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Let $H^s$, $0\leq s<\infty$ be the $L^2$ based Sobolev spaces such that

$$ \hat{f}(\xi)(1+|\xi|^2)^{s/2} \in L^2. $$

Let $r_1,r_2,p_1,p_2>0$ be given parameters. Assume that a linear operator $T$ satisfies that $$ T:H^{r_1}\rightarrow H^{r_2} $$ with operator norm $$ \|T\|_{H^{r_1},H^{r_2}}=C_1 $$ and $$ T:H^{p_1}\rightarrow H^{{p_2}} $$ with operator norm $$ \|T\|_{H^{p_1},H^{p_2}}=C_2. $$

Question 1I would like to have a precise reference for the following result (that I'm confident it's true)

Lemma: Assume $r_2=r_1$ and $p_2=p_1$. Assume also that $T$ satisfies the assumptions above and let $$ s_1=\theta r_1+(1-\theta)p_1. $$
Then

$$ T:H^{s_1}\rightarrow H^{{s_1}} $$ with operator norm $$ \|T\|_{H^{s_1},H^{s_1}}\leq C_1^\theta C_2^{1-\theta}. $$

Question 2 Is there some result in the case $r_1\neq r_2$, $p_1\neq p_2$.?

Thanks!!

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    $\begingroup$ This should be contained exactly in the Wikipedia article on interpolation spaces: en.wikipedia.org/wiki/… $\endgroup$ Commented Dec 28, 2016 at 10:31
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    $\begingroup$ Lions and Magenes, Non-homogeneous boundary value problems and applications, Vol. 1, Theorem 5.1 $\endgroup$ Commented Dec 28, 2016 at 15:23
  • $\begingroup$ Do you really want to have an equality for the operator norm, or only an estimate? $\endgroup$ Commented Dec 28, 2016 at 21:08
  • $\begingroup$ Only an estimate. I already edited the post. Thank you. $\endgroup$ Commented Dec 29, 2016 at 16:10

1 Answer 1

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A precise statement in the language of interpolation theory (in more generality, but this is the best way to state the result in my opinion) can be found in Theorem 6.4.5(7) of Bergh and Löfström's Interpolation Spaces. If you haven't learned some abstract interpolation theory, now's your chance! This also answers your Question 2: there is indeed a result in the case $r_1 \neq r_2$, $p_1 \neq p_2$. You have $T : H^{s_1} \to H^{s_2}$ with $s_2$ defined analogously to $s_1$.

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