits_{n \to \infty} |f-f_n| = \limsup \limits_{n \to \infty} |\frac{x^{2n}}{1+x^{2n}}-1| = \limsup \limits_{n \to \infty} |\frac{-1}{1+x^{2n}}|=0$ iff $|x|>1$. This would mean that the sequence is uniformly convergent on D2, but not on D1.

This is the answer I submitted for my class assignment, but it was marked incorrect with the note that the sequence is also uniformly convergent on D1. However, I am failing to see how that is. Did the grader make a mistake, or is my answer wrong? In any case, is my approach appropriate for determining whether a sequence is uniformly convergent?

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  • $\begingroup$ Just wanted to add that the grader is on vacation at the moment, so I unfortunately cannot ask him. $\endgroup$
    – Flow
    Sep 3 '15 at 13:13
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    $\begingroup$ For $x<1$, $f_n\to 0$. $\endgroup$
    – Mark Viola
    Sep 3 '15 at 13:21
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    $\begingroup$ What you wrote does not really tell anything on the behaviour of $f_n$ over $D_1$. Over $D_1$ we have $0\leq \left|f_n\right| \leq x^n$ so $f_n$ converges uniformly to zero. $\endgroup$ Sep 3 '15 at 13:28
  • $\begingroup$ Yes, I see now. Knowing the limit over $D1$, I have used the same method to see that the sequence is uniformly convergent over $D1$ as well. Thanks for your help. $\endgroup$
    – Flow
    Sep 3 '15 at 13:32
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For $|x| \leq q <1$ , $\limsup \limits_{n \to \infty} f_n = 0$, and convergence is uniform, $m>n$

$$f_n(x) - f_m(x) = \frac{x^{2n}}{1+x^{2n}} - \frac{x^{2m}}{1+x^{2m}} \leq x^{2n} - \frac{x^{2m}}{2} \leq q^{2n} - \frac{q^{2m}}{2} \to 0$$ $$f_n(x) - f_m(x) = \frac{x^{2n}}{1+x^{2n}} - \frac{x^{2m}}{1+x^{2m}} \geq \frac{x^{2n}}{2} - x^{2m} \geq \frac{q^{2n}}{2} - q^{2m} \to 0$$

independently of $x$

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