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Grafakos in his $\textit{Classical Fourier Analysis}$ formulates (see Exercise 4.5.2 therein) the following vector-valued version of the Riesz-Thorin interpolation theorem.

$\textbf{Theorem}$

Let $1\le p_0, q_0,p_1,q_1, r_0, s_0, r_1, s_1\le\infty$ and $\theta\in(0,1)$ satisfy \begin{align*} \frac{1-\theta}{p_0}+\frac{\theta}{p_1}&=\frac{1}{p},\qquad \frac{1-\theta}{q_0}+\frac{\theta}{q_1}=\frac{1}{q},\\ \frac{1-\theta}{r_0}+\frac{\theta}{r_1}&=\frac{1}{r},\qquad \frac{1-\theta}{s_0}+\frac{\theta}{s_1}=\frac{1}{s}, \end{align*} and let $T$ be a linear operator mapping $L^{p_0}(\mathbb{R}^n, \ell^{r_0})$ to $L^{q_0}(\mathbb{R}^n, \ell^{s_0})$ and $L^{p_1}(\mathbb{R}^n, \ell^{r_1})$ to $L^{q_1}(\mathbb{R}^n, \ell^{s_1})$. Then $T$ maps $L^{p}(\mathbb{R}^n, \ell^{r})$ to $L^{q}(\mathbb{R}^n, \ell^{s})$.

My $\textbf{question}$ is whether one can weaken the assumption that $T$ is linear? More precisely, does an analogous result hold for an operator of the form $$ T: \{f_j\}_{j\in\mathbb{N}}\rightarrow\{M f_j\}_{j\in\mathbb{N}}, $$ where $M$ is a sublinear operator i.e. $|M(f+g)(x)|\le |Mf(x)|+|Mg(x)|,\quad x\in\mathbb{R}^d$?

In the scalar-valued there is an interpolation theorem for sublinear operators and it goes by the name of Marcinkiewicz-Zygmund.

I would appreciate any hints or perhaps a reference to suitable literature.

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    $\begingroup$ How do you define sublinearity in the vector-valued case? $\endgroup$
    – Bazin
    Commented Jul 3, 2020 at 14:20
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    $\begingroup$ In general let's say $\|T(f+g)\|_{\ell^s}\le\|Tf\|_{\ell^s}+\|Tg\|_{\ell^s}$, $s\in[s_0,s_1]$. In the particular application I have in mind $T$ is of the form $T(\{f_j\}_{j\in\mathbb{N}})=\{Mf_j\}_{j\in\mathbb{N}}$, for some sublinear operator $M$. Thanks for a good question! $\endgroup$
    – Tony419
    Commented Jul 3, 2020 at 14:46
  • $\begingroup$ Is your first condition sufficient in the scalar case? It looks as a very weak requirement compared to sublinearity. $\endgroup$
    – Bazin
    Commented Jul 4, 2020 at 21:42
  • $\begingroup$ I'm not sure if I understand what you mean. In the scalar case the norm $\|\cdot\|_{\ell^s}$ is replaced by the standard absolute value and then this condition becomes exactly the sublinearity condition for $T$. $\endgroup$
    – Tony419
    Commented Jul 4, 2020 at 22:10

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