# $\| fg \|_2 \leqslant c ( \| f \|_\infty \| \nabla g \|_2 + \| \nabla f \|_2 \| g \|_\infty )$

If $f, g \in W^{1,2} ( \mathbb R^n ) \cap L^\infty ( \mathbb R^n)$, then prove that there exists $c >0$ such that $$\| fg \|_2 \leqslant c ( \| f \|_\infty \| \nabla g \|_2 + \| \nabla f \|_2 \| g \|_\infty ),$$ where $\| \nabla f \|_2 := ( \| \partial_1 f \|_2^2 + \cdots + \| \partial_n f \| _2^2 )^{1/2} .$

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## 1 Answer

Is it true? Let $f,g \in C^{\infty}_0(\mathbb{R}^n)$. Let $f_\lambda(x) = f(\lambda x)$.

$$\|f_\lambda g_\lambda\|_2^2 = \int f^2(\lambda x) g^2(\lambda x) \mathrm{d}x = \int f^2(y)g^2(y) \frac{1}{\lambda^n} \mathrm{d}y = \lambda^{-n} \|fg\|_2^2$$

Clearly we also have

$$\|f\|_\infty = \|f_\lambda\|_\infty$$

and

$$\|\nabla f_\lambda \|_2^2 = \int \lambda^2 (\nabla f)^2(\lambda x) \mathrm{d}x = \int \lambda^2 (\nabla f)^2(y) \frac{1}{\lambda^n}\mathrm{d}y = \lambda^{2-n} \| \nabla f\|_2^2$$

So if the inequality were to hold, in general, we need to have

$$\lambda^{-n/2} \|fg\|_2^2 \leq c \lambda^{1 - n/2}\left( \|f\|_\infty \|\nabla g\|_2 + \|g\|_\infty \|\nabla f\|_2\right)$$

which if you take $\lambda \to 0$ requires that $\|fg\|_2 = 0$. By the above dimensional analysis, your inequality is not true as stated.

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In fact, using dimensional analysis, perhaps what you are interested in is the bilinear Sobolev inequality $$\|fg\|_p \leq c( \|f\|_\infty \|\nabla g\|_2 + \|g\|_\infty \|\nabla f\|_2)$$ for $p = 2n / (n-2)$? – Willie Wong Jul 9 '12 at 9:41
Oh it was wrong as well. Thank you Willie Wong. – KiaSure Jul 9 '12 at 11:42