2
$\begingroup$

Do I understand the interpolation of bounded multilinear operators on Banach spaces correct ? For simplicity, let us Consider bilinear operators on $L^p$ spaces.

Suppose the bilinear operator $T$ is bounded from $L^{p_0}\times L^{q_0}$ to $L^{r_0}$. This fact is usually written:

$T:L^{p_0}\times L^{q_0}\longrightarrow L^{r_0}$

Suppose also that

$T:L^{p_1}\times L^{q_1}\longrightarrow L^{r_1}$.

Then

$T:L^{p_{\theta}}\times L^{q_{\theta}}\longrightarrow L^{r_{\theta}}$ where

$0<\theta<1$, $\frac{1}{p_{\theta}}= \frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}}$, $\frac{1}{q_{\theta}}= \frac{1-\theta}{q_{0}}+\frac{\theta}{q_{1}}$, and $\frac{1}{r_{\theta}}=\frac{1-\theta}{r_{0}}+\frac{\theta}{r_{1}}$.

Moreover, if

$$\parallel T(f,g)\parallel_{L^{r_{0}}} \leq M_0 \parallel (f,g) \parallel_{L^{p_0}\times L^{q_0}}, \\ \parallel T(f,g)\parallel_{L^{r_{1}}} \leq M_1 \parallel (f,g) \parallel_{L^{p_1}\times L^{q_1}}. $$

Then

$$\parallel T(f,g)\parallel_{L^{r_{\theta}}} \leq M^{1-\theta}_{0} M_{1}^{\theta}\parallel (f,g) \parallel_{L^{p_{\theta}}\times L^{q_{\theta}}}$$

Now, consider a trilinear operator $T$ such that

$T:L^{p_0}\times L^{q_0}\times L^{r_0} \longrightarrow L^{s_0}$,

$T:L^{p_1}\times L^{q_1}\times L^{r_1} \longrightarrow L^{s_1}$

Question (1):

Is it true that

$T:L^{p_\theta}\times L^{q_\theta}\times L^{r_\theta} \longrightarrow L^{s_\theta}$

with

$\frac{1}{p_{\theta}}= \frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}}$,

$\frac{1}{q_{\theta}}= \frac{1-\theta}{q_{0}}+\frac{\theta}{q_{1}}$,

$\frac{1}{r_{\theta}}=\frac{1-\theta}{r_{0}}+\frac{\theta}{r_{1}}$, and

$\frac{1}{s_{\theta}}=\frac{1-\theta}{s_{0}}+\frac{\theta}{s_{1}}$

for every fixed

$\theta \in\,]0,1[$.

The "boundedness constant" is $M_{\theta}=M^{1-\theta}_{0}M_{1}^{\theta}$ ?

What if, in addition,

$T:L^{p_2}\times L^{q_2}\times L^{r_2} \longrightarrow L^{s_2}$.

Is it true that

$T:L^{p_{\theta}}\times L^{q_{\theta}}\times L^{r_{\theta}} \longrightarrow L^{s_{\theta}}$

where

$\frac{1}{p_{\theta}}= \frac{\alpha_1}{p_{0}}+\frac{\alpha_2}{p_{1}} +\frac{\alpha_3}{p_{2}}$,

$\frac{1}{q_{\theta}}= \frac{\alpha_1}{q_{0}}+\frac{\alpha_2}{q_{1}} +\frac{\alpha_3}{q_{2}}$,

$\frac{1}{r_{\theta}}= \frac{\alpha_1}{r_{0}}+\frac{\alpha_2}{r_{1}} +\frac{\alpha_3}{r_{2}}$, and

$\frac{1}{s_{\theta}}= \frac{\alpha_1}{s_{0}}+\frac{\alpha_2}{s_{1}} +\frac{\alpha_3}{s_{2}}$

for every $(\alpha_{1},\alpha_{2},\alpha_{3})\in\,]0,1[\times ]0,1[\times]0,1[$ such that $\sum_{i}^{3}\alpha_{i}=1$

and the constant is given by

$M_{\theta}=M^{\alpha_{1}}_{0} M^{\alpha_{2}}_{1} M^{\alpha_{3}}_{2}.$

Correct?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.