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I was wondering if the following result is true:

Let $\{f_n\}$ be a sequence of functions on $(0,\infty)$ and $f$ also has the same domain, $(0,\infty)$. Suppose that $f_n$ is uniformly convergent to $f$ on $[\epsilon, N]$ for every $\epsilon>0$ and $N \in \mathbb{N}$ satisfying $\epsilon<N$.

A hint will be appreciated. At the moment, all I can see is that given $\epsilon > 0$ and given $x \in (0,\infty)$, $x\in [x,\lfloor x \rfloor +1]$ and so there exists some $N \in \mathbb{N}$ such that $\forall n \ge N, |f_n(x)-f(x)|<\epsilon$. This does prove that $f_n$ goes to $f$ pointwise but I don't see how this implies uniform convergence. Thanks.

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Uniform convergence on all compacts does not imply uniform convergence on $(0,\infty)$.

Consider for instance, for $0<x\leq1$

$$f_n(x)=\sum_{k=0}^n (1-x)^k$$

Anf $f_n(x)=1$ for all $x>1$.

The series converges to $f(x)=\dfrac1x$ for $0<x\leq1$, but it diverges to $+\infty$ for $x=0$, hence you can't have uniform convergence on $(0,+\infty)$: for a given $\epsilon>0$, then for any $n$ you can always find an interval near zero such that $|f(x)-f_n(x)|>\epsilon$, since $f_n$ is bounded near $0$, whereas $f$ is not.

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No. Given those assumptions on $f_{n}$ and $f$, it does not necessarily follow that $f_{n} \to f$ uniformly on $]0, +\infty[$. A handy example is $f_{n}: x \mapsto \frac{1}{nx}$ on $]0, +\infty[$ for all $n$ and $f:= 0$ on $]0, +\infty[$. I claim that $f_{n}$ does not converge to $f$ uniformly. For, there is some $\varepsilon > 0$, say $\varepsilon := 1$, such that for every $N$ there are some $n \geq N$, say $n := N$, and some $x \in ]0, +\infty[$, say $x := 1/2N$, such that $$ \frac{1}{nx} = 2 > \varepsilon. $$

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Note: When I speak of "$\epsilon$" I'm referring to the one used in $\epsilon-N$ definitions.

It's not true. Suppose $f_n$ is a sequence of functions which pointwise (but not uniformly) converges to some continuous $f$ on $(0,\infty)$. (Choose any such sequence, it's up to you.) Then you can show that $f_n$ converges uniformly on any closed interval $[a, b]$, essentially because you can take the smallest $\epsilon$ from each point considered, and such an $\epsilon$ exists because of the extreme value theorem.

They key is the order in which things are defined. If a function is uniformly convergent for each closed interval to which it is restricted, you first define the interval and then the $\epsilon$.

If instead, for each $\epsilon >0$, you could find an $N$ such that $n>N \Rightarrow |f_n(x)-f(x)|$ for each $x\in[a,b]$ for each $[a,b]$, this would imply uniform convergence on $(0,\infty)$. However this is obvious and boring, because the union of all closed intervals greater than zero is precisely $(0,\infty)$.

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    $\begingroup$ For your first statement, you have to assume that the pointwise limit is continuous. Otherwise, you don't necessarily have uniform convergence on every compact. $\endgroup$ Apr 30, 2017 at 9:34
  • $\begingroup$ yes you're right, I'll fix that. Thanks. $\endgroup$
    – Harambe
    Apr 30, 2017 at 9:38

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