1

While working through Helemskii's Lectures and Exercises in Functional Analysis, I have been asked to show by elementary methods that pointwise convergence in $C[0,1]$ is not metrizable. I can do this abstractly by topological methods, but am not sure what to do with the provided hint:

Hint. If the pointwise convergence implies the metric convergence in some metric $d$, then one can show that for every $t \in [0,1]$ and $\epsilon > 0 $ we have $d(y,0) < \epsilon$ when $y \in C[0,1]$ vanishes outside an interval $[t, t+h]$.

Assuming I am interpreting this correctly, the proof is immediate given the hint. I am not sure how one shows the claim in the hint though.

Might I request a nudge in the right direction?

Note: I can find a related question here for showing $C[0,1]$ is not a normed space, but this both bypasses the hint I would like to understand and would require showing the given metric is induced by a norm

Brevan Ellefsen
  • 15,317
  • To be clear, when you refer to "pointwise convergence", are you talking about convergence of nets or just sequences? – Eric Wofsey Jan 20 '20 at 23:35
  • @EricWofsey Hard to say since the book reviews nets in this same chapter and is not explicit here, but it appears to be talking about sequences. I presume here the author refers to the "pointwise convergence on $C[0,1]$" as being the case where $f \to g$ means $f(x) \to g(x)$ as sequences pointwise. – Brevan Ellefsen Jan 20 '20 at 23:40
  • 1
    To prove the hint: show that for every sequence $h_n$ converging to zero such that $t+h_n \leq 1$, for every continuous function $f$ vanishing outside $[t,t+h_n]$, $f_n$ converges pointwise to zero. – Aphelli Jan 20 '20 at 23:43

1 Answers1

1

Suppose there exist $t\in(0,1]$ and $\epsilon>0$ such that for all $h>0$ there exists $y_h\in C[0,1]$ which vanishes outside $[t,t+h]$ but $d(y_h,0)\geq \epsilon$. Taking a sequence $h_n\to 0$, can you show $y_{h_n}\to 0$ pointwise?

(Note that here I am excluding the case $t=0$. I don't see any direct way to prove the statement of the hint for $t=0$. In any case, the way I see to finish the problem using the hint still works fine without the case $t=0$.)

More details are hidden below.

Note that in fact by continuity, $y_h$ vanishes outside $(t,t+h]$ (here we use the assumption that $t>0$). Thus $y_{h_n}$ converges to $0$ pointwise, since any point is eventually outside $(t,t+h_n]$. Thus $d(y_{h_n},0)$ must go to $0$ as well which is a contradiction.

Eric Wofsey
  • 342,377