Proof of an inequality in Sobolev space.

Question

I want to show the next inequality:

$$\| D^a u D^b v \|_{L^2(\mathbb{R}^n)} \leq C \| u \|_{H^s(R^n)} \| v \|_{H^s(R^n)}$$ (for $s>n/2$) Where $D$ is a differential operator.

What I did so far is to write the next stuff:

$$\| D^a u D^b v \|_{L^2} = \| \hat{(D^a u D^b v)} \|_{L^2} = \| \hat{D^a u} * \hat{D^b v} \|_{L^2}$$

But how to proceed from there? (I used Parseval identity and the fourier transform of multiplication equals the convolution of fourier transforms).

$D$ is the differential operator, i.e: $D^b = \frac{\partial^{|b|}}{\partial x_1^{a_1}\cdots \partial x_n^{a_n}}$, where $|b|=a_1+\cdots+ a_n$. — MathematicalPhysicist, Dec 16 '14 at 17:16

score 0 · Answer 1 · answered Dec 17 '14 at 05:43

0

An outline:

If $s-|a|>n/2$ or $s-|b|>n/2$ then we're done since, in the first case, we get $D^a u\in L^{\infty}$ with appropriate control by Sobolev embedding.

If $s-|a|, s-|b|<n/2$, then again by Sobolev embedding we get that $D^a u\in L^p$ for every $p\in [2, 2^*_a]$, where $2^*_a:=2n/(n-2(s-|a|))$, and similarly for $b$. The desired inequality then follows from Hölder's inequality since $1/2^*_a+1/2^*_b<1/2$ (since $|a|+|b|\leq s$).

If $s-|a|=n/2$ then $D^a u\in L^p$ for every $p\in [2,\infty)$, and we proceed as before.

answered Dec 17 '14 at 05:43

Jose27

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Do you happen to know how to prove this claim with a similar argument as in the proof that sobolev space is a banach algebra which timur provided in math.stackexchange? – MathematicalPhysicist Dec 17 '14 at 06:58
Which proposition are you quoting as in "Sobolev embedding"? – MathematicalPhysicist Dec 19 '14 at 10:00

Proof of an inequality in Sobolev space.

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