Welcome to our community

Be a part of something great, join today!

Contraction mapping (Maryam Ishfaq's question at Yahoo! Answers)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Maryam Ishfaq,

The result is more general. Suppose $(E,d)$ is a metric space and $T:E\to E$ is a Lipschitz mapping, i.e. there is a positive constant $K$ such that $d(T(x),T(y))\le Kd(x,y)$ for all $x,y\in E$. Then, for all $\epsilon >0$ and choosing $\delta=\epsilon/K:$ $$d(x,y)<\delta \Rightarrow d(T(x),T(y))\le Kd(x,y)<K\frac{\epsilon}{K}\Rightarrow d(T(x),T(y))<\epsilon$$ This means that every Lipschitz mapping is uniformly continuous, as a consequence continuous. But a contraction map is a Lipschitz mapping with $K<1$, hence every contraction is a continuous mapping.