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Does there exist a pointwise convergent series on $\mathbb{R}$, but not uniformly convergent at any interval?

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closed as off-topic by Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel Jul 23 '18 at 7:54

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    $\begingroup$ What does it mean for a series to be uniformly convergent 'at a point'? $\endgroup$ – Fimpellizieri Jul 22 '18 at 20:55
  • $\begingroup$ Sorry , it just a mistake , I have modified it $\endgroup$ – ArMaRm Jul 22 '18 at 21:14
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Let $\mathbb{Q} = \{ r_1, r_2, \ldots\}$ be an enumeration of the rational numbers and take

$$f_n(x) = \begin{cases}1, &x = r_n \\ 0, & \text{otherwise} \end{cases}$$

We have pointwise convergence of the series

$$\sum_{n=1}^\infty f_n(x) = f(x) =\begin{cases}1, &x \in \mathbb{Q} \\ 0, & \text{otherwise} \end{cases}$$

The rational numbers in any interval $[a,b]$ can be enumerated by a subsequence $(r_{n_k})$. The series does not converge uniformly on the interval because for any $k \in \mathbb{N}$ we have $n_k \geqslant k$ such that

$$\left|\sum_{n=1}^{n_k} f_n(r_{n_{k+1}}) - f(r_{n_{k+1}})\right| = 1, $$

and it is not possible to find for any $0 < \epsilon < 1$ an integer $k$ such that for all $m >k$ and all $x \in [a,b],$

$$\left|\sum_{n=1}^{m} f_n(x) - f(x)\right| < \epsilon$$

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If you mean "Does there exist a sequence of functions $f_n:\Bbb R\to\Bbb R$ which converges pointwise, but doesn't converge uniformly on any interval?", then yes, there are such sequences.

For instance, let $f$ be the Conway base 13 function. It is a function which attains all real numbers as value on any (non-trivial) interval. Then set $f_n(x)=\frac1n f(x)$. Each $f_n$ still has the property that on any interval it attains any real number as value, so the sequence cannot possibly converge uniformly on any interval. But the sequence clearly converges pointwise to the zero function.

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  • $\begingroup$ Let $f_n$ be a sequence of functions , and $S_n$ = $f_1$ + $f_2$ + .... be the partial sum of the series $Z$ . If the limit of $S_n$ converges P.W then $Z$ converges P.W , similarly for U.C . $\endgroup$ – ArMaRm Jul 22 '18 at 21:37
  • $\begingroup$ From any sequence, you can easily make a series with that sequence as partial sums. So the existence of one implies the existence of the other. $\endgroup$ – Arthur Jul 22 '18 at 22:04
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This answer constructs a sequence $(f_n)_n$ of continuous functions $f_n : \mathbb{R} \to \mathbb{R}$ which converges pointwise to Thomae's function $f: \mathbb{R} \to \mathbb{R}$ given by

$$f(x)= \begin{cases} \frac1q &: \text{ if $x = \frac{p}{q}$ with $p \in\mathbb{Z}, q \in \mathbb{N}$ coprime}\\ 0 &: \text{ if $x$ is irrational} \end{cases}$$

The set of discontinuities of $f$ is precisely $\mathbb{Q}$.

If $f_n|_{[a,b]} \to f|_{[a,b]}$ uniformly for some interval $[a,b]$ then it would follow that $f|_{[a,b]}$ is continuous, which is impossible because the set of discontinuities $\mathbb{Q}$ is dense in $\mathbb{R}$.

Therefore the convergence $f_n \to f$ is not uniform on any interval.

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