Let $X$ be a Banach space and let $T:X \rightarrow X$ be a Lipschitz-continuous map. Show that for $\mu$ sufficiently large, the equation $$Tx + \mu x = y$$ has, for any $y \in X$, a unique solution.
Here is what I did: Suppose there are two solutions $x_2, x_1$, then consider $$d(y,y) = 0 = d(Tx_2 + \mu x_2 ,Tx_1 + \mu x_1)$$ Since we are in Banach space, the metric is simply the norm: $$0 = \vert \vert Tx_2 + \mu x_2 - Tx_1 + \mu x_1 \vert \vert \leq \vert \vert Tx_2 - Tx_1 \vert \vert + \vert \vert \mu x_2 - \mu x_1 \vert \vert $$ By definition of Lipschitz-continuous map, there exist a $M \in \mathbb{R}$ such that $$d(Tx_2 , Tx_1) \leq M d(x_2 , x_1)$$ Thus we have $$0 \leq (M+ \mu)\vert \vert x_2 - x_1 \vert \vert$$ And now let $\mu < -M$, we reach the contradiction. is this somehow valid? I am not that confortable that $\mu < -M$ is sufficient large, what if $M>0$....
Also, I dont know how to prove there is a solution exists.