K(X,\bar{x}). \label{eq:TsubK}
            \end{aligned}$$

(ii) If there is some $ \hat{x} \in X$ such that $ f(\hat{x} )=0$, then $\hat{x} =  T(\hat{x}) \in K(X,\bar{x})$.

(iii) Since $ T(X) \subseteq K(X,\bar{x})$ by *(i)* and $K(X,\bar{x}) \subseteq X$ by assumption, therefore $ T(X) \subseteq X$. As $T$ is continuous and $X$ is convex and compact, then by the Schauder fixed point theorem there exists some $ \hat{x} \in X$ such that $\hat{x} =T(\hat{x})$. Since $A$ is injective, the zeros of $f$ are in bijective correspondence with the fixed points of $ T$, thereby $ f(\hat{x})=0$.

(iv) Inductively define: $X_0=X$, $x_0 = \bar{x}$, and $X_{n+1} = T(X_n)$, $x_{n+1} = T(x_n) $. Note that as $T(X) \subseteq X$ then $X_{n+1} \subseteq X_n$ for all $n$. We show that $ X_n \subseteq x_n + \lambda^n (X_0 - x_0)$. This is clearly true for $ n=0$. For $ n \geq 1$ then: $$\begin{aligned}
            X_{n+1} &\subseteq K(X_n,x_n) \\
            &=  x_n - A^\dagger f(x_n) + (I -A^\dagger Df(X_n)) \cdot (X_n -x_n) \\
            &\subseteq x_{n+1} + (I -A^\dagger D f(X_0)) \cdot \lambda^n (X_0 -x_0) \\
            &\subseteq x_{n+1} + \lambda^{n+1} (X_0 - x_0). 
            \end{aligned}$$ Since $ \lambda^{n}  \| X_0-x_0 \|$ can be made arbitrarily small and $ \{ x_n\}_{n=N }^\infty\subseteq X_N$, 