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Consider $f \in L^{\infty}(\mathbb{R}^n)$ and $g \in H^{-2}(\mathbb{R}^n) = \{r \in S'(\mathbb{R}^n); \| \langle \xi\rangle^{-2}F(r)\|_{L^2}<\infty \}$, is $fg \in H^{-2}(\mathbb{R}^n)$?

Some partial progress under different hypotheses

  • When $f, \nabla f, \nabla^2f \in L^{\infty}(\mathbb{R}^n)$, use the dual norm $$ \| fg\|_{H^{-2}}=\sup\{|\langle fg, p\rangle_{L^2}|; p \in H^{2}(\mathbb{R}^n), \|p\|_{H^2}=1\}. $$ Then $$ |\langle fg, p\rangle_{L^2}|=|\langle g,fp\rangle_{L^2}| \leq \|g\|_{H^{-2}} \|fp\|_{H^2} \leq C \|g\|_{H^{-2}} \|p\|_{H^2}, $$ where $C$ depends on the $L^{\infty}$ norms of $f, \nabla f$ and $\nabla^2 f$ and so in this case $\|fg\|_{H^{-2}} \leq C \|g\|_{H^{-2}}$.
  • If on the other hand $g \in H^{-1}(\mathbb{R}^n)$ and we're just trying to show $fg \in H^{-1}(\mathbb{R}^n)$, then again using the dual norm $$ \| fg\|_{H^{-1}}=\sup\{|\langle fg, p\rangle_{L^2}|; p \in H^{1}(\mathbb{R}^n), \|p\|_{H^1}=1\}. $$ Then, since $p\in H^{1}$ implies $|p| \in H^{1}$ (which I believe is not true for $H^2$) and if $g \in H^{-1}$ implies $|g| \in H^{-1}$ (which I'm not 100% sure about) $$ |\langle fg, p\rangle_{L^2}| \leq \int_{\mathbb{R}^n} |fgp| dx \leq \|f\|_{L^{\infty}} \int_{\mathbb{R}^n} |gp| dx \leq C \|g\|_{H^{-1}} \|p\|_{H^1}, $$ and so in this case $\|fg\|_{H^{-1}} \leq C \|g\|_{H^{-1}}.$
Perry Kleinhenz
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1 Answers1

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Some remarks to your attempt:

(1) If $p\in H^k$, $k\ge 1$, $f\in L^\infty$, then $fp \not\in H^k$ in general. So the attempt to define $fg$ as an element of $H^{-2}$ is about to fail.

Consider the case $n\le 3$, then $H^2$ functions are continuous. The Dirac functional $\delta : u \mapsto u(0)$ is in $H^{-2}$. Now take a function $f$ that is discontiuous at $x=0$. How to define $f\cdot \delta$?

(2) If $g\in H^{-1}$ then its positive and negative part and so also $|g|$ are not in $H^{-1}$ in general, see this answer Decomposition of measures acting on sobolev spaces

daw
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