# Composition operator

For $A,B \subsetneq \mathbb{R}$ let $f_j: A \to B$, $j\in \{1,2\}$ be two functions that are close in the $W^{s,\infty}$-norm, i.e. $$\max_{k \in \{0,\ldots,s\}} \|f_1^{(k)} - f_2^{(k)}\|_{L^\infty(A)} < \epsilon.$$

Now for some (suffciently smooth), monotone function $t: B \to \mathbb{R}$, let $T(f)(x) := t (f(x))$.

What is known about the distance of $Tf_1$ and $T f_2$ in the Sobolev-norm. I would expect a result of the type $$\|T f_1 - T f_2 \|_{W^{s,\infty}} \leq C \|f_1 - f_2 \|_{W^{s,\infty}} ,$$ where $C$ depends on $T$ and $A,B$, $\|f_1\|_{W^{s,\infty}}, \|f_2\|_{W^{s,\infty}}$ but not on $\| f_1 - f_2\|_{W^{s,\infty}}$ [edit according to the remark from Julián Aguirre].

Is there a general theory on this topic?

The constant $C$ will depend oh $f_1$ (or $f_2$.) Consider the simples case $s=1$ and suppose that all the functions are smooth. Then, by the Mean Value Theorem $$|(Tf_1)(x)-(Tf_2)(x)|\le\|t'\|_\infty\|f_1-f_2\|_\infty.$$ Let's see what happens with the derivative. \begin{align} |(Tf_1)'(x)-(Tf_2)'(x)|&=|t'(f_1(x))\,f_1'(x)-t'(f_2(x))\,f_2'(x)|\\ &\le |t'(f_1(x))|\,|(f_1'(x)-f_2'(x))|+|f_2'(x)|\,|t'(f_1(x)-t'(f_2(x)|\\ &\le\|t'\|_\infty\|f_1'-f_2'\|_\infty+\|f_2'\|_\infty\|t''\|_\infty\|f_1-f_2\|_\infty \end{align} Observe that in the last summand of the last equation you have a dependence on $\|f_2'\|_\infty$.
• you are right. C has to depend on $f_1$ and $f_2$ as well as on their derivatives, but not on their sobolev distance. – User133713 Feb 16 '17 at 16:04