A variant of Nadler's fixed point theorem

Question

I am trying to prove an implicit function type of result. The following problem will do the trick for me:

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Suppose $X$ is a Banach space, $B(\mathbf{0};r)=\{x\in X:\|x\|<r\}$. Let $F:B(\mathbf{0};r)\rightarrow 2^X\setminus\{\emptyset\}$ be such that

1. $F(x)$ is closed for every $x\in B(\mathbf{0};r)$,
2. $d_H(F(x),F(y))\leq k\|x-y\|$ for all $x,y\in B(\mathbf{0};r)$ where $0<k<1$ is a contant, and $d_H$ us the Hausdorff distance ($d_H(A,B)=\max\{\sup_{x\in A}d(x,B),\sup_{x\in B}d(x,B)\}$)
3. $d(\mathbf{0},F(\mathbf{0}))<r(1-k)$.\

Then, there exists a fixed point, i.e., there is $x\in B(\mathbf{0};r)$ such that $x\in F(x)$.

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This is a variant of Nadler's fixed point theorem and is given as an exercise in Klaus Diemling's Nonlinear Functional Analysis book. My fist attempt was to construct directly a sequence of points $x_n$, $x_n\in F(x_{n-1})\cap B(0;r')$ for some $0<r<'r$, that converges. But I can't seem to get control over the sizes (diameters) of balls.

I am asking for some hints, or ideally a solution to this seemingly trivial problem.

Answer 1

I found a solution that uses a construction similar to that of Banach-Cacciopoli's fixed point theorem and also using the idea described in the OP.

(a) Let $0<r'<r$ such that $d(\mathbf{0},F(\mathbf{0}))<(1-k)r'$. This will allows to work with the closed ball $\overline{B(\mathbf{0};r')}$ which is complete.

(b) By assumption (3) and (a), there is $x_1\in F(\mathbf{0})$ such that $\|x_1-\mathbf{0}\|<(1-k)r'$. By assumption (2), $$ d(x_1,F(x_1))\leq d_H(F(\mathbf{0}),F(x_1))\leq k\|x_1-\mathbf{0}\|<k(1-k)r' $$ Hence, there exists $x_2\in F(x_1)$ such that $\|x_2-x_1\|<k(1-k)r'$, and $$ \begin{align} \|x_2-\mathbf{0}\|&\leq\|x_2-x_1\|+\|x_1-\mathbf{0}\|<k(1-k)r'+(1-k)r'\\ &=(1-k)r'(1+k)=r'(1-k^2) \end{align} $$

(c) By induction, we obtain a sequence $\{x_n:n\in\mathbb{N}\}\subset B(\mathbf{0};r')$ such that

• $x_n\in F(x_{n-1})$.
• $\|x_n-x_{n-1}\|<(1-k)r'k^{n-1}$.
• $\|x_n-\mathbf{0}\|< r'(1-k^n)$.

Clearly $\{x_n:n\in\mathbb{N}\}$ is a Cauchy sequence in $\overline{B(\mathbf{0};r')}$. It follows that there is $x_*\in X$ with $\|x^*\|\leq r'$ such that $x_n\xrightarrow{n\rightarrow\infty}x_*$ in $X$.

It remains to show that $x_*\in F(x_*)$. This follows from $$ d(x_{n+1},F(x_*))\leq d_H(F(x_n),F(x_*))\leq k\|x_n-x_*\|\xrightarrow{n\rightarrow\infty}0 $$ which in turn leads to $$ d(x_*,F(x_*))=0$$ By assumption (1) in the OP, $x_*\in F(x_*)$.