There are many of fixed point theorems but two of them are:

Krasnoselskii theorem:

Let K be a non-empty, bounded, closed and convex subset of Banach space E. Moreover let $f: K \to E$ be a contraction and let $g: K \to E$ be a compact mapping and $f(x) + g(y) \in K$ for all $x,y \in K$. Then the mapping $F(x) = f(x) + g(x)$ has a fixed point.

Darbo theorem

Let assume that $f$ is continuous mapping of a non-empty closed, convex and bounded subset $K$ of a Banach space $E$ into inself and there exist constant $0 < k < 1$ such that $$ \alpha(f(A)) \leq k \alpha(A) $$ for every set $A \subset K$. Then the mapping $f$ possesses a fixed point. Where $\alpha$ is Kuratowski measure of non-compactness.

It is easy to see that if $f$ satisfies assumptions of Krasnoselskii theorem, then also satisfies assumptions of Darbo theorem. Is there any counterexample of function that satisfies assumptions of Darbo theorem but not Krasnoselskii?


1 Answer 1


I think "yes". In the following let $\alpha$ and $\beta$ denote the Kuratowski and the Hausdorff measure of non-compactness, respectively. Note that $$ \beta(A) \le \alpha(A) \le 2 \beta(A) $$ for each bounded subset $A$ of a Banach space $E$. The following example is in $E:=c_0(\mathbb{N},\mathbb{R})$ (endowed with the maximum norm $\|\cdot\|$). In this space we have the following representation of $\beta$: For each bounded set $A \subseteq E$: $$ \beta(A)= \lim_{n \to \infty} (\sup_{x \in A} (\max \{|x_k|: k \ge n\})), $$ see Banas J., Goebel K. Measures of noncompactness in Banach spaces, 6.1.

Let $\varphi: \mathbb{R} \to \mathbb{R}$ be the function $$ \varphi(t)=\frac{t}{4} \sin(1/t) ~ (t \not=0), \quad \varphi(0)=0, $$ and note that $|\varphi(t)| \le |t|/4$. Let $K$ denote the closed unit ball in $E$ and set $f:K \to K$ as $$ f(x)=f((x_k)_{k=1}^\infty) :=(\varphi(x_k))_{k=1}^\infty. $$ Now, for each set $A \subseteq K$ we have $$ \beta(f(A)) = \lim_{n \to \infty} \sup_{y \in f(A)} \max \{|y_k|: k \ge n\} = \lim_{n \to \infty} \sup_{x \in A} \max \{|\varphi(x_k)|: k \ge n\} \le \frac{1}{4} \beta(A), $$ thus $$ \alpha(f(A)) \le \frac{1}{2} \beta(A) \le \frac{1}{2}\alpha(A). $$ Now we show that $f$ is not the sum of a contraction (even a Lipschitz map) and a compact map: Assume by contradiction that there are functions $g, h:K \to E$ such that

1.) $\forall x \in K$: $f(x)=g(x)+h(x)$,

2.) $\exists L \ge 0$ $\forall x,y \in K$: $\|g(x)-g(y)\| \le L\|x-y\|$,

3.) $\overline {h(K)}$ is compact.

Since the derivative of $\varphi$ is unbounded (in both directions) near $0$ we find $-1 \le \eta < \xi \le 1$ such that $$ (\ast) \quad \varphi(\xi)-\varphi(\eta) > L (\xi-\eta). $$

For $m \in \mathbb{N}$ set $e^{(m)}:=(\delta_{k,m})=(0,\dots,0,1,0,\dots)$, and $$ x^{(m)}:= \xi e^{(m)}, ~~ y^{(m)}:= \eta e^{(m)}, ~~ u^{(m)}:= g(x^{(m)})-g(y^{(m)}), ~~ v^{(m)}:= h(x^{(m)})-h(y^{(m)}) $$ (note that $x^{(m)},y^{(m)} \in K$).

By 1.) and 2.) we have for each $m$: $$ (\ast\ast) ~~ \varphi(\xi)-\varphi(\eta)=u^{(m)}_m + v^{(m)}_m \le L\|x^{(m)}-y^{(m)}\| + v^{(m)}_m = L(\xi-\eta)+ v^{(m)}_m. $$ Next, by 3.), the sequence $(v^{(m)})$ has a convergent subsequence $(v^{(m_j)})$ with limit $w=(w_k) \in E$, say. We have $$ |v^{(m_j)}_{m_j}| \le |v^{(m_j)}_{m_j}-w_{m_j}| + |w_{m_j}|\le \|v^{(m_j)}-w\| + |w_{m_j}| \to 0 \quad (j \to \infty). $$ Now $(\ast\ast)$ with $m=m_j$ and $j \to \infty$ yields $\varphi(\xi)-\varphi(\eta) \le L (\xi-\eta)$, a contradiction to $(\ast)$.

Hope everything is correct, please check.

  • 1
    $\begingroup$ Seems fine to me, don't see any mistakes. Thank you a lot, such a elegant counterexample $\endgroup$
    – patryko519
    Commented May 22, 2023 at 18:45

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