Proving that the set of fixed points of a nonexpansive operator on a bounded closed convex subset of a Hilbert space is convex

Question

I'm struggling to understand the proof of Lemma $3.1.$ from the book Iterative Approximations of Fixed Points by Vasile Berinde, which I'll transcribe here

Lemma 3.1. Let $C$ be a bounded closed convex subset of a Hilbert space $H$ and $T:C \rightarrow C$ be a nonexpensive and demicompact operator. Then the set $F_T$ of fixed points of $T$ is a nonempty convex set.

The proof starts by using a previous result (Theorem $3.1.$) in which he shows that $F_T$ is nonempty, and this is fine. I'm struggling in verifying the convexity of $F_T$. The original argument starts by letting $x,y \in F_T, \lambda \in [0,1]$, and writing $u_\lambda = (1-\lambda)x+\lambda y$. Then observe that $$ \|T(u_\lambda) - x\| = \|T(u_\lambda) - T(x)\| \le \|u_\lambda - x\| $$ $$ \|T(u_\lambda) - y\| = \|T(u_\lambda) - T(y)\| \le \|u_\lambda - y\|. $$ And it follows that \begin{align} \| x- y \| &= \| x - T(u_\lambda) + T(u_\lambda) - y\| \\ &\le \|x-T(u_\lambda)\| + \| T(u_\lambda) - y\| \\ &\le \| u_\lambda -x\| + \| u_\lambda - y\| \\ &= \lambda\|x-y\| + (1-\lambda)\|x-y\| \\ &= \| x-y\|, \end{align} and this is fine too. But at this point, the author claims that this last fact guarantees the existence of some $a\ge 0$ and $b \le 1$ such that $$ x-T(u_\lambda) = a(x-u_\lambda) $$ $$ y-T(u_\lambda) = b(y-u_\lambda), $$ from which it follows that $T(u_\lambda)=u_\lambda \in F_T$.

I am not understanding this very last part of the argument, in which he invokes the existence of constants $a$ and $b$. My first guess was that he was somehow using the strict convexity of $H$, but this would only provide $c,d>0$ such that \begin{align} x-T(u_\lambda) &= c (T(u_\lambda) - y) \\ x-u_\lambda &= d (u_\lambda-y). \end{align} What am I missing here?

Answer 1

The equality $$\| x - T(u_\lambda) + T(u_\lambda) - y\| \\ =\|x-T(u_\lambda)\| + \| T(u_\lambda) - y\|$$ implies that $T(u_\lambda)$ belongs to the interval connecting $x$ and $y.$ Also $u_\lambda$ belongs to that interval (by definition). Therefore $$x-T(u_\lambda)=a(x-u_\lambda),\ a\ge 0 \\ y-T(u_\lambda)=b(y-u_\lambda),\ b\ge 0$$ Since $T$ is nonexpansive we get $a,b\le 1.$

We have two points $T(u_\lambda)$ and $u_\lambda$ in the segment connecting $x$ and $y.$ The distances from $T(u_\lambda)$ to $x$ and $y$ are shorter or equal to the distances of $u_\lambda$ to $x$ and $y,$ respectively. This is possible only when $T(u_\lambda)=u_\lambda.$

Answer 2

From your inequalities you get $$ \|T(u_\lambda) - x\| = \|u_\lambda - x\|,\qquad\qquad \|T(u_\lambda) - y\| = \|u_\lambda - y\|, $$ Then you deduce (since you can discard the cases $u_\lambda=x $ and $u_\lambda=y $ from the start) that $c=d$. Thus \begin{align}\tag1 x-T(u_\lambda) &= c (T(u_\lambda) - y) \\ \tag2 x-u_\lambda &= c (u_\lambda-y). \end{align}

Subtracting $(2)$ from $(1)$,
$$ u_\lambda-Tu_\lambda=c(Tu_\lambda-u_\lambda). $$ Looking at the norms, $c=1$ and then $Tu_\lambda=u_\lambda$.