Firmly nonexpansive mapping with the fixed point set same as for given nonexpansive mapping

Question

I found PAMS publication vol. 113, no. 3, 1991 by Ryszard Smarzewski called "On firmly nonexpansive mappings".

It is written that "to each nonexpansive T on set C one can associate a firmly nonexpansive mapping whenever C is closed and convex". It is stipulated that C is subset of the Banach space. Furthermore, author clearly refers in this context to the book by Goebel and Reich, "Uniform convexity hyperbolic geometry and nonexpansive mappings", but unfortunatelly I do not have access to it.

I thought that this fact is not necessarily true for every Banach space, but e.g. on Hilbert spaces. Could someone clear the issue and - if possible- provide the sketch of the proof?

Answer 1

The same claim appears on page 350 of the article Iterating holomorphic self-mappings of the Hilbert ball by Goebel and Reich, which, unlike the book, is in open access.

If $T$ is a nonexpansive self-mapping of a closed convex subset $C$ of a Banach space, then for each $0\le k<1$ there is a firmly nonexpansive mapping $g_k: C\to C$ that satisfies $$g_k(x)=(1-k)x+k Tg_k(x)$$ for all $x \in C$.

The existence of such $g_k$ follows from the fact that for each fixed $x$, the map $z\mapsto (1-k)x+kTz$ is a strict contraction of $C$ to itself; its unique fixed point is what we call $g_k(x)$.

The proof that $g_k$ is firmly nonexpansive should follow along the lines of Theorem 3 in the paper, but I admit I could not make it work.

Answer 2

I will only comment on the Hilbert space case (I'm not familiar with the more general Banach space setting).

If $T \colon C \to C$ is nonexpansive and $C$ is convex, then the operator $$ \frac{T+\operatorname{id}}2 $$ maps $C$ into $C$ (this is clear), firmly nonexpansive, and has the same fixed points as $T$.

The reverse is also true to a certain degree. A nonexpansive operator $S$ is called alpha-averaged, in symbols $S \in \mathcal A(\alpha)$, if it can be written as $$ S = (1-\alpha) \operatorname{id} + \alpha T$$ with a nonexpansive operator $T$ and $\alpha \in [0,1]$. The class $\mathcal A(1/2)$ is not only a subset of but identical to the class of firmly nonexpansive maps, which is to say: For a firmly nonexpansive operator $S$, the operator $2S - \operatorname{id}$ is nonexpansive.