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Let $N\colon \mathrm{H}^s(\mathbb{R}) \to (\mathrm{H}^s(\mathbb{R}))^*$, where $s > \frac{1}{2}$, be an operator given by $N(u) = \langle u^p, \cdot \rangle_{\mathrm{L}^2(\mathbb{R})}$ for a fixed integer $p \geq 2$. Recall that $\mathrm{H}^s(\mathbb{R}) = \big \{f \in L^2(\mathbb{R}) : \xi \mapsto\left(1+ |\xi|^2 \right )^{\frac{s}{2}}\mathscr{F}f(\xi) \in L^2(\mathbb{R}) \big \}$.

(Note: $N$ is well-defined by the fractional-order Sobolev embedding theorem: $\mathrm{H}^s(\mathbb{R}) \hookrightarrow \mathrm{BC}(\mathbb{R})$ for $s > \frac{1}{2}$, where $\mathrm{BC}(\mathbb{R})$ is the space of bounded and continuous functions $\mathbb{R} \to \mathbb{C}$.)

Is $N$ continuous on $\mathrm{H}^s(\mathbb{R})$ or continuous when restricted to some open ball $B_R(0) \subset \mathrm{H}^s(\mathbb{R})$?

I suspect some results on composition operators on Sobolev spaces might be useful, but have not found an applicable version.

F. H.
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You are essentially asking whether the map $u\mapsto u^p$ is continuous from $H^s$ to $H^s$. Yes, it is. For $s>1/2$ the Sobolev space $H^s(\mathbb{R})$ is an algebra: $$ \|u v\|_s \le C\|u\|_s\|v\|_s $$ This implies that the multiplication map $(u,v)\mapsto uv$ is continuous, since on every ball it is Lipschitz in each argument separately.

For $p\ge 2$, the multiplication map $(u_1,\dots,u_p)\mapsto u_1\cdots u_p$ is continuous from $H^s\times\cdots\times H^s$ to $H^s$, because it can be written as the composition of $p-1$ binary multiplications. Restricting to the diagonal $u_1=\cdots u_p$ shows that $u\mapsto u^p$ is continuous.

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