1

Background to the question: I want to find dense subsets $U, V$ of the Hilbert Space $L^{2}([0,1])$ so that $U\cap V=0$. One candidate is obviously $U:=$the space of polynomials on $[0,1]$. And now I have seen that the other candidate would be $V=(f_{n})_{n}$

where $f_{n}(x)=\sin{(nx)}$ and $x \in [0,1]$. But I struggle to see why in this case $U \cap V=0$ and further why $V$ is in any case dense in $L^{2}([0,1])$

Any ideas?

SABOY
  • 1,828
  • 1
    Do you really struggle to see why $U \cap V=0$? Also, your $V$ isn't dense, in the same way ${x^n}$ isn't dense. – Arthur Jul 08 '19 at 08:40

3 Answers3

1

$V$ is not dense. A better solution to finding disjoint dense sets would be to consider your $U$ and the set $\{e^{x}p(x):p \,\text {is a polynomial} \}$. Can you verify both the facts in this case?

Kavi Rama Murthy
  • 311,013
1

Consider instead your $U$ and let $V$ be the space of nonconstant simple functions on $[0,1]$. Then $V$ is dense in $L^2[0,1]$ and $U \cap V = \emptyset$.

mechanodroid
  • 46,490
1

You could also let $U$ be the set of polynomials with rational coefficients, $V$ the set of polynomials with irrational coefficients.

zhw.
  • 105,693