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Let $A$ and $B$ be nonempty, closed and convex subsets of a Hilbert space $H$. Let $\alpha, \beta \in (0,1)$ such that $\alpha + \beta <1 $. Define $T:H \rightarrow H$ by $$ Tx = \alpha P_A x + \beta P_B x .$$ Show that $T$ is a contraction.

Here, $P_A$ and $P_B$ are the projection operators onto $A$ and $B$, respectively.

I need to show that there exists $\gamma \in [0,1)$ such that $$ d(Tx,Ty) \leq \gamma d(x,y) \text{ for every $x,y\in H$}. $$

Let $x,y \in H$. Then $$d(Tx,Ty) = d(\alpha P_A x + \beta P_B x, \alpha P_A y + \beta P_B y)$$

Please help. Thank you.

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since you are in a Hilbert space, you can you the norm : $$||\alpha P_Ax+\beta P_Bx-\alpha P_Ay-\beta P_By||$$

$$\le \alpha|| P_Ax-P_Ay||+\beta|| P_Bx-P_By||$$ Since $P_A$ and $P_B$ are projection on convex closed set in a Hilbert space, they are contractant (see this answer for a demonstration)

Finally : $$||\alpha P_Ax+\beta P_Bx-\alpha P_Ay-\beta P_By||\le (\alpha+\beta)|| x-y||$$

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  • $\begingroup$ The projection operator does not have to be linear. Also, it does not have to hold $\|Px\| \le \|x\|$. Consider the projection on the closed subset $\{x_0\} \subseteq H$. It is the constant function $x_0$. You are probably thinking of projections onto linear subspaces, but then the proof is easy: $$\|T\| = \|\alpha P_A + \beta P_B\| \le \alpha \|P_A\| + \beta \|P_B\| = \alpha + \beta$$ so for every $x, y \in H$ we have $$\|Tx - Ty\| = \|T(x-y)\| \le \|T\|\|x-y\| \le \underbrace{(\alpha + \beta)}_{<1}\|x-y\|$$ $\endgroup$ – mechanodroid Dec 11 '17 at 16:33
  • $\begingroup$ @mechanodroid I was indeed thinking of linear spaces but I edited my answer $\endgroup$ – stity Dec 11 '17 at 17:59

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