Contraction mapping and linear continous operator

40 Views Asked by At

I'm working on contraction mapping theorem with parameter, and this leads me to Appendix D of G. Da Prato, Introduction to stochastic analysis and Malliavin calculus.

In the book, it says enter image description here enter image description here

enter image description here

whereas I have literally no idea why $G$ defined above is a contraction. Please help me with it

1

There are 1 best solutions below

0
On BEST ANSWER

Assuming the problem is that you don't see why the final inequality holds, note that if you set $\mu =0$ in D.3, and take norms you get $$|F_x(\lambda, x)y| = \lim_{h\to 0} \frac 1{|h|} |F(\lambda, x+hy) - F(\lambda, x)|$$ By D.1, this becomes $$|F_x(\lambda, x)y| \le \lim_{h\to 0} \frac 1{|h|} k|hy| = k|y|$$ This holds for any $\lambda, x, y$, so we can rewrite it as $$|F_x(\alpha, u)y| \le k|y|$$

Now, letting $\alpha= \lambda + \xi h\mu,\ u = x(\lambda) + \xi\big(x(\lambda + h\mu) - x(\lambda)\big),\ y = z$, we have $$|Gz| \le \int_0^1\left|F_x\bigg(\lambda + \xi h\mu,\ x(\lambda) + \xi\big(x(\lambda + h\mu) - x(\lambda)\big)\bigg)z\right|\,d\xi\le\int_0^1 k|z|\,d\xi = k|z|$$