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Let $X$ be a Banach space. Denote for $x_0\in X$ and $r>0$ the closed ball centered at $x_0$ by $B(x_0,r)=\lbrace x\in X:\|x-x_0\|\le r\rbrace$.

Suppose $f:X\to X$ a bounded map with a fixed point $x_*$ (that is, $f(x_*)=x_*$) and assume there is an constant $r>0$ and $M\in[0,1)$ such that $$\|f(x)-f(y)\|\le M\|x-y\|$$ for any $x,y\in B(x_*,r)$.

Put for $\alpha\in[0,\infty)$, $$\phi_\alpha(A)=\overline{\{f(x):x\in A\}}+\alpha C,$$where $A,C\subset X$ are closed and bounded subsets and the overline denotes the closure. The map $A\mapsto\overline{\{f(x):x\in A\}}$ is locally Lipschitz continuous.

Let $BC[B(x_*,r)]$ denote the collection of all bounded closed subsets in $X$. Is there a $\alpha_0>0$ such that $\phi_\alpha(BC[B(x_*,r)])=BC[B(x_*,r)]$ for all $\alpha\in[0,\alpha_0]$? and does $\phi_\alpha$ have fixed point $A_*\in BC[B(x_*,r)]$ that contains $x_*$?

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Assuming that $BC$ denotes the set of closed and bounded subsets of $B(x_*, r)$ and $\phi_\alpha(BC) = \{A: A = \phi_\alpha(B)\text{ for some }B\in BC\}$, it is not in general true that $\phi_\alpha(BC) = BC$. To prove this, with $M\in (0, 1)$, take $X = [-1/M, 1/M]$ (of course this is not a Banach space, so view it as sitting inside $\mathbb{R}$, and $f$, to be defined next on $X$, can be extended to $\mathbb{R}$), let $f(x) = Mx$. Take $r = 1$, and $x_* = 0$. Take $C = \emptyset$. It is easily seen that for sufficiently small $\epsilon > 0$, the interval $[-1 + \epsilon, 1 - \epsilon]\subset (-1,1)$ is not an image of any compact $B\subset (-1,1)$ under $\phi_\alpha$ (for any $\alpha$, since we are taking $C = \emptyset$). This can easily be generalized to $C\neq \emptyset$.

On the other hand, it is in general true that $\phi_\alpha(BC)\subset BC$ for $\alpha\in[0, \alpha_0]$ with appropriately chosen $\alpha_0 > 0$. Indeed, say $B\in BC$. Observe that for each $x\in B$, $\|x - x_*\| \leq r$. Since $f$ is a contraction on $B(x_*, r)$, for each $y\in \overline{f(B)}$, $\|y - x_*\| \leq Mr < r - \delta < r$ for some $\delta > 0$ which doesn't depend on $B$ or $y$. Since $C$ is closed and bounded, for all sufficiently small $\alpha > 0$, for all $z\in \alpha C$, $\|z \| < \delta/2$. Hence for all such $z$, $\|(y + z) - x_*\|\leq \|y - x_*\| + \|z\| < r$.

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