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I know that for product of function, Sobolev's inequality tell us that, for any $s>\tfrac{1}{2}$ the following holds: $$ \Vert uv\Vert_{H^s(\mathbb{R})}\leq c\Vert u\Vert_{H^s(\mathbb{R})} \Vert v\Vert_{H^s(\mathbb{R})}, $$ where, of course, $u,v\in H^s(\mathbb{R})$. Now, while reading a book I have found something like the following (this is what I think they did, there is no explanation actually): Consider three functions $u,v,w\in H^{1/6}(\mathbb{R})$. Is it true that $$ \Vert uvw\Vert_{L^1(\mathbb{R})} \leq c \Vert u\Vert_{H^{1/6}(\mathbb{R})}\Vert v\Vert_{H^{1/6}(\mathbb{R})}\Vert w\Vert_{H^{1/6}(\mathbb{R})}? $$

Neldrock
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1 Answers1

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Here it's really the Sobolev embedding and generalized Hölder that are doing the work. The former shows that since $$ \frac{1}{3} = \frac{1}{2} - \frac{1}{1} \left(\frac{1}{6} \right) $$ we have the continuous embedding $H^{1/6}(\mathbb{R}) \hookrightarrow L^3(\mathbb{R})$. Then, since $$ 1= \frac{1}{3} + \frac{1}{3}+ \frac{1}{3} $$ the general Hölder inequality says that $$ \Vert f g h \Vert_{L^1(\mathbb{R})} \le \Vert f \Vert_{L^3(\mathbb{R})} \Vert g \Vert_{L^3(\mathbb{R})} \Vert h \Vert_{L^3(\mathbb{R})}. $$ Chaining these two bounds together then shows that $$ \Vert f g h \Vert_{L^1(\mathbb{R})} \le c \Vert f \Vert_{H^{1/6}(\mathbb{R})} \Vert g \Vert_{H^{1/6}(\mathbb{R})} \Vert h \Vert_{H^{1/6}(\mathbb{R})} $$

Glitch
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