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

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.

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 $(a,b)$.

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 $r_n$.

• Do you mean a broken line going through $(r_n-2^{-n},0),(r_n,1),(r_n+2^{-n},0)$ ? How do i know they don't overlap (for instance $r_k\in (r_n-2^{-n},r_n+2^{-n})$ )? – Kulisty Sep 26 '15 at 20:48
• I think i know how to cope with it. Define $g_k$ as a broken line going through $(r_k-2^{-k},0), (r_k,1), (r_k+2^{-k},0)$ and zero on $\mathbb{R}\setminus (r_k-2^{-k},r_k+2^{-k})$. Then define $f_n=\max\{g_1,\ldots, g_n\}$. Is this the sequence of functions you meant? It's easy to show that it's not uniformly convergent on any $(a,b)$, but how to show $f_n\rightarrow 0$? – Kulisty Sep 26 '15 at 21:06
• I am wrong. I must define $f_n(x)=\sup\{g_k(x):k\ge n\}$ – Kulisty Sep 26 '15 at 21:26

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.

• I forgot to mention that $f_n$ must be continuous. Sorry. – Kulisty Sep 26 '15 at 20:21

This has occurred here recently.

Let $f_n(x) =x(1-x)^n$. Then, on $[0, 1]$, $0 \le f_n(x) \le 1$.

$\begin{array}\\ f_n'(x) &=(1-x)^n-nx(1-x)^{n-1}\\ &=(1-x)^{n-1}(1-x-nx)\\ &=(1-x)^{n-1}(1-(n+1)x)\\ \end{array}$

so $f_n'(x) = 0$ at $x_n = \frac1{n+1}$. There,

$\begin{array}\\ f_n(x_n) &=\frac1{n+1}(1-\frac1{n+1})^n\\ &=\frac1{(n+1)(1+1/n)^n}\\ &\approx\frac1{e(n+1)}\\ \end{array}$

Therefore, if $F_n(x) =nf_n(x)$, then $F_n(x)$ goes pointwise to zero on $[0, 1]$ and $F_n(x_n) \to \frac1{e}$.