Is this a valid proof re: Banach Fixed Point Theorem and Lipschitz continuous function?

Question

The question posed is the following: Let $X$ be a Banach Space and let $T:X\to X$ be a Lipschitz-Continuous map. Show that, for $\mu$ sufficiently large, the equation \begin{equation} Tx+\mu x=y \end{equation} has, for any $y\in X$, a unique solution.

Note that $x,y$ are vectors, since our book (Mathematical Analysis by Mariano Giaquinta and Giuseppe Modica) generally ignores vector indicators, since it's all multivariable.

My proof is based on the Banach Fixed Point Theorem: Since $T$ is Lipschitz-continuous, we have $\|Tx\|\leq k\|x\|$ for $0<k\leq1$. So $\|Tx-\mu x\|\leq k\|x\| - \mu \|x\|$.

Then we can say

\begin{equation} \|Tx-\mu x\|\leq (k-\mu)\|x\| \end{equation}

So, if $\mu$ is large enough that $|k-\mu|<1$, we have a contractive map, and by the Banach Fixed Point theorem, there exists a unique fixed point $x_0$ for $(T-\mu)x$. Then, $Tx-\mu x=y$ has a unique solution, namely, $x_0$.

My question is whether this is a valid proof. I'm mostly foggy on if I applied the theorem correctly, and if I am allowed to say $Tx-\mu x=(T-\mu)x$, since $T$ is a map and $\mu$ is a constant (I think).

Answer 1

Rewrite the equation as $$ x=\frac1{\mu}\,(y-Tx)=Gx, $$ and show that $G$ is a contraction for large enough $\mu$.

Answer 2

Here was my final answer for reference:

Let $Gx=\frac{1}{\mu}\left(y-Tx\right)$. Then \begin{align*} |Gx_1-Gx_2|&=\left|\frac{1}{\mu}\left(y-Tx_1\right)-\frac{1}{\mu}\left(y-Tx_2\right)\right|\\ &=\frac{1}{\mu}\left|Tx_1-Tx_2\right|\\ &\leq \frac{1}{\mu} k |x_1-x_2|\tag{For $0<k\leq 1$} \end{align*} Then for $\mu > k$, $G$ is a contractive map, and by the Banach Fixed Point theorem, has a fixed point $\bar{x}$ such that $G\bar{x}=\bar{x}$.

Then $G\bar{x}=\bar{x}=\frac{1}{\mu}(y-T\bar{x})$, and thus the equation $Tx-\mu x=y$ has the unique solution $\bar{x}$.