# How do functions operate in a Sobolev space $H^{s}$?

Let $s>\frac{1}{2};$ and define a Sobolev space as follows: $$H^{s}(\mathbb R)=\{f\in L^{2}(\mathbb R):[\int_{\mathbb R} |\hat{f}(\xi)|^{2}(1+|\xi|^{2})^{s}d\xi]^{1/2}<\infty \}.$$ Fact: Let $m$ be an integer greater than $s+1.$ Assume that $F\in C^{m}(\mathbb R)$ and $F(0)=0.$ Then $F(f)\in H^{s}(\mathbb R)$ for all real-valued $f\in H^{s}(\mathbb R).$ ( Note that $F(f)$ denotes the composition of $F$ and $f$ )

My Question is: What can we say about the converse of the above fact, that is, If $F(f)\in H^{s}(\mathbb R)$ for all $f\in H^{s}(\mathbb R)$, then does the $m$-th derivative of $F$ exist, is it continuous ($F\in C^{m}(\mathbb R)$), and $F(0)=0$ ?

In the case s>3/2, the answer looks to be given quite comprehensively as Theorem 1 (see also Remark 1) of

Bourdaud, G., Moussai, M., and Sickel, W. Composition operators on Lizorkin-Triebel spaces. J. Funct. Anal. 259 (2010), no. 5, 1098–1128.

As far as I am aware, results giving necessary and sufficient conditions of this type go back to work of Marcus and Mizel in the late 1970s (although not in the fractional setting). There is of course quite a bit of fascinating literature on this and related topics -- if you aren't aware of them already, you may want to also look at the following references:

Brezis, H. and Mironescu, P. Gagliardo-Nirenberg, composition and products in fractional Sobolev spaces. J. Evol. Equ. 1 (2001), no. 4, 387–404.

and

Mazʹya, V. and Shaposhnikova, T. An elementary proof of the Brezis and Mironescu theorem on the composition operator in fractional Sobolev spaces. J. Evol. Equ. 2 (2002), no. 1, 113–125.

The property $F(0)=0$ follows from applying $F$ to the zero function, but the regularity is not as good as you ask for.

As a partial counterexample, let $F(x)=|x|$ and $s=1$. Now $F(f)\in H^1$ for all $f\in H^1$ but $F\notin C^1$. Although $F$ is not classically differentiable, it is weakly differentiable, and for example $F\in W^{1,\infty}_{\text{loc}}$. I assume something similar can be cooked up for other values of $s$ as well, but I suppose this example shows the general idea.