Let $V$ be a vector space with a inner product $\langle\,,\rangle$. The orthogonal complement of a subspace $U\subseteq V$ is defined to be $ U^\perp = \{ x \in V \,|\, \langle x,y\rangle =0 \text{ for all }y\in U \}$.
My question is:
Does $( U_1 \cap U_2 ) ^\perp = U_1^\perp +U_2 ^\perp$ hold in infinite dimensions? And if not, does anyone have a counterexample?
The inclusion $( U_1 \cap U_2 ) ^\perp \supseteq U_1^\perp +U_2 ^\perp$ follows from the definition of the orthogonal complement.
To get the other inclusion I proved that $( U_1 + U_2 ) ^\perp = U_1^\perp \cap U_2 ^\perp$ again just by following the definitions of the orthogonal complement. Then I want to apply that $U=U^{\perp ^\perp}$ holds in a finite dimensional vector space $V$, to get that:
$(U_1 \cap U_2 )^\perp =\left( U_1^{\perp \perp} \cap U_2^{\perp \perp} \right)^\perp = \left( \left(U_1^\perp + U_2^\perp \right)^\perp \right)^\perp = U_1^\perp + U_2^\perp $
But I do not see a way to avoid using a fact like $U=U^{\perp ^\perp}$, which only holds for a finite dimensional vector space.
Best Answer
Take $V = \ell^2(\mathbb{N})$, $U_1$ the linear span of the vector $(1,\frac{1}{2},\frac{1}{4},\ldots)$ and $U_2$ the set of vectors with finite support. Then clearly $U_1 \cap U_2 = \{0\}$ so that $(U_1 \cap U_2)^{\perp} = V$. But $U_2^{\perp} = \{0\}$ and $U_1^{\perp} \neq V$.