Consider $\ell^2(\Bbb C)=\{x=(x_n)_{n\in \Bbb N}|\, \sum_{n\ge 1}|x_n|^2<+\infty\}$ endowed with the usual inner product and the subspace $V:=\{x=(x_n)_{n\in \Bbb N}\in \ell^2 |\, x_1=2x_2-x_3\}$. Find an explicit formula for the projection on $V$.
My attempt
First of all, I considered the map $f:\ell^2\to \Bbb C$ defined by $f(x)=x_1-2x_2+x_3$, since it's continuous, then $V=f^{-1}(\{0\})$ is a closed set. Given $x=(x_n)_{n\in \Bbb N}$, we have $x=(x_1,x_2,\dots)=(2x_2-x_3,x_2,\dots)=(2x_2,x_2,\dots)-(x_3,0,\dots)$, I tried to use this property of the elements of $V$ to have more information about the inner product: $$\langle x,y\rangle=\sum_{n\ge 1} x_n \overline{y_n}=2x_2\overline{y_1}+x_2 \overline{y_2}+\dots -x_3 y_1 $$ Other than that, I've no idea on how to find an explicit formula for the projection.
Note that here I'm using the convention $\Bbb N=\{1,2,\dots\}$.
The subspace can be described as $V=v^\perp$ where $v=(1,-2,1,0,\ldots,0,\ldots).$ In general the orthogonal projection on $V=v^\perp$ is given by $$Px=x-{\langle x,v\rangle\over \langle v,v\rangle}\,v$$ So in the OP case we get $$Px=x-{1\over 6}(x_1-2x_2+x_3)\,v$$