0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $r_n$.

$\endgroup$
  • $\begingroup$ 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 $r_k\in (r_n-2^{-n},r_n+2^{-n})$ )? $\endgroup$ – Kulisty Sep 26 '15 at 20:48
  • $\begingroup$ 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 $f_n\rightarrow 0$? $\endgroup$ – Kulisty Sep 26 '15 at 21:06
  • $\begingroup$ I am wrong. I must define $f_n(x)=\sup\{g_k(x):k\ge n\}$ $\endgroup$ – Kulisty Sep 26 '15 at 21:26
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I forgot to mention that $f_n$ must be continuous. Sorry. $\endgroup$ – Kulisty Sep 26 '15 at 20:21
0
$\begingroup$

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 $F_n(x_n) \to \frac1{e}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.