Construct a sequence of functions which is pointwise convergent to zero and not uniformly convergent on any interval.

Question

Construct a sequence of functions $f_n:\mathbb{R}\rightarrow \mathbb{R}, n\in \mathbb{N}$ , which is pointwise convergent to $f(x)=0 , x\in \mathbb{R}$ and not uniformly convergent on any interval $(a,b)$.

I noticed that it suffices to construct a sequence of functions $f_n:[0,1] \rightarrow \mathbb{R}, n\in \mathbb{N}$ that fulfills the requirements and just copy it.

EDIT. I forgot to mention that $f_n$ must be continuous.

Answer 1

Let $\{r_n\}_{n\geq 1}$ be an enumeration of the rationals, and define $f_n(x)$ as follows: $$ f_n(x)=\begin{cases} 1,& x\in \{r_k\}_{k\geq n}\\ 0,& else \end{cases} $$ Then $f_n\to 0$ pointwise because for every $x$ and all sufficiently large $N$, we have $f_N(x)=0$. On the other hand, $\{f_n\}$ does not converge uniformly on any interval $(a,b)$. If it did, the limit would have to be the zero function (since this is the pointwise limit). However there are infinitely many rationals in $(a,b)$, so therefore $\sup_{x\in (a,b)}|f_n(x)|=1$ for all $n$. Consequently there is no uniform convergence on any interval $(a,b)$.

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If you want to make the $\{f_n\}$ continuous, you can modify this example by using bump functions supported on $(r_n-2^{-n},r_n+2^{-n})$ in place of the spikes to $1$ at $r_n$.

Answer 2

Start with your favourite countably infinite partition of $\mathbb{R}$ into dense sets $R_k$, $k\in\mathbb{N}$.

Then set $f_n=0$ on all $R_k$ with $k\leq n$ and $f_n=1$ otherwise.