Understanding multilinear interpolation

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 $$M_{\theta}=M^{1-\theta}_{0}M_{1}^{\theta}$$ ?

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

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?