6

If we consider $PC[0,1]$ as a subset of $L^2[0,1]$, is it complete when equipped with the $L^2$ norm? I have been trying to prove this for some time but did not get very far. The search for a counterexample has also proved fruitless. I would therefore be grateful for some help.

Thank you.

JohnK
  • 6,444
  • 4
  • 28
  • 54

3 Answers3

5

This is false. Let $C$ be a 'fat' Cantor set of positive measure. Since $C$ is closed we can find continuous functions $f_n$ with values in $[0,1]$ such that $f_n \to I_C$ pointwise. By Bounded Convergence Theorem $f_n \to I_C$ in $L^{2}[0,1]$. But $I_C$ is not piecewise continuous.

Kavi Rama Murthy
  • 311,013
  • That's a great counterexample, thank you. – JohnK Apr 05 '19 at 10:15
  • @Kavi I rly like your example! Would greatly appreciate if you could help me with my problem :) https://math.stackexchange.com/questions/3178840/functional-analysis-inner-product-and-haar-rademacher-walsh-functions – Homaniac Apr 07 '19 at 22:06
3

If you now that $C[0,1]$ is dense in $L^2(0,1)$, there is another simple approach: Since $PC[0,1]$ contains $C[0,1]$, the closure of $PC[0,1]$ is $L^2(0,1)$. From $PC[0,1] \ne L^2(\Omega)$, this implies that $PC[0,1]$ is not closed. Thus, $PC[0,1]$ cannot be complete.

gerw
  • 31,359
2

If $P[0,1]$ were not dense in $L^2[0,1]$, then there would exist $f\in L^2[0,1]$, $f\ne 0$, such that $\langle f, \chi_{[0,a]}\rangle = 0$ for all $0 \le a \le 1$. But that would imply that $\int_0^a f(t)dt=0$ for all $a$. By the Lebesgue Differentiation Theorem, $\frac{d}{da}\int_0^1 f(t)dt=f(a)$ a.e., which would imply that $f=0$ in $L^2[0,1]$ and give a contradiction. Therefore $P[0,1]$ is dense in $L^2$.

Disintegrating By Parts
  • 87,459
  • 5
  • 65
  • 149
  • @ Disintegrating I rly like your example! Would greatly appreciate if you could help me with my problem :) https://math.stackexchange.com/questions/3178840/functional-analysis-inner-product-and-haar-rademacher-walsh-functions – Homaniac Apr 07 '19 at 22:05