My colegues and me was trying to solve the following exercise:
Let $\mathcal{B}=\{e_k;k\in I\}$ an orthonormal subset of the Hilbert space $X$. Let consider $Z=\overline{\text{span}(\mathcal{B})}$. Show that the orthogonal projection $P\colon X\to Z$ is given by $$Px=\sum_{k\in I}\langle x,e_k\rangle e_k.$$
Some ideas: Since $Z$ is a closed subspace of a Hilbert space, we have that $Z$ is complete. Then, for each $x\in X$, $Px$ is the unique element of $Z$ which satisfies $$x-Px\perp Z\qquad\Longleftrightarrow\qquad\langle x-Px,z\rangle=0,\ \forall z\in Z.$$ In particular, we have that $$\langle x-Px,e_k\rangle=0,\ \forall k\in I.$$ The next step is try to write $Px$ as some combination of the $e_k$'s and make the computations. But we can't finish this, because if $Px\in Z$, then there exists $(y_n)\subseteq\text{span}(\mathcal{B})$ such that $y_n\to Px$ in the norm induced by the inner product.
Maybe a important fact that we could use here is that:
- $\{k\in I;\langle x,e_k\rangle\ne0\}$ is countable.
I have some computations about the $y_n$'s, but since I don't know if it's fine, I will not write here right now.
Thanks for any help!
Let $y = \sum\limits_{k=1}^{\infty} \langle x, e_k \rangle e_k$. Then $y \in Z$. We have to show that $\langle (x- y), z \rangle =0$ for all $z \in Z$. By continuity of the inner product in the second variable it is enough to show that this holds when $z$ is a finite linear combination of $e_k$'s. By linearity of inner product in the second variable it is enough to show that this holds when $z=e_j$ for some $j$. But in this case $\langle (x- y), z \rangle=\langle x, e_j \rangle-\langle y, z \rangle=\langle x, e_j \rangle-\langle x, e_j \rangle=0$.
[$\langle y, e_j\rangle =\lim_{N \to \infty} \langle \sum\limits_{k=1}^{N} \langle x, e_k \rangle e_k ,e_j \rangle$ and $\langle \sum\limits_{k=1}^{N} \langle x, e_k \rangle e_k ,e_j \rangle=0$ if $N < j$ and $\langle x, e_j \rangle$ if $N\geq j$].