Prove the existence of a fixed point for the sum of two mappings

Question

Let $X$ be a real valued Banach-space and $A \subseteq X$ which is bounded, closed and convex. Furthermore let $f,g,h : A \mapsto X$ be continuous with $f=g+h$ and $g(A)+h(A) \subseteq A$. Lastly $g$ ist a contraction mapping and $h$ compact map. I'm supposed to show that $f$ has a fixed point $x^* \in A$. Sadly I am getting nowhere with this. I know by Banach's theorem, that $g$ has a unique fixed point, and by Schauder's theorem, that $h$ has a non-unique fixed point. Obviously these fixed point need not coincide; even if they did it would do me no good. Furthermore I've been told as a hint to look at the map $id_A - g: A \mapsto X$, where $id_A$ refers to the Identity. I'd appreciate a nudge in the right direction, since I'm getting nowhere with this.

Answer 1

Let $y \in A$. Since $g(A)+h(A) \subseteq A$ we have that $x \mapsto g(x)+h(y)$ maps $A$ to $A$ and since $g$ is a contraction (with constant $q < 1$, say) there is a uniqe fixed point $w(y) \in A$ of this mapping, that is $w(y)=g(w(y))+h(y)$. We have obtained a function $w:A \to A$ and we show that $w$ is continuous and compact:

1. Let $y_1,y_2 \in A$. Then $\|w(y_1)-w(y_2)\| \le q\|w(y_1)-w(y_2)\| + \|h(y_1)-h(y_2)\|$, hence $$ (\ast) \quad \quad \|w(y_1)-w(y_2)\| \le \frac{1}{1-q}\|h(y_1)-h(y_2)\|, $$ and $w$ is continuous since $h$ is continuous.

2. Let $(y_n)$ be any sequence in $A$. Since $h$ is compact, the sequence $(h(y_n))$ has a convergent subsequence $(h(y_{n_k}))$, which therefore is a Cauchy sequence. By means of $(\ast)$ also $(w(y_{n_k}))$ is a Cauchy sequence, hence convergent in $A$. Thus $w$ is compact

Finally by Schauder's fixed point theorem $w$ has a fixed point $z \in A$, that ist $$ z=w(z)=g(w(z))+h(z)=g(z)+h(z). $$