I am able to prove the following generalization of Schur's test using the Riesz-Thorin interpolation theorem, however I have been stuck for days now trying to prove it using Young's inequality:
Let the integral operator $T$ from functions $f: X \to \Bbb C$ to functions $Tf: Y \to \Bbb C$ be defined via the kernel $K: X \times Y \to \Bbb C$ , which is some measurable function. Moreover, let
$$ \| K(x,\cdot) \| _{L^{q_0}(Y) } \leq 1 $$
$$\| K(\cdot,y)\|_{L^{p_1'}(X) } \leq 1$$
for all $x\in X$ and all $y\in Y$.
Then for every $0<\theta<1$ and all $f\in L^{p_\theta}$
$$\| Tf \| _{L^{q_\theta}(Y) } \leq \| f\|_{L^{p_\theta}(X) } $$
where
$${1\over p_\theta} = {1-\theta\over p_0} + {\theta\over p_1} $$
$${1\over q_\theta} = {1-\theta\over q_0} + {\theta\over q_1} $$
$${1} = {1\over p_1} + {1\over p_1'} $$
and $p_0=1$ and $q_1=+\infty$.
I am able to prove the special case $p_1'=q_0=1$ via Hölder's as well as Young inequalities. However, I am making zero progress trying to prove the general case. I am struggling with this now for almost a week and I would greatly appreciate help! From online sources I know that a proof based on Young's inequality is possible. Young's inequality for non-negative reals $x,y$ is $$ xy \leq x^r/r + y^s/s$$ for dual exponents satisfying $1/r+1/s=1$ for $1<r<\infty$. Thanks in advance.
[EDIT: This answer, in fact, is wrong. See comments.]
I was finally able to use a three-way Young's inequality to prove the theorem. However, I was able to see a very slightly more direct proof using Hölder's inequality. Both theorems exploit convexity, whereas Riesz-Thorin relies on complex analyticity.
Here is the sketch of the proof using Hölder's inequality:
Below, $\mu$ and $\nu$ are the measures on $X$ and $Y$ respectively.
The idea is to prove the theorem for simple functions $f$ of finite measure support and then use monotone convergence.
$$ \|~T f~\|_{L^{q_\theta} } \leq \sup_{\|{h}\|_{L^{q_\theta'}(Y) }\leq 1} | \int_Y \int_X |K(x,y)| |f(x)| ~d\mu(x) ~h(y) ~d\nu(y) ~| $$
so that it suffices to show that
$$ | \int_Y \int_X |K(x,y)f(x) h(y)| ~d\mu(X) ~d\nu(y) ~| \leq \|{f}\|_{L^{p_\theta}(X) } \|{h}\| _{L^{q_\theta'}(Y) } ~. $$
We can exploit homogenization symmetry in the claim. Let us normalize $\|{f}\|_{L^{p_\theta}(X) }=\|{h}\| _{L^{q_\theta'}(Y) }=1$.
Denoting $L^p(X\times Y)$ as $L^p$ for convenience, we can use Hölder's inequality for multiple exponents as follows:
\begin{align} \|{ K f h }\|_{L^1} &\leq \|{ K^{p_1'/r_1} f^{p_\theta/r_1} } \|_{L^{r_1}} \|{ K^{q_0/r_2} h^{q_\theta'/r_2} }\|_{L^{r_2}} \|{ f^{p_\theta/r_3} h^{q_\theta'/r_3} } \|_{L^{r_3}} \\ 1&= {1\over r_1} +{1\over r_2} +{1\over r_3} \\ 1&= {p_1'\over r_1} + {q_0\over r_2}\\ 1&= {p_\theta\over r_1} + {p_\theta\over r_3} \end{align}
which has solution the solution
\begin{align} {1\over r_1} &={1\over q_0}\\ {1\over r_2} &= {q_\theta-q_0 \over q_\theta p_1' }\\ {1\over r_3} &= {1\over q_0 } - {1\over p_\theta} ~. \end{align}
Since $q_\theta >q_0$ exponents $r_1,~r_2$ are finite. If $1/r_3$ becomes zero then the three-way Hölder's inequality becomes the standard Hölder's inequality.
Each of the $L^r$ norms can be evaluated by the Fubini-Tonelli theorem. Since each of 3 the norms on the RHS equals 1, we finally get, after putting everything together, $$ \| Tf \|_{q_\theta} \leq 1$$ and the claim follows.