Adjoint operator and orthogonal projection

[mathzero]

Hello,

I want to show $ T^{*}(Pr_{C}(y)) = Pr_{T^{*}(C)}(T^{*}y)$

where $T \in B(H)$ and $TT^{*}=I$ , $H$ is Hilbert space and $C$ is a closed convex non empty set.

but i don't know how to start, or what tricks needed to solve this type of problems.

also i want know how to construct $T$ to satisfying

$ Pr_{x + C}(y) = Pr_{C}(x + y)$ and $ Pr_{\lambda C}(\lambda y) = \lambda Pr_{C}(y)$

Thanks!

 

[Euge]

Hello,

I want to show $ T^{*}(Pr_{C}(y)) = Pr_{T^{*}(C)}(T^{*}y)$

where $T \in B(H)$ and $TT^{*}=I$ , $H$ is Hilbert space and $C$ is a closed convex non empty set.

Since $TT^* = I$, then $\|T^*x\| = \|x\|$ for all $x\in H$. Indeed, given $x\in H$, $\|T^*x\|^2 = \langle T^*x, T^*x\rangle = \langle TT^*x, x\rangle = \langle x,x\rangle = \|x\|^2$. Show that
$$\inf_{c\in C} \|T^*c - T^*y\| = \inf_{c\in C} \|c - y\|$$
and deduce $T^*(\operatorname{Pr}_C(y))$ is a point of $T^*(C)$ closest to $T^*y$. Then the result follows.

also i want know how to construct $T$ to satisfying

$ Pr_{x + C}(y) = Pr_{C}(x + y)$ and $ Pr_{\lambda C}(\lambda y) = \lambda Pr_{C}(y)$

How does $T$ relate to those equations?

 

[mathzero]

Thank you!

How does $T$ relate to those equations?

i want apply this result $T^{*}(Pr_{C}(y)) = Pr_{T^{*}(C)}(T^{*}y)$, to

$Pr_{x + C}(y) = Pr_{C}(x + y)$ and $Pr_{\lambda C}(\lambda y) = \lambda Pr_{C}(y)$

by creating $T$ first and later deducing $T^{*}$ (i don't know if this is possible!),

i imagine maybe, just by letting $Ty = y - x$ then $T^{*}y= y + x$? this can be possible?

 

[Euge]

Thank you!
i want apply this result $T^{*}(Pr_{C}(y)) = Pr_{T^{*}(C)}(T^{*}y)$, to

$Pr_{x + C}(y) = Pr_{C}(x + y)$ and $Pr_{\lambda C}(\lambda y) = \lambda Pr_{C}(y)$

by creating $T$ first and later deducing $T^{*}$ (i don't know if this is possible!),

i imagine maybe, just by letting $Ty = y - x$ then $T^{*}y= y + x$? this can be possible?

In the case $\operatorname{Pr}_C(\lambda C) = \lambda \operatorname{Pr}_C(y)$, you can take $T(y) = \lambda y$, for all $y\in H$, which defines an element of $B(H)$ with $\|T\| = \lvert \lambda \rvert$. For the other equation, you suggest letting $Ty = y - x$, but this does not define a linear operator unless $x = 0$.