fixed point theory: need help with this proposition $\|x-T(x)\| \leq \|x-P_{Fix(T)}(x)\| $

48 Views Asked by At

can someone help me with this problem:

Let T:$ \ \mathbb{R}^n \rightarrow \mathbb{R}^n$ a non linear mapping.
We say T is $\gamma$-strongly quasi-nonexpansive with $\gamma \geq0$ if: $\|T(x)-z\|^2 \leq \|x-z\|^2-\gamma\|T(x)-x\|^2 , \ \forall x \in\mathbb{R}^n,z\in Fix(T) $

Another equal definition is that $\frac{1+\gamma}{2}*\|x-T(x)\|^2 \leq\langle x-T(x),x-z\rangle$, for $z \in Fix(T)$ and $\forall x \in\mathbb{R}^n$.

Proof: Let T be a $\gamma$-strongly quasi-nonexpansive mapping with $ \gamma \geq1.$ Then,

$\|x-T(x)\| \leq \|x-P_{Fix(T)}(x)\| $,

where $P_{Fix(T)}(x)$ represents the projection onto the fixed-point set of T.