Projection of a vector on orthogonal subspaces

Question

Let $H$ be a Hilbert space, $G\subset H$ a subspace, i.e. a complete linear manifold, and $$F = \{ f'\in H\mid\forall g'\in G:(f',g')=0 \}$$ the orthogonal complement of G.

Consider a vector $h\in H$. Since $G$ is a subspace, there is one and only one projection $g\in G$ of $h$ onto $G$. I have already proved that $F$ is a subspace, too. So there is one and only one projection $f\in F$ of $h$ onto $F$.

I want to prove that $$h=g+f.$$ So, I have already shown that $h-g\in F$. Now, I have to prove that $h-g$ is the projection of $h$ on $F$. How can I prove it?

Answer 1

What you have shown is that $h = g + f'$ for some $f'\in F$. You can similarly show that $h = g'+f$ for some $g'\in G$ (since $G$ is the orthogonal complement of $F$). Then $g+f'=g'+f $, which means that

$$ g-g' = f-f' \in G\cap F$$ so $(\underbrace{g-g'}_{\in G},\underbrace{g-g'}_{\in F}) = 0$, by the definition of $F$. Hence, $g-g'=0=f-f'$ and $f'=f$, and thus $h=g+f$.

Answer 2

Projection $Ph$ of $h$ onto $F$ is characterized by the properties

• $Ph \in F$
• $h - Ph \perp F$, that is $h - Ph \in G$

For $h-g$ we indeed have $h - g \in F$ and $h - (h-g) = g \in G$ so $Ph = h-g$.