Let $X$ be a Banach space, $f: X \to \mathbb{R}$ a bounded linear functional, $y \in X$ fixed and $\alpha \in \mathbb{R}$. Prove there exists a constant $C > 0$ such that if $|\alpha| < C$, then the non-linear equation $x + \alpha f(x) x = y$ has a unique solution $x$ in the ball $B = \left\{x \in X : ||x - y|| \leq 1\right\}$.
Let $Tx = x + \alpha f(x) x$. My strategy is to show that $T$ is a contraction; that is, $||Tx - Tz|| < ||x - z||$ for $x, z \in X$. Then I could apply the Banach fixed-point theorem to reach the conclusion.
Unfortunately I am having trouble showing that $T$ is a contraction. I get as far as showing $||Tx - Tz|| = ||(x - z) + \alpha (f(x)x - f(z)z)||$. I've tried the adding zero trick by adding then subtracting $\alpha f(x) z$ but this yields nothing that will show that $T$ is a contraction.
This is where I would like help.
Note that if $T$ as you have defined it has a fixed point $x$, then $$x=x+\alpha f(x)x,$$ or $\alpha f(x)x=0,$ which doesn't give us the equation you're looking for. What you should try to show is that the map $T$ given by $$Tx=y-\alpha f(x)x$$ is a contraction on $B$. Now we calculate: \begin{align*} \|Tx-Tz\|&=|\alpha|\|f(x)x-f(z)z\|\\ &\leq|\alpha|\left(\|f(x)x-f(x)z\|+\|f(x)z-f(z)z\|\right)\\ &\leq|\alpha|\|f\|\left(\|x\|+\|z\|\right)\|x-z\| \end{align*} Now, since $B$ is bounded, there is some $M>0$ such that $\|x\|,\|z\|\leq M$. This gives us $$\|Tx-Tz\|\leq2M|\alpha|\|f\|\|x-z\|,$$ So if we let $C=\frac{1}{2M\|f\|}$, if $|\alpha|<C$ then $T$ is a contraction on $B$ and the result is true.