TITLE: Proof of Schur's test via Young's inequality
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: Actually two-way Young's inequality is good enough:
Notice that since $ p_0 =1$ and $ q_1  = \infty $ we have $ p_\theta ( 1 -\theta ) + \frac{\theta p_\theta }{ p_1 } =1$ and $q_\theta' \theta +\frac{(1 -\theta) q_\theta' }{q_0 ' } =1 $. So write $$|f(x) h(y) | = |f(x)|^{p_\theta ( 1 -\theta )} |h(y)|^{\frac{(1 -\theta) q_\theta' }{q_0 ' }} |f(x)|^{\frac{\theta p_\theta }{ p_1 }}|h(y)|^{q_\theta' \theta}$$Thus by Young's inequality we have $$ |f(x) h (y) | \le (1 - \theta ) |f(x) |^{p _\theta } |h (y) |^{ \frac{q_\theta ' }{ q_0 ' }} + \theta |f(x) |^{ \frac{p_\theta}{p_1}} |h(y) |^{ q_\theta ' } $$ Then $$\int_Y \int_X |K(x,y ) f(x) h (y)| d \mu (x) d \mu (y) \le ( 1 - \theta ) \int_Y \int_X |K (x,y )|f(x) |^{p _\theta } |h (y) |^{ \frac{q_\theta ' }{ q_0 ' }} d\mu (x) d \mu (y) + \theta \int _ Y \int_X K(x,y ) |f(x) |^{ \frac{p_\theta}{p_1}} |h(y) |^{ q_\theta ' }  d\mu (x) d \mu (y) $$
Since $ K(x, \ast )$ is in $L^{q_0 }$ for almost every $x$ and $ |h(y) |^{ \frac{q_\theta ' }{ q_0 ' }} $ is in $L^{ q_0 ' } $, by holder's inequality we have $$\int_Y |K (x,y )| |h (y) |^{ \frac{q_\theta ' }{ q_0 ' }} d \mu (y) \le 1$$ for almost everywhere $x$. And similarly we have $$\int_X K(x,y ) |f(x) |^{ \frac{p_\theta}{p_1}}   d\mu (x)  \le 1$$ for almost everywhere $y$.
Thus by Fubini's theorem we have $$ (1 - \theta ) \int _Y \int_X |K (x,y )|f(x) |^{p _\theta } |h (y) |^{ \frac{q_\theta ' }{ q_0 ' }} d\mu (x) d \mu (y)  \le (1 - \theta )  $$ and $$\theta \int _ Y \int_X K(x,y ) |f(x) |^{ \frac{p_\theta}{p_1}} |h(y) |^{ q_\theta ' }  d\mu (x) d \mu (y) \le \theta $$ 
Sum up the two parts we get $$ \int_Y \int_X |K(x,y ) f(x) h (y)| d \mu (x) d \mu (y)   \le 1- \theta + \theta =1$$ as wanted.