In my homework I have a Hilbert space with orthonormal basis $(e_n)_{n=1}^\infty$. Let $E=\{e_1,e_2\}^\perp$
then I have to show that: $$P_Eh=h- \langle h,e_1 \rangle_1 - \langle h,e_2 \rangle e_2$$
I know that $E$ is closed but not sure how that helps. I also know that for all $e_n$ with $n>2$ they are perpendicular to $E$.
Any hint would be appreciated
On
As the previous answer states the $P_E(h)$ is the unique element such that
$P_E(h)\in E$ and $(h-P_E(h))\in E^{\perp}$ holds.
In our case $E=\{e_1,e_2\}^{\perp}$.
Lets first notice that
$$\{e_1,e_2\}\in\{e_1,e_2\}^{\perp \perp}$$ which implies that
$$span\{e_1,e_2\}\subset \{e_1,e_2\}^{\perp \perp}$$
Notice then that
$$\{e_1,e_2\}\subset span\{e_1,e_2\}$$ $$\implies span\{e_1,e_2\}^{\perp} \subset \{e_1,e_2\}^{\perp}$$ $$\implies \{e_1,e_2\}^{\perp \perp} \subset span\{e_1,e_2\}^{\perp \perp}=span\{e_1,e_2\}$$
We then conclude that
$$span\{e_1,e_2\}=\{e_1,e_2\}^{\perp \perp}=E^{\perp}$$
Now, remembering that since we are dealing with an Orthonormal System we can write:
$$h=\sum_{j=1}^{\infty}\langle h,e_j\rangle e_j $$ we see that $$P_Eh=h- \langle h,e_1 \rangle_1 - \langle h,e_2 \rangle e_2=\sum_{j=1}^{\infty}\langle h,e_j\rangle e_j - \langle h,e_1 \rangle_1 - \langle h,e_2 \rangle e_2$$ $$=\sum_{j=3}^{\infty}\langle h,e_j\rangle e_j\in span\{e_j,j\in\mathbb N,j>2\}=span\{e_1,e_2\}^{\perp}$$
$$P_Eh\in E\supset span\{e_1,e_2\}^{\perp}$$
And now $$h-P_Eh=\langle h,e_1 \rangle_1 +\langle h,e_2 \rangle e_2$$ which clearly belongs to the $span\{e_1,e_2\}=E^{\perp}$
We conclude that
$$P_E(h)=(h- \langle h,e_1 \rangle_1 - \langle h,e_2 \rangle e_2) \in E$$ and $$h-P_E(h)=(\langle h,e_1 \rangle_1 +\langle h,e_2 \rangle e_2)\in E^{\perp}$$
And with that we are done.
By definition $P_E(h)$ is the unique element of $E$ such that $h=P_E(h)+u_h$ where $u_h$ is orthogonal to $E$.
$\langle h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2,e_i\rangle=0, i=1,2$ implies that $P_E(h)\in E$ and $u_h=-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2$ is orthogonal to $E$.