# Complex function inequality

Let $\lambda \in (0, 1)$ . Suppose, that function $f : \mathbb{C} \rightarrow \mathbb{C}$ satisfy the inequality

$|f(u) - f(v)| \leq \lambda|u-v|$

Prove, that for all $a \in \mathbb{C}$ $z = f(z) + a$ have a unique solution

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Oh, you wanted only to proof that the solution is unique, not that there is a solution for every $a$? –  wisefool Jan 6 '13 at 15:13

## 2 Answers

Let $u$ and $v$ be two distinct solutions. Then $f(u)=u-a$ and $f(v)=v-a$. So $\mid u-v\mid =\mid f(u)- f(v)\mid \leq \lambda \mid u-v\mid$. Hence $\lambda\geq 1$ which is contradiction.

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Fix $a\in\mathbb{C}$ and consider $g(z)=f(z)+a$, then $|g(z)-g(w)|=|f(z)-f(w)|\leq \lambda|z-w|$.

So we can suppose $a=0$.

Now, take $z_0\in\mathbb{C}$ and define $z_n=f(z_{n-1})$ for $n\geq1$. We have $$|z_n-z_{n-1}|=|f(z_{n-1})-f(z_{n-2})|\leq\lambda|z_{n-1}-z_{n-2}|$$ so by induction $|z_n-z_{n-1}|\leq\lambda^{n-1}|z_1-z_0|$.

Therefore $$|z_n-z_0|\leq\sum_{k=0}^{n-1}\lambda^k|z_1-z_0|\leq\frac{|z_1-z_0|}{1-\lambda}$$ which is finite and independent of $n$. So, all the $z_n$'s are contained in a disk, therefore there is a subsequence $z_{n_k}$ converging to some $\tilde{z}$. We have $$|f(\tilde{z})-\tilde{z}|=\lim_{k}|f(z_{n_k})-z_{n_k}|\leq \lim_k\lambda^{n_k}|z_1-z_0|=0$$ so $f(\tilde{z})=\tilde{z}$. Uniqueness follows easily from the fact that $\lambda<1$: if $f(z)=z$ and $f(w)=w$, then $|f(z)-f(w)|=|z-w|>\lambda|z-w|$ for every $\lambda\in (0,1)$.

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