finding the orthogonal projection of function in $L^2((0,1)^2)$ over subspace

Question

Let $H = L^2((0,1)^2)$, and let $$V = \{f\in H: \exists g\in L^2(0,1): f(x,y)=g(x) \}$$ I've already prooved that V is a closed subspace of H.

Now i want to find the orthogonal projection of $f(x,y)=xy$ in V, the problem is that i don't know how to aproach the problem. I've seen similar exercises, but for subspaces with finite dimension, wich is not this case. Any ideas on how to start solving this? Thanks

Answer 1

The orthogonal projection of $f$ onto $V$ is the unique $v\in V$ such that $(f-v) \perp V$. That is, $$ \int\int (f(x,y)-v(x))w(x)dydx=0,\;\; \forall w\in L^2(0,1). $$ That is, $\int(\int f(x,y)dy - v(x))w(x)dx = 0$ for all $w\in L^2$, which forces $\int f(x,y)dy=v(x)$. So, $$ Pf=\int f(x,y)dy. $$ In particular, $P(xy)=\int xy dy = \frac{1}{2}x$.

Answer 2

Given $f\in H$, let $g(x)=\int_0^1f(x,y)dy$, which turns out to be an element of $L^2(0,1)$ by Cauchy Schwarz. Then the projection of $f$ onto $V$ is $(P_Vf)(x,y)=g(x)$.

One can get an intuition of sorts by making an analogy with matrices: the projection of a matrix $(f_{ij})$ on the space of matrices of the form $f_{ij}=g_j$ is given by replacing each row in $f$ with the average value of the entries in that row.