Prove that (orthogonal) projections $P_{L_1}x=P_{L_1}(P_{L_2}x)$ for closed subspaces $L_1\subset L_2\subset H $

Question

Let $L_1\subset L_2\subset H $ closed subspaces of $H$ and let $x_2=P_{L_2}x$ (*orthogonal projection of $x$ onto $L_2$). Prove $P_{L_1}x=P_{L_1}x_2$ ($P_{L_1}x=P_{L_1}(P_{L_2}x)$)

According to this functional-analysis book by Eidelman, $P_{L}x$ for a closed supspace $L\subset H$ satisfies $||x-P_Lx||=\inf_{y\in L}||x-y||$ and it is unique.

It's clear that $P_{L_1}x_2\in L_1$ and therefore $||x-P_{L_1}x||\le||x-P_{L_1}x_2||$. But how do I proceed showing that $||x-P_{L_1}x_2||\le||x-P_{L_1}x||$ as well? While $x_2-P_{L_1}x_2$ is orthogonal to $L_1$, I am not sure how to work with $x-P_{L_1}x_2$. I know it must be simple and direct, but still, I am stuck. Maybe I am not approaching it the right way.

Answer 1

I don't think that the $\inf$ definition is the most illuminating here. If $H$ is a Hilbert space and $M$ is a closed subspace of $H$, we have $M\oplus M^\perp=H$, i.e. $$\forall x\in H,\exists ! (y,z)\in M\times M^\perp, x=y+z$$ Then the projection $P_M$ is characterized by $P_M(x)=y$. In fact, $$x = \underbrace{P_M(x)}_{y\in M} + \underbrace{\left(x-P_M(x)\right)}_{z\in M^\perp}$$

Now let's turn back to your problem with $L_1\subset L_2$. Using the same decomposition as above, $$P_{L_1}(x) = P_{L_1}\left(P_{L_2}(x) + x - P_{L_2}(x)\right) = P_{L_1}\circ P_{L_2}(x) + P_{L_1}\left(x - P_{L_2}(x)\right)$$ But $x-P_{L_2}(x)\in L_2^\perp\subset L_1^\perp$ so $P_{L_1}\left( x-P_{L_2}(x)\right)=0$. It follows that $$P_{L_1} = P_{L_1}\circ P_{L_2}$$ Note that this reasoning holds for all projections, not just orthogonal projections.

Answer 2

Orthogonal projections in an inner product space are characterised by orthogonality. Namely, $P_{L_i}(x)=x'$ when $x'\in L_i$ and $(x-x')\perp L_i$.

Now, let $x_3:= P_{L_1}(x_2)$. Then clearly $x_3\in L_1$, so you only need to show that $(x-x_3)\perp L_1$. But you now that $(x_2-x_3)\perp L_1$ and $(x-x_2)\perp L_2\supseteq L_1$, and these two easily imply what you need.