Prove or disprove the following statement:
Let H be a Hilbert space and M a closed subspace of H. For any $h \in H$ there exists some $w \in M$ such that $h - w \in M^{\perp}$ (the orthogonal complement).
My own attempt at a counter example (I believe it to be false since M is not linear):
Consider the hilbert space $H = \mathcal{L}_{[0,1]}^2({\mu})$ (that is, the functions on [0,1] with integrable second moment wrt. Lebesgue measure).
I then define $M := \{1_{[0,1]} \}$ and consider $h := x \in H$ and $w := 1_{[0,1]}$ Then we have:
$\langle h-w,w \rangle = \langle h,w \rangle - \langle w,w \rangle = \int_{[0,1]}x d\mu - \int_{[0,1]}1d\mu = -\frac{1}{2} $
Which shows that $h - w \notin M^{\perp}$, but is this correct?
Before you start looking for counterexamples consider that every closed subspace $M\subseteq H$ has an orthogonal complement $N\subseteq H$, which means that every vector $h\in H$ has a decomposition $h=m+n$ where $m\in M$ and $n\in N$, solving your problem. See https://en.wikipedia.org/wiki/Orthogonal_complement#Properties_2