If $U\subset W\subset V$ and $W\cap U^\perp = \{0\}$ then $U=W$

Question

Statement :

Let $V$ be a Hilbert space. Let $U\subset W\subset V$ be closed subspaces. Suppose that $W\cap U^\perp = \{0\}$. Then $U=W$.

I know this is true in the finite dimensional case (see proof below). Is this true in the infinite dimensional case? Under which added hypothesis is it true? Or is there a counter-example?

Proof :

Because $W$ is closed in the Hilbert space $V$, we have $V=W\oplus W^\perp$. To show that $W=U$, it suffices to show that $V=U\oplus W^\perp$. That is, we have to show two things :

1. $U\cap W^\perp = \{0\}$
2. $U + W^\perp = V$

The first equality is direct : $$ U\subset W \implies U\cap W^\perp \subset W\cap W^\perp = \{0\} $$ It remains to show the second equality. We have : $$ W\cap U^\perp = \{0\} \\ \implies (W\cap U^\perp)^\perp = \{0\}^\perp \\ \implies W^\perp + U^{\perp\perp} = V \quad \text{(if $V$ finite dimensional)} \\ \implies W^\perp + U = V \quad \text{(because $U$ closed)} \\ \implies U + W^\perp = V $$ QED

Answer 1

Since $U$ is a closed subspace, any element $v \in V$ may be written in the form $$v = u + u^\perp$$ where $u \in U$ and $u^\perp \in U^\perp$.

Let $w \in W$ and express $w = u + u^\perp$. Since $U \subset W$ implies $u \in W$ you have $u^\perp = w - u \in W$.

Thus $u^\perp \in U^\perp \cap W$ so that $u^\perp = 0$. Thus $w = u \in U$ so that $W \subset U$.

Answer 2

Here is a proof. Let $w \in W$ be given. Let $u$ be the projection of $w$ onto $U$.

Then, $w - u \in W$ since $u,w \in W$ and $w - u \in U^\perp$ by the properties of the projection. Hence, $w - u = 0$ and this shows $w \in U$.