1
$\begingroup$

We have a function $f$ and a sequence of functions $f_n$, both on $[a,b] \to \mathbb{R}$. $f_n$ is continuous for each $n \in \mathbb{N}$, and $f_n$ converges pointwise to $f$. I am asked to give an example to show this does not imply that $f$ is continuous.

My thinking so far is that the example must break down on the boundaries, since within $(a,b)$, we pick the same $\epsilon>0$ for both definitions of continuity and pointwise convergence, we get an epsilon-delta rectangle about each point which both $f_n(x)$ and $f(x)$ must be in, so as epsilon shrinks we can define a slightly larger epsilon to get continuity of $f$. I can't get an example where it breaks though.

Is my reasoning sound, and any hints for getting an example that breaks this?

$\endgroup$
  • 1
    $\begingroup$ $$f_n:[0,1]\ni x\mapsto x^n$$ $\endgroup$ – Vim Apr 11 '17 at 11:02
  • 1
    $\begingroup$ The issue doesn't need to happen on the boundary, but it can. Think about $[0,1]$ and the function $f_n(x)$ is $1$ for $x<1/n$, the function is $0$ for $x>2/n$, and connected by a line between those two regions to make the function continuous. $\endgroup$ – Michael Burr Apr 11 '17 at 11:06
  • $\begingroup$ It could also break at points other than the boundary. An example can be easily constructed from Vim's. Define the functions on $[0, 1]$ as Vim's and reflect them into $[1, 2]$. $\endgroup$ – badjohn Apr 11 '17 at 11:06
0
$\begingroup$

As pointed out in the comments, the problem may be at boundaries or it may not.

  • Considering $a=0$, $b=1$ and $f_n\colon x\mapsto x^n$, the sequence converges pointwise to the function $f$ such that $f(x)=0$ for $0\leqslant x\lt 1$ and $f(1)=1$, which is not continuous.

  • Let $a=0$, $b=1$, $f_n$ equal to $1$ on $[0,1/2-1/n)$, $-1$ on $(1/2+1/n,1] $ and linear on $(1/2-1/n,1/2+1/n)$. Then $f_n$ is continuous and converges pointwise to $f$, which equals $1$ on $[0,1/2)$, $-1$ on $(1/2,1]$ and $0$ at $1/2$, which is not continuous.

$\endgroup$

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.