1

Let $V_j=\{f\in L^2(\mathbb{R}): f$ is constant on $[\frac{n}{2^j},\frac{n+1}{2^{j}}) $for all $ n \in\mathbb{Z}\}$ , $j\in\mathbb{Z}$ be a sub set of $L^2(\mathbb{R}).$ Prove that $V_j$ is an closed linear subspace of $L^2(\mathbb{R})$.

Attempt: I proved the set is linear space. To prove the closed. I took the sequence $f_n$ in $V_j$ such that $f_n$ converges to $f$. I want to prove $f$ is in $V_j$. That is i want to prove $f$ is constant on $[\frac{n}{2^j},\frac{n+1}{2^{j}}).$ I don't know to how to proceed help me!

Murugan
  • 54

1 Answers1

0

The question should really say "constant almost everywhere" (a.e.) rather than "constant", because members of $L^2$ are equivalence classes rather than actual functions.

Hint: if $f$ is not a.e. constant on that interval (call it $I_n$), there are two disjoint intervals (if the scalars are real) or balls (if the scalars are complex) $J_1$ and $J_2$ such that $\{x \in I_n: f(x) \in J_1\}$ and $\{x \in I_n: f(x) \in J_2\}$ both have positive measure.

Robert Israel
  • 448,999