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Suppose $(X,*)$ is a pre-hilbert real space. Is it true that a linear projection $P:X\rightarrow X, P(X)=Y$, self-adjoint respect $*$, is the identity on $Y$? this means that $Px$ realize $\min_{y \in Y} \Vert x-y \Vert$

i know that $$(x-Px)*Px=(P(x-Px))*x=(Px-P^2x)*x=0$$ so if $Py=y$ on $Y$ $$(x-Px)*y=(x-Px)*Py=(P(x-Px))*y=0$$

then $\forall y \in Y$ $$\Vert x-Px \Vert \leq \Vert x-y \Vert$$

but idk if $Py=y, \space \forall y \in Y$ is true.

#edited after the answers

I use the definition of @Shaqinho about projection

For me, if we assume that exists $P$ linear proj self-adjoint respect inner product, both completeness and $Py =y$ on $Y$ are not necessary to realize the $\min$, in fact $x-Px$ and $Px$ are orthogonal. Moreover $x-Px$ is orthogonal with every $y \in Y$ because i can write $y=Pz$ for some $z \in X$ and $$(x-Px)*y= (x-Px)*Pz=(P(x-Px))*z=...=0$$ so we can easily show $$\Vert x-Px \Vert \leq \Vert x - y \Vert$$ for every $y \in Y$

Someone could check it?

anto_zoolander
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  • That minimum/infimum can fail to be achieved by any $y\in Y$, if $Y$ is not complete. That is, there can fail to be a projection map to a non-complete subspace $Y$ ... even with a Hilbert space $X$. – paul garrett Oct 04 '20 at 19:18
  • @paulgarrett that is untrue. Yes, the orthogonal does not need to exist. But there always exists a linear projection in a sense that $P^2 = P$. You prove that like you prove Hahn-Banach's theorem. – Hyperbolic PDE friend Oct 04 '20 at 19:25
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    @Shaqinho, we must be misunderstanding each other: for me, an orthogonal projection $P:X\to Y$ with subspace $Y$ would be such that $Px$ is orthogonal to $x-Px$. If $Y$ is not complete, this may not make sense. That is, for me, the natural minimizing thing would often be in the closure of $Y$, but possibly not in $Y$. – paul garrett Oct 04 '20 at 19:37
  • Okay, then our defintions diverge. For me, a projection onto $Y$ is a linear map $P:X \rightarrow Y$ such that $P^2 = P$. But from your standpoint, you are right of course. – Hyperbolic PDE friend Oct 04 '20 at 19:39

1 Answers1

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If such a linear projection $P$ exists, (that is, a linear operator such that $Px$ minimizes $\min_{y\in Y}\|x-y\|$, or, equivalently, $\|x-Px\| \leq \|x-y\|$ for all $y\in Y$) then yes, $Pz=z$ is indeed true for all $z\in Y$.

Let $z\in Y$ be given. Then we have $$ \|z-Pz\| \leq \|z - y\| \qquad \forall y\in Y. $$ If we choose $y=z$ here we get $\|z-Pz\| \leq \|z-z\|=0$, which implies $\|z-Pz\|=0$ and $z=Pz$.

However, as paul garrett notes in the comments, for a given subspace $Y$ such an operator $P$ does not always exist in a pre-Hilbert space.

supinf
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  • thanks! i was blinded lol this problem comes out from an exercise about conditional expectation: the operator $$ \mathbb{E}[X|\mathcal{G}]$$ is a self-adjoint projection respect $<X|Y> = \mathbb{E}[X*Y]$ in some probability space, i had to demonstrate that $\mathbb{E}[X|\mathcal{G}]$ minimize in this sense the distance all over the $\mathcal{G}$-measurable functions only using the properties of a self-adjoint projection – anto_zoolander Oct 08 '20 at 17:29