Proof of Schur's test via Young's inequality 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.
 A: [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.
A: I don't have enough points to comment, but the above solution doesn't seem to work, for example, one can just take $\theta = 1/7$, $q_{0} = 3$, $p_{1} = 5$, $q_{1} = \infty$, $p_{0} = 1$. This makes $q_{\theta} = 7/2$, $p_{1}' = 5/4$, $p_{\theta}' = 35/4$, $p_{\theta} = 35/31$, and $q_{\theta}' = 7/5$. Given these numbers, we cannot find $r_{1}, r_{2}, r_{3}$ as above.
A: 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.
A: We want to show that
$$\int_Y \int_X |K(x, y)| |f(x)| |h(y)| \, dx \, dy \le 1$$
for $f \in L^{p_\theta}(X)$ with $||f||_{L^{p_\theta}(X)} = 1$ and $h \in L^{q_\theta'}(Y)$ with $||h||_{L^{q_\theta'}} = 1$.
To show this, we use Young's inequality for three variables.  That is, for any $x, y, z \ge 0$ and $1 < r_1, r_2, r_3 < \infty$ such that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = 1$, we have
$$xyz \le \frac{1}{r_1}x^{r_1} + \frac{1}{r_2}y^{r_2} + \frac{1}{r_3}z^{r_3}.$$
Pick $r_1, r_2, r_3$ by
$$\frac{1}{r_1} = \frac{1}{q_\theta}, \hspace{0.2in} \frac{1}{r_2} = \frac{1}{p_\theta'}, \hspace{0.2in} \frac{1}{r_3} = \frac{1}{p_\theta} - \frac{1}{q_\theta} = \frac{1}{q_\theta'} - \frac{1}{p_\theta'}.$$
Note
$$\frac{1}{r_3} = 1 - \frac{1 - \theta}{q_0} - \frac{\theta}{p_1'} \ge 1 - (1 - \theta) - \theta = 0.$$
If $\frac{1}{r_3} = 0$, then we proceed as below but using Young's inequality for the two variables $r_1$ and $r_2$ and ignoring all terms with $r_3$.
Also let
$$r = \frac{q_0}{r_1}, \hspace{0.2in} s = \frac{p_1'}{r_2}, \hspace{0.2in} t = \frac{p_\theta}{r_1},$$
$$u = \frac{p_\theta}{r_3}, \hspace{0.2in} v = \frac{q_\theta'}{r_2}, \hspace{0.2in} w = \frac{q_\theta'}{r_3}.$$
By construction, these satisfy
$$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = 1,$$
$$r + s = 1, \hspace{0.2in} t + u = 1, \hspace{0.2in} v + w = 1,$$
$$rr_1 = q_0, \hspace{0.1in} sr_2 = p_1', \hspace{0.1in} tr_1 = p_\theta, \hspace{0.1in} vr_2 = q_\theta', \hspace{0.1in} ur_3 = p_\theta, \hspace{0.1in} wr_3 = q_\theta'.$$
Thus by Young's inequality for three variables (or two variables if $r_3$ is infinite),
$$\left|K(x, y) f(x) h(y)\right| = 
 (|K(x, y)|^{r}|f(x)|^t)(|K(x, y)|^s |h(y)|^v)(|f(x)|^u |h(y)|^w)
\le \frac{1}{r_1} |K(x, y)|^{rr_1}|f(x)|^{tr_1} + \frac{1}{r_2} |K(x, y)|^{sr_2} |h(y)|^{vr_2} + \frac{1}{r_3} |f(x)|^{ur_3} |h(x)|^{wr_3}$$
$$\le \frac{1}{r_1} |K(x, y)|^{q_0} |f(x)|^{p_\theta} + \frac{1}{r_2} |K(x, y)|^{p_1'} |h(y)|^{q_\theta'} + \frac{1}{r_3}|f(x)|^{p_\theta} |h(y)|^{q_\theta'} \hspace{0.2in} (1).$$
Using Fubini-Tonelli and the assumption that $||K(x, \cdot)||_{L^{q_0}(Y)} \le 1$ for almost every $x \in X$
\begin{eqnarray*}
\int_{Y} \int_X |K(x, y)|^{q_0} |h(y)|^{q_\theta'} \, dx \, dy & = & \int_X \int_Y |K(x, y)|^{q_0} |h(y)|^{q_\theta'} \, dy \, dx\\
& \le & \int_X ||K(x, \cdot)||_{L^{q_0}(Y)}^{q_0} \, dx \int_Y |h(y)|^{q_\theta'} \, dy \le ||h||_{L^{q_\theta'}(Y)}^{q_\theta'} \le 1.
\end{eqnarray*}
Likewise, using the assumption that $||K(\cdot, y)||_{L^{p_1'}(X)} \le 1$ for almost every $y \in Y$,
$$\int_Y \int_X |K(x, y)|^{p_1'} |f(x)|^{p_\theta} \, dx \, dy \le 1.$$
Finally,
$$\int_Y \int_X |f(x)|^{p_\theta} |h(y)|^{q_\theta'} \, dx \, dy \le ||f||_{L^{p_\theta}(X)}^{p_\theta}||h||_{L^{q_\theta'}(Y)}^{q_\theta'} \le 1.$$
Integrating Inequality (1) and substituting these three estimates,
$$\int_Y \int_X |K(x, y)| |f(x)| |h(y)| \, dx \, dy \le \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = 1.$$
There appears to be enough symmetry in the system of ten equations for the nine variables $r_1, r_2, r_3, r, s, t, u, v, w$ that one equation is redundant and we obtain a unique solution.
