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I'm looking for an example of a sequence of real-value functions $\left\{ f_{n}:\mathbb{R}\to\mathbb{R}\ |\ n\geq1\right\}$ which converges locally uniformly to some function $f$ on some interval $\left(a,b\right)\subseteq\mathbb{R}$ but does not converge uniformly there. Just for clarity, the definitions involved are:

A sequence of functions $f_{n}$ is said to converge uniformly to a function $f$ on $\left(a,b\right)$ if $$\sup\limits _{x\in\left(a,b\right)}\left|f_{n}\left(x\right)-f\left(x\right)\right|\overset{n\to\infty}{\longrightarrow}0$$

The sequence is said to converge locally uniformly to $f$ on $\left(a,b\right)$ if for each $x\in\left(a,b\right)$ there is a neighborhood $U$ of $x$ such that $f_{n}$ converges uniformly to $f$ on $\left(a,b\right)\cap U$ .

I did some searching and I found a couple of places that suggest looking at $f_{n}\left(x\right)=x^{n}$ on $\left(0,1\right)$ but it seems to me this sequence simply converges uniformly to $f\equiv0$ on the entire open interval.

JamesM
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    It doesn't converge uniformly on $(0,1)$. You have $\sup_{x \in (0,1)} f_n(x) = 1$ for all $n$. – Umberto P. Sep 09 '15 at 15:36
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    You want $f_n : \mathbb R \to \mathbb R$ not to be uniformly convergent and also the restricted functions $f_n \big| _{(a,b)} : (a,b) \to \mathbb R$ to be uniformly convergent for every interval $(a,b)$, right? Then $f_n(x)= \frac x n$ works. – ThePortakal Sep 09 '15 at 15:40

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Consider the following sequence of functions $f_n:\mathbb{R}\to\mathbb{R}$.$$f_n(x)=\left\{\begin{array}{rc}-n&x<n\\x&-n\leq x\leq n\\n&n<x\end{array}\right..$$This sequence converges uniformly locally to $f(x)=x$, but does not converge uniformly to any function.

Apply any homeomorphism $\mathbb{R}\to(0,1)$, and get your desired sequence.

Amitai Yuval
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$f_n(x)=x^n$ does not converge uniformly to zero on $(0,1)$. It does converge pointwise to zero on the entire interval, but the convergence is very slow near $1$.

More precisely, to have $x^n<\varepsilon$ it is equivalent to have $n \ln(x)<\ln(\varepsilon)$ so $n>\frac{\ln(\varepsilon)}{\ln(x)}$. As $x \to 1^-$ the denominator goes to zero, so you do not have globally uniform convergence. On the other hand, on compact subsets $\ln(x)$ is bounded away from zero, and so you have locally uniform convergence.

Note that if it did converge uniformly, then we would have

$$\lim_{n \to \infty} \lim_{x \to 1^-} x^n = \lim_{x \to 1^-} \lim_{n \to \infty} x^n.$$

But that equality reads $1=0$.

Ian
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