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Prove that for every real number $r > 0$ the sequence of functions $f_n(x) = \frac{e^x}{1 + n + x^2}$ converges uniformly on $[−r, r]$ to the zero function.

I'm not sure if I did this right.

Work:

The definition of uniform convergence for a sequence of functions is:

Let $E$ be a nonempty subset of $\mathbb{R}$. Sequence $f_n:E \to \mathbb{R}$ converges uniformly iff there is $\epsilon > 0$ and $N \in \mathbb{N}$ such that $n \ge N \implies |f_n(x) - f(x)| < \epsilon$ for all $x \in E$.

So, in this case we have $E = [-r, r]$ and basically we need to show that for $n \ge N$ for $N \in \mathbb{N}$ and $\epsilon >0$ that $|f_n(x)| < \epsilon$.

$\lim_{m \to \infty} f_mx = 0$, so clearly $|f_nx| < \epsilon$, which means the sequence $f_n(x)$ converges uniformly to the zero function on $[-r,r]$.

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I'm not sure if I tackled it right, or did it even remotely correctly. Did I? How should one actually solve it?

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  • $\begingroup$ Well, you haven't done anything. You need to show that for any $\epsilon>0$ there is some $N$ such that if $n \ge N$ and $|x| \le r$ that $|f_n(x)| < \epsilon$. Hint: It is fairly straightforward to find a convenient upper bound $|f_n(x)|$ $\endgroup$
    – copper.hat
    Feb 15 '15 at 23:31
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    $\begingroup$ Theres a difference between uniform and pointwise convergence. $\endgroup$
    – Batman
    Feb 15 '15 at 23:33
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For any $x ∈ [−r,r]$, $e^x \leqslant e^r$, and let $N = [e^r/ϵ]$. Then for any $x ∈ [−r,r]$ and $n>N$,

$$|f_{n}(x)|\leqslant \dfrac{{e^r}} {{1+n}}<\dfrac{{e^r}} {{n}}<\dfrac{{e^r}} {{N}}<ϵ$$

So $f_n(x)→0 $ uniformly on $[−r,r]$.

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Advance as

$$ \sup\bigg|\frac{e^{x}}{1+n+x^2}-0\bigg| \leq \frac{e^{r}}{1+n} < \frac{e^r}{n} < \epsilon. $$

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