Let $s>\frac{1}{2};$ and we define 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 can expect $m$-th derivative of $F$ exists and it is continuous ($F\in C^{m}(\mathbb R)$), and $F(0)=0$ ?

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 $F(0)=0$ ?