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Suppose H is a Hilbert space, M is a closed linear subspace, and $T:H \rightarrow H$ is a bounded linear operator with bounded inverse.

Define a projection operator $P:H \rightarrow M$ as $Px = z$, where $$\inf_{y \in M} ||T(y-x)|| = ||T(z -x)||.$$

I've shown that $P$ is well defined (for each $x \in H$ there is a unique $z \in M$ where the infinum is attained) and that $P$ is a projection operator.

Additionally, I've shown that $(T(Px-x),Tm) = 0$ for all $m \in M$. Equivalently, $(T^*T(Px-x),m) = 0$, so $T^*T(P-I)$ maps to $M^{\perp}$.

Now I want to show that $||P|| \le ||T||\ ||T^{-1}||$. I've attempted to write some sort of expression in terms of $P$ and $P^{\perp}$, but I'm stuck. Any help appreciated. Thank you.

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  • $\begingroup$ It's not true that for each $x\in H$ there is a unique $z\in M$ where the minimum is attained. For example if $T=0$ the minimum is attained by any $z$. $\endgroup$
    – David C. Ullrich
    Commented Oct 29, 2017 at 3:03
  • $\begingroup$ Yes, but T in the problem has an inverse. $\endgroup$
    – stackedtritones
    Commented Oct 29, 2017 at 3:09
  • $\begingroup$ I missed that, sorry. $\endgroup$
    – David C. Ullrich
    Commented Oct 29, 2017 at 3:10
  • $\begingroup$ @DavidC.Ullrich:Which property of projection operator fails if we take $M$ to be an open subspace? $\endgroup$
    – Styles
    Commented Jul 3, 2018 at 13:44
  • $\begingroup$ @PKStyles If $X$ is a topological vector space and $M$ is an open subspace then $M=X$. $\endgroup$
    – David C. Ullrich
    Commented Jul 3, 2018 at 13:51

1 Answer 1

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The $T$ norm, $\|x\|_T = \|Tx\|$, is a Hilbert norm under the conditions you state. The projection $P$ is an orthogonal projection with respect to the $T$ norm. Therefore, $\|Px\|_T \le \|x\|_T$ holds for all $x\in H$. Expanding, $$ \|TPx\| \le \|Tx\| \\ \|TPT^{-1}x\| \le \|x\| \\ \|TPT^{-1}\| \le 1. $$ So, $\|P\| = \|T^{-1}(TPT^{-1})T\| \le \|T^{-1}\|\|T\|$, which is the desired result.

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  • $\begingroup$ I'm not sure what a Hilbert norm is. Why can we say $||x|| = ||Tx||$? $\endgroup$
    – stackedtritones
    Commented Oct 29, 2017 at 15:08
  • $\begingroup$ @stackedtritones : $\langle x,y\rangle_T = \langle Tx,Ty\rangle$ defines an inner product $\langle\cdot,\cdot\rangle_T$ on $H$, and $H$ is a Hilbert space with respect to this inner product, because $T$ is bounded with a bounded inverse. The corresponding norm is $\|x\|_T = \|Tx\|$. The definition of your projection $P$ makes your projection an orthogonal projection on the Hilbert space $(H,\|\cdot\|_T)$. So, $\|Px\|_T \le \|x\|_T$ for all $x\in H$. Note: I did not write $\|x\|=\|Tx\|$. I wrote $\|x\|_T = \|Tx\|$. $\endgroup$
    – Disintegrating By Parts
    Commented Oct 29, 2017 at 17:03

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