If $u_n \to u$ in $H^{\frac 12}(\partial\Omega)$, does $f(u_n) \to f(u)$ in $H^{\frac 12}(\partial\Omega)$ for $f$ Lipschitz?

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be a smooth function with $|f'(x)| \leq C$ for all $x$ and $f(0)=0$.

Suppose $u_n \to u$ in $H^{\frac 12}(\partial\Omega)$, where $\Omega$ is a bounded domain of class $C^2$. Here, I use the norm $$|u|_{H^{\frac 12}(\partial \Omega)}^2 = |u|^2_{L^2(\partial\Omega)} + \int_{\partial\Omega}\int_{\partial\Omega} \frac{|u(x)-u(y)|^2}{|x-y|^n}.$$

Does it follow that $f(u_n) \to f(u)$ in $H^{\frac 12}(\partial\Omega)$ (at least for a subsequence)?

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Note that $f(u) \in H^{\frac 12}$ by the Lipschitz nature of $f$. The only issue is in the convergence of the seminorm, which I can't seem to prove.

Answer 1

Yes. It might be more convenient to think of $H^{1/2}(\partial\Omega)$ as a quotient space: $$ H^{1/2}(\partial\Omega)=H^1(\Omega)/H^1_0(\Omega). $$ Let $T:H^1(\Omega)\to H^{1/2}(\partial\Omega)$ be the corresponding quotient map (the trace map). Since $u_n\to u$ in $H^{1/2}(\partial\Omega)$, there is are functions $v_n,v\in H^1(\Omega)$ so that $v_n\to v$ in $H^1(\Omega)$ and $T(v_n)=u_n$ and $T(v)=u$. The trace map and composing with $f$ commute with each other and $T$ is continuous, so it suffices to show that $f\circ v_n\to f\circ v$ in $H^1(\Omega)$. It follows from theorems 1 and 5 in this paper that $H^1(\Omega)\ni w\mapsto f\circ w\in H^1(\Omega)$ is continuous, from which the convergence follows. It suffices that $f$ is Lipschitz; smoothness is not required.