Interpolation of a linear operator acting on a sequence of functions Let $\mathbf{f} = \{f_{n}\}$ be a sequence of Schwarz functions and suppose $T$ is a linear operator which sends a given sequence of Schwarz functions to a given function in $L^{p}(\mathbb{R}^n)$ for all $p$. Suppose for some $1 \leq q_{1}< q_{2} \leq \infty$, we had $\|T\mathbf{f}\|_{L^{q_i}} \leq C_{i}(\sum_{n}\|f_{n}\|_{L^{q_i}}^{q_i})^{1/q_i}$. Is it true that for $q_{1} \leq q \leq q_{2}$, $$\|T\mathbf{f}\|_{L^{q}} \lesssim (\sum_{n}\|f_{n}\|_{L^{q}}^{q})^{1/q}?$$
I imagine this is an application of some sort of interpolation since I can write the right hand side as $\|T\mathbf{f}\|_{L^{q}} \leq C_{i}\|\mathbf{f}\|_{L^{q}_{x}\ell_{n}^{q}}$ where the $L^{q}_{x}\ell_{n}^{q}$ norm is defined by we first take the $L^{q}_{x}$ norm of each term then take the $\ell_{n}^{q}$ norm.
 A: Yes, the result is true--unless I'm making an error below--and even holds under the weaker hypothesis
$$\|T\mathbf{f}\|_{L^{q_{i},\infty}}\leq C_{i}\|\mathbf{f}\|_{L_{x}^{q_{i}}\ell_{n}^{q_{i}}},\quad\forall i=1,2$$
To prove this, we mimic the proof of the Marcinkiewicz interpolation theorem. For $\delta>0$ be a parameter to be determined later. For $\lambda>0$, we write $\mathbf{f}=\mathbf{f}_{1}^{\lambda}+\mathbf{f}_{2}^{\lambda}$, where
$$\mathbf{f}_{1}^{\lambda}:=\{f_{n}1_{|f_{n}|>\delta\lambda}\}_{n}=\{f_{1,n}^{\lambda}\}_{n}\enspace \mathbf{f}_{2}^{\lambda}:=\{f_{n}1_{|f_{n}|\leq\delta\lambda}\}_{n}=\{f_{2,n}^{\lambda}\}_{n}$$
By the triangle inequality together with Chebyshev and the hypotheses, we have the estimate
\begin{align*}
|\{|T\mathbf{f}|>\lambda\}|&\leq|\{|T\mathbf{f}_{1}^{\lambda}|>\lambda/2\}|+|\{|T\mathbf{f}_{2}^{\lambda}|>\lambda/2\}|\\
&\leq \dfrac{2^{q_{1}}\|T\mathbf{f}_{1}^{\lambda}\|_{L^{q_{1}}}^{q_{1}}}{\lambda^{q_{1}}}+\dfrac{2^{q_{2}}\|T\mathbf{f}\|_{L^{q_{2}}}^{q_{2}}}{\lambda^{q_{2}}}\\
&\leq\dfrac{2^{q_{1}}(C_{1}\|\mathbf{f}_{1}^{\lambda}\|_{L_{x}^{q_{1}}\ell_{n}^{q_{1}}})^{q_{1}}}{\lambda^{q_{1}}}+\dfrac{2^{q_{2}}(C_{2}\|\mathbf{f}_{2}^{\lambda}\|_{L_{x}^{q_{2}}\ell_{n}^{q_{2}}})^{q_{2}}}{\lambda^{q_{2}}}
\end{align*}
Using the layer cake formula, we have the estimate
\begin{align*}
\|T\mathbf{f}\|_{L^{q}}^{q}&=q\int_{0}^{\infty}\lambda^{q-1}|\{|T\mathbf{f}|>\lambda\}|d\lambda\\
&\leq (2C_{1})^{q_{1}}q\int_{0}^{\infty}\lambda^{q-q_{1}-1}\sum_{n}\int_{\mathbb{R}^{n}}|f_{1,n}^{\lambda}|^{q_{1}}dxd\lambda\\
&+(2C_{2})^{q_{2}}q\int_{0}^{\infty}\lambda^{q-q_{2}-1}\sum_{n}\int_{\mathbb{R}^{n}}|f_{2,n}^{\lambda}|^{q_{2}}dxd\lambda\\
&=(2C_{1})^{q_{1}}q\int_{0}^{\infty}\lambda^{q-q_{1}-1}\sum_{n}\int_{\{|f_{n}|>\delta\lambda\}}|f_{n}|^{q_{1}}dxd\lambda\\
&+(2C_{2})^{q_{2}}q\int_{0}^{\infty}\lambda^{q-q_{2}-1}\sum_{n}\int_{\{|f_{n}|\leq\delta\lambda\}}|f_{n}|^{q_{2}}dxd\lambda
\end{align*}
By monotone convergence, we can pull the sum outside the outer integrals. By Fubini-Tonelli,
\begin{align*}
\int_{0}^{\infty}\lambda^{q-q_{1}-1}\int_{\{|f_{n}|>\delta\lambda\}}|f_{n}(x)|^{q_{1}}dxd\lambda&=\int_{\mathbb{R}^{n}}|f_{n}(x)|^{q_{1}}\int_{0}^{\delta^{-1}|f_{n}(x)|}\lambda^{q-q_{1}-1}d\lambda dx\\
&=\dfrac{\delta^{-q+q_{1}}}{q-q_{1}}\int_{\mathbb{R}^{n}}|f_{n}|^{q}dx\\
\end{align*}
Similarly,
\begin{align*}
\int_{0}^{\infty}\lambda^{q-q_{2}-1}\int_{\{|f_{n}|\leq\delta\lambda\}}|f_{n}(x)|^{q_{2}}dx d\lambda&=\int_{\mathbb{R}^{n}}|f_{n}(x)|^{q_{2}}\int_{\delta^{-1}|f_{n}(x)|}^{\infty}\lambda^{q-q_{2}-1}d\lambda dx\\
&=\dfrac{\delta^{-q+q_{2}}}{(q_{2}-q)}\int_{\mathbb{R}^{n}}|f_{n}|^{q}dx
\end{align*}
Combining these two results, we obtain that
\begin{align*}
\|T\mathbf{f}\|_{L^{q}}^{q}&\leq \sum_{n}\left[\dfrac{(2C_{1})^{q_{1}}q}{q-q_{1}}\delta^{-q+q_{1}}+\dfrac{(2C_{2})^{q_{2}}q}{q_{2}-q}\delta^{-q+q_{2}}\right]\|f_{n}\|_{L^{q}}^{q}\\
&=\left[\dfrac{(2C_{1})^{q_{1}}q}{q-q_{1}}\delta^{-q+q_{1}}+\dfrac{(2C_{2})^{q_{2}}q}{q_{2}-q}\delta^{-q+q_{2}}\right]\|\mathbf{f}\|_{L_{x}^{q}\ell_{n}^{q}}^{q}
\end{align*}
Choosing $\delta$ so that both terms inside the brackets are equal and then taking $q^{th}$ roots completes the proof.
