2
$\begingroup$

Consider for instance the sequence of continuous functions \begin{equation*} f_n(x)= \begin{cases} x^n,\quad x\in[0,1)\\ 1,\quad x=1 \end{cases}. \end{equation*} It converges pointwise to \begin{equation*} f_n(x)= \begin{cases} 0,\quad x\in[0,1)\\ 1,\quad x=1 \end{cases}, \end{equation*} which is not continuous. In other words, this examples shows us that pointwise convergence does not preserve continuity. On the other hand, one can prove that uniform convergence does preserve continuity.

This is why, when proving that the spaces of continuous functions $\mathcal{C}([a,b])$ with the $\mathcal{C}([a,b])$ norm are complete, we need uniform convergence. In particular, in that proof we identify the pointwise limit $f$ as the "natural candidate" and then prove that $f_n$ converges uniformly to $f$ on $[a,b]$. From this, it follows that $f\in\mathcal{C}([a,b])$ (a uniform limit of continuous functions is continuous) and that $f_n$ converges to $f$ in $\mathcal{C}([a,b])$. So in the end it's the pointwise limit $f$ (for which continuity might fail as in the example above) which happens to be continuous.

So I don't seem to understand how to put together the example above with the fact that $\mathcal{C}([a,b])$ is complete, because we have that $f_n\xrightarrow{||\cdot||_{\mathcal{C}([a,b])}} f$ which needs to be continuous but also $f_n\xrightarrow{|\cdot|} f$ with $f$ as above which is not continuous, and the limits are supposed to be the same. I wonder what my misunderstanding here is.

$\endgroup$
3
  • $\begingroup$ We do NOT have that $f_n\to f$ in the $\|\cdot \|$ metric. $\endgroup$ – DanielWainfleet Feb 17 '20 at 7:18
  • $\begingroup$ @DanielWainfleet In what sense we don't? I mean that $f_n$ converges to $f$ pointwise in the metric space $(\mathbb{R},|\cdot|)$. $\endgroup$ – Edu Feb 17 '20 at 9:46
  • $\begingroup$ The script above the 1st arrow in your 3rd-last line implies that $\sup_x \{|f(x)-f_n(x)|\}=\|f-f_n\|\to 0$, which is false. $\endgroup$ – DanielWainfleet Feb 17 '20 at 15:11
1
$\begingroup$

One short answer is that the two spaces $(C([a,b]),\|\cdot\|_{\infty})$ and (C([a,b]), p.w. convergence)

Will have different topologies. Remember a topology on a space will dictate which sequences will converge and which will not.

$\endgroup$
2
  • $\begingroup$ Then it has only to do with the metric? The sequence of functions that I mentioned in the example converges to the pointwise limit but only under the metric of the space of continuous functions? $\endgroup$ – Edu Feb 14 '20 at 18:48
  • $\begingroup$ yes it depends on the metric (which defines the topology) $\endgroup$ – guy3141 Feb 14 '20 at 19:13

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.