What does it mean when it says a series of $f_n$ does not converge uniformly, but it will for any fixed $x \geq \delta$?

Question

A (maybe) related question: Is this a way to show a series does not converge uniformly?

I used the above fact to show that the series (for $x \in \mathbb{R}$)
$\sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$ does NOT converge uniformly to $f(x) = 1+x^2$ for $x\neq0$ and $0$ if $x=0$, since $f$ is a discontinuous function. This is validated by my answer sheet, but then it goes on to say: "However, for any fixed $\delta > 0$, the convergence is in fact uniform for $|x|\geq \delta$.
I'm not sure if I'm missing something here, it seems that it's basically saying "for any $x \leq -\delta$ or $x \geq \delta$, then it does converge uniformly.... but isn't this just saying "for $x\neq 0$, it converges uniformly"?

Answer 1

You have a series of functions $\sum_{n=0}^\infty f_n(x)$ where each $f_n$ is a continuous function from $\Bbb R$ to $\Bbb R$. This series is not uniformly convergent on $\Bbb R$.

For each positive $\delta$ the series $\sum_{n=0}^\infty f_n(x)$ is uniformly convergent on the set $\{x\in\Bbb R:|x|\ge\delta\}$. (Strictly speaking $\sum_{n=0}^\infty F_n(x)$ is uniformly convergent on $\{x\in\Bbb R:|x|\ge\delta\}$ where $F_n$ is the restriction of $f_n$ to $\{x\in\Bbb R:|x|\ge\delta\}$.)

But the series $\sum_{n=0}^\infty f_n(x)$ is not uniformly convergent on $\{x\in\Bbb R:x\ne0\}$.

Answer 2

Uniform convergence isn't a property of a function as a point, rather, it matters what the domain is, since something has to hold for all elements of the domain. Recall that $f_n$ converges uniformly to $f$ on $S\subseteq \mathbb R$ provided that for all $\varepsilon > 0$ there exists $N\in\mathbb N$ such that whenever $n>N$ and whenever $x\in S$ we have $\vert f(x)-f_n(x)\vert < \varepsilon$. Consider the series $$\sum_{n=0}^\infty x^n$$ which converges pointwise to $\frac 1 {1+x}$ on $(-1,1)$, but not uniformly, since this function is unbounded on $(-1,1)$ but none of the partial sums are. This is to say, we can find an $\varepsilon > 0$ such that for any $N\in \mathbb N$ we can find an $x\in (-1,1)$ so that for some $n>N$ we have $\vert f_n(x)-f(x)\vert \geq \varepsilon$. Intuitively, the bigger we make $N$, the closer the associated $x$ has to be to $1$ in order to find a value for $f$ that's far away from $f_n$ (for some $n>N$). If we limit the domain of $f_n$ and $f$ even a little bit, say, $S=(-\delta, \delta)$ where $0<\delta <1$, eventually we can make $N$ so large that, in order to violate uniform convergence, the associated $x$ would have to be between $\delta$ and $1$. But then this $x$ would not be in $S$, so $f_n$ is uniformly convergent on $S$.