Let $W_j\subseteq L²(ℝ)$ also Hilbertspaces. If I would like to show $$L²(ℝ)=\overline{\oplus_{j\in ℤ}W_j},$$ can I instead show that $$L²(ℝ)=\overline{span\left(\bigcup\limits_{j\in \mathbb{Z}}^{}W_j\right)}?$$
2026-03-27 14:57:05.1774623425
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Notation orthogonal direct sum
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No. The direct sum sign $\oplus$ usually includes some kind of unique decomposition of the base space into the subspaces. That is, if $V = U_1\oplus U_2$ then each element in $V$ can be uniquely decomposed into a sum of elements of $U_1$ and $U_2$. Just proving $V=span( U_1\cup U_2)$ is not sufficient.
Yes. A subspace is dense iff any function orthogonal to each element of the subspace is necessarily $0$. And a function is orthogonal to a union of subspaces iff it is orthogonal to their direct sum.