# prove uniqueness Lipschitz-continuous map in Banach space

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.

• Your proof of uniqueness is not right, since when you look at $\|\mu x_2 - \mu x_1\|$ and pull out the $\mu$ from the norm, you are assuming $\mu \geq 0$. Oct 9, 2018 at 3:25
• @WillieWong ok I see that now Oct 9, 2018 at 3:31
• @WillieWong still have no idea how to do this tho. Oct 9, 2018 at 3:31
• Oct 9, 2018 at 3:34

Banach fixed point theorem If $$F:X\to X$$ is such that there exists $$\lambda\in [0,1)$$ such that for every $$x_1, x_2\in X$$, $$\|F(x_1) - F(x_2)\| \leq \lambda \|x_1 - x_2\|$$, then there exists a unique fixed point of $$F$$.
Consider the function $$f(x) = \frac{1}{\mu} (y - Tx)$$ A solution to your original equation is also a fixed point of $$f$$. So if we can show that $$f$$ is a contraction mapping (equivalently it is Lipschitz with Lipschitz constant $$<1$$), then by Banach's fixed point theorem we are done.
But $$f(x_1) - f(x_2) = \frac{1}{\mu} [T(x_2) - T(x_1)]$$ so as long as you choose $$\mu$$ bigger than the Lipschitz constant of $$T$$ you have that $$f$$ is a contraction mapping.