Non-expansiveness implies strict pseudo‑contraction in a Hilbert space

Question

I know that non-expansiveness implies pseudo‑contraction while strict‑contraction implies strict pseudo‑contraction. However, the authors of this paper (page 4) stated that strict pseudo‑contraction is more general than non-expansiveness.

Question: Does this mean that non-expansiveness implies strict pseudo‑contraction in a Hilbert space? If yes, how can it be? I would be grateful if someone can show me how.

For the definition of these terms, please see Browder and Petryshyn, page 2 .

Answer 1

An operator is nonexpansive if it is Lipschitz with constant $1$.

Let $U$ be a nonexpansive operator. By nonexpansiveness we have,

$$\|Ux-Uy\|\leq \|x-y\|\implies \|Ux-Uy\|^2\leq \|x-y\|^2$$

since all quantities involved are positive. We then have,

$$\|Ux-Uy\|^2 \leq \|x-y\|^2\leq \|x-y\|^2 + k \|(I-U)x-(I-U)y\|^2$$

for any $k\geq 0$. So nonexpansive implies strict pseudo-contractiveness. Indeed, it seems that strict pseudo-contractiveness is "more general" than nonexpansiveness in the sense that there may be maps which are strictly pseudo-contractive but not necessarily nonexpansive.