Sequence of functions converges

Question

I'm searching for sequence of functions that converge pointwise on $\mathbb{R}$ but not uniformly on any interval of $\mathbb{R}$.

Any ideas?

Answer 1

Contrary to my original suspicion, a satisfactory sequence of continuous functions exists; see this link from the comment below

One example of a sequence of discontinuous functions is as follows: define $$ f(x) = \begin{cases} q &\text{if }x=\tfrac{p}{q}\quad (x \text{ is rational), with }p \in \mathbb Z \text{ and } q \in \mathbb N \text{ coprime}\\ 0 &\text{if }x\text{ is irrational.} \end{cases} $$ and take the sequence $f_n(x) = \frac{f(x)}{n}$. The important fact about $f$ here is that it is unbounded on every interval.

Answer 2

Since $\Bbb Q$ is countable, there is a bijection $f:\Bbb Q \to \Bbb N$.

Define a sequence of functions $$ g_n(x) = \begin{cases} 0,&x\notin\Bbb Q,\\ 1, & x\in\Bbb Q,\quad f(x)\ge n,\\ 0, & x\in\Bbb Q,\quad f(x)< n \end{cases} $$

The sequence $\{g_n\}_{n\in\Bbb N}$ converges to $0$ pointwisely on $\Bbb R$, but not uniformly on any interval.

Answer 3

It is actually possible to define a sequence of continuous functions with this property. One example is as follows.

We aim to get a sequence of functions which converge to the function given by $f(p/q)=1/q$ if $p/q$ is in lowest terms with $q>0$, $f(x)=0$ if $x$ is irrational.

To do this, let $f_n$ be the function defined by $f(p/q)=1/q$ if $p/q$ in lowest terms with $0<q\leq n$, and $f_n$ is piecewise linear between these points. Each $f_n$ is continuous, and certainly $\lim_{n\to\infty}f_n(p/q)=1/q$ for each $p,q$ coprime.

Now we just need to show that $\lim_{n\to\infty}f_n(x)=0$ if $x$ is irrational. Suppose not, so there is some $x$ irrational with $f_n(x)>\delta>0$ infinitely often. Now let $m=\lfloor 1/\delta\rfloor$, and suppose the nearest rational with denominator at most $m$ to $x$ is at distance $\epsilon$ from $x$. Now for any $n>1/\epsilon$ we have rationals with denominator $n$ which are closer to $x$ (on both sides) than $\epsilon$. Thus the closest rationals on both sides with denominator $\leq n$ are closer than any rational of denominator $\leq m$; it follows by piecewise linearity that $f_n(x)<1/m<\delta$, contradiction.