An acting condition for a superposition operator from $H^1(\Omega)$ to $H^1(\Omega)$

If i take $v\in H^1(\Omega)$ where $$H^1(\Omega)=\{u\in L^2(\Omega), \frac{\partial u}{\partial x_i}\in L^2(\Omega), i=1,\ldots,N\}$$

$\Omega$ is bounded open set from $\mathbb{R}^N$

What is the sufficient condition for a real function $f:\mathbb{R}\rightarrow \mathbb{R}$ to have that $f\circ v\in H^1(\Omega)$

Thank you

The following is stated in the paper

Moshe Marcus and Victor J. Mizel, MR 531975 Complete characterization of functions which act, via superposition, on Sobolev spaces, Trans. Amer. Math. Soc. 251 (1979), 187--218.

as Theorem 1 (note that $T_f$ is the superposition operator corresponding to $f$ and that the paper assumes $\Omega$ to satisfy the cone condition):

Suppose that $\Omega$ is bounded. Let $f \colon \mathbb R^m \to \mathbb R$ be a Borel function and let $p$, $r$ be two such that $1 \le r \le p < N$. Then $T_f$ maps $W^{1,p}(\Omega)^m$ into $W^{1,r}(\Omega)$ if and only if the following conditions hold:

1. $f$ is locally Lipschitz in $\mathbb R^m$
2. The first order partial derivatives of $f$ satisfy the inequality $$\left| \frac {\partial f}{\partial \xi_i}(\xi) \right| \le a_0(1+|\xi|^\nu)\ \text{a.e. in \mathbb R^m, i = 1,\dotsc,m}$$ where $a_0$ is a constant and $\nu = N(p-r)/(r(N-p))$.

If $N < p$ (or $N = 1$ and $1 \le p$) and $1 \le r \le p$ then $T_f$ maps $W^{1,p}(\Omega)^m$ into $W^{1,r}(\Omega)$ if and only if condition 1 holds.

The theorem goes on to state that under these circumstances $T_f$ is a continuous operator; bounds are established for its image, too. The case of unbounded $\Omega$ is considered in the same paper, see Theorem 3.

The obvious sufficient condition for $v\mapsto f(v)$ to map $H^1(\Omega)$ to itself ($\Omega$ bounded) is $f$ globally Lipschitz, i.e. $f'\in L^\infty(\mathbb R)$. Add $f(0)=0$ for general $\Omega$.

This condition is also necessary for $N\ge2$, I think (but don't ask me for a reference, that's folk wisdom, although it must be proven somewhere).

• this answer is on contradiction with the answer of @Bazin no? Dec 27, 2016 at 16:47
• No, Bazin is right: for polynomials, only affine functions (polynomials of degree 1) work for $N\ge2$. Dec 27, 2016 at 17:11
• and how we find that $f'\in L^{\infty}$ please Dec 27, 2016 at 17:14
• Lipschitz continuity: $\sup_{s<t}|\frac{f(s)-f(t)}{s-t}|<\infty$ Dec 27, 2016 at 17:22
• I man how we see that we must choose this condition Dec 27, 2016 at 19:49

Maybe just a long comment: if you want this property for any $f$ polynomial (or any smooth $f$), you will need $H^1$ to be an algebra, which is true only in 1D ($N=1$). More generally, $H^s$ is an algebra iff $s>N/2$.

Now the previous comment is also pointing out that, excluding the trivial case where $f$ is an affine function, you will essentially need your Sobolev space to be an algebra, which requires $s>N/2$. Note that there is a nice proof, using Bernstein polynomials, that these Sobolev spaces ($H^s$ with $s>N/2$) are stable by composition by a general smooth function $f$.

• So we must consider $N<2$ ! it means $N=1$. and what about $f$ please and where i can found the proof ? Dec 27, 2016 at 15:39
• @Bazin Could you give a reference for the proof using Bernstein polynomials? Dec 27, 2016 at 15:53
• Look at Guy Métivier's "Para-Differential Calculus ... systems", Edizioni della Normale, 2008. Dec 27, 2016 at 16:05