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.


1 Answer 1


This is a classic application of Banach's fixed point theorem.

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.

  • $\begingroup$ Thanks, much appreciated. $\endgroup$
    – Frost
    Oct 9, 2018 at 3:37

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