Let $H=L^2(\mathbb{R}^3, \mathbb{R}^3)$ be a Hilbert space and $V=\{u \in H \mid \langle x, u(x) \rangle = 0 \text{ for a.e. } x\in \mathbb{R}^3\}$ a subspace.
Is $V$ closed? What is the orthogonal projection of $H$ onto $V$?
I think V consists of "rotations"? At least that's how it would work in the $\mathbb{R}^2$ case.
Now consider: $\varphi : H \rightarrow \{\text{I don't know}\}, f \mapsto \langle f(\cdot), \cdot \rangle$
If this is a continuous well defined function then $V$ is the preimage of $0$ and hence closed. However I have no idea how to show this. And how can one approach finding the projections?
Regarding the codomain of $\varphi$, the inner product is continous,so also integrable?
It is closed. Indeed, assume $u_n \in V$ and $u_n \to u \in H$. Then there exists a subsequence $u_{n_m} \to u$ a.e. But then, for almost every $x \in \mathbb{R}^3$,
$$\langle x, u(x) \rangle = \lim_m \langle x, u_{n_m}(x) \rangle = 0$$
That is, $u \in V$.
The projection $P$ onto $V$ is given by the following map: For $u \in H$, $x \in \mathbb{R}^3$,
$$[P(u)](x) = u(x) - \frac{\langle x, u(x)\rangle x}{\langle x, x \rangle}$$
Indeed, since $u(x) \mapsto u(x) - \frac{\langle x, u(x)\rangle x}{\langle x, x \rangle}$ is simply the orthogonal projection of $u(x)$ onto $\mathrm{span}\{x\}^\perp$, we see that $\|[P(u)](x)\| \leq \|u(x)\|$, so,
$$\|P(u)\|^2 = \int \|[P(u)](x)\|^2 \, dx \leq \int \|u(x)\|^2 \, dx = \|u\|^2$$
Thus, $P$ is well-defined and in fact a contraction. It is easy to verify that $P(u) \in V$ for all $u \in H$ and also $P(u) = u$ for all $u \in V$. Hence, $P$ must be the orthogonal projection onto $V$.
Note that this also gives an alternative proof that $V$ is closed.
Remark: Observe that nowhere in the proof is any property specific to $\mathbb{R}^3$ used. Thus, the proof will actually work for any (separable) inner product space equipped with a (Borel) measure.