# Counterexample for fixed point theorems

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

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.

• Seems fine to me, don't see any mistakes. Thank you a lot, such a elegant counterexample May 22 at 18:45