I was looking at some notes of Real Analysis and I encountered this exercise :
Prove that the series of functions $ f_n(x) = \frac{x2^n}{1 +n2^nx^2}$ is not uniformly convergent on $\Bbb{R}$ but is uniformly convergent on the intervals $(-\infty, - a] \cup[a, \infty) $ with $ a \gt 0 $.
The function is pointwise convergent to $f(x) = 0$, but I cannot understand why the series of functions is uniformly convergent on $(-\infty, -a] \cup[a, \infty) $ and not uniformly convergent on $\Bbb{R}$.
I would appreciate some help in trying to understand the rationale behind it. When say “the rationale” I mean using the negation of the formal definition of uniform convergence. I can understand the non-uniform convergence graphically but not analytically. Thanks!
For any $f:D\to\Bbb R$ we may define $\|f\|=\min (1,\sup_{x\in D}|f(x)|).$ In other words, if $|f(x)|>1$ for some $x\in D$ then $\|f\|=1$ ; otherwise $\|f\|=\sup_{x\in D}|f(x)|.$
A sequence $(f_n)_{n\in\Bbb N}$ of functions from $D$ to $\Bbb R $ converges to $0$ uniformly iff $$\lim_{n\to\infty}\|f_n\|=0.$$ For the $f_n$ in your Q, with $D=\Bbb R,$ we have $f_n(2^{-n})>1/2$ so $$\|f_n\|=\min(1,\sup_{x\in D}|f(x)|)\ge \min (1,f_n(2^{-n})) >1/2.$$ So we do not have $\lim_{n\to\infty}\|f_n\|=0.$
Note: This def'n of $\|f\|$ is not the same as others commonly seen in functional analysis but it is convenient when $\sup_{x\in D}|f(x)|=\infty.$ And there is no $\infty$ in $\Bbb R$ so $\sup_{x\in D}|f(x)|=\infty$ is a convenient way of saying that $\{|f(x)|:x\in D\}$ has no upper bound in $\Bbb R.$