This question is somewhat related to Gradient Estimate - Question about Inequality vs. Equality sign in one part.
That question was related to part (c) of a problem I am working on, and this question is related to part (b).
Specifically, I need to show the following:
Let $u$ and $v$ be two functions on $\Omega$ related as follows: $$ |u(x)| \leq \int_{\Omega}K(x,y)|v(y)|dy, \quad x \in \Omega, $$ for some kernel $K(x,y)$. Show that for any $1 \leq p \leq q \leq \infty$, we have $$ ||u||_{L^{q}(\Omega)}\leq A||v||_{L^{p}(\Omega)} $$ where $A = \max \left( \sup_{x}||K(x,\cdot)||_{L^{\frac{pq}{pq+p-q}}(\Omega)},\sup_{y}||K(\cdot,y)||_{L^{\frac{pq}{pq+p-q}}(\Omega)} \right)$.
So, starting with $u$, $v$ on $\Omega$ satisfying $$|u(x)| \leq \int_{\Omega}K(x,y)|v(y)|dy \quad (*)$$
I have that $$(*) \leq \int_{\Omega} |K(x,y) |v(y)|| dy \\ = \int_{\Omega}|K(x,y)v(y)| dy \\ = \int_{\Omega}|K(x,y)^{\alpha+1-\alpha}v(y)|dy\\ \leq \int_{\Omega}|K(x,y)|^{\alpha}|K(x,y)|^{1-\alpha}|v(y)|dy \\ = \int_{\Omega} |K(x,y)|^{\alpha} |K(x,y)|^{1-\alpha}|v(y)|^{1-\beta + \beta} dy\\ \leq \int_{\Omega}|K(x,y)|^{\alpha}(|K(x,y)|^{1-\alpha}|v(y)|^{1-\beta})|v(y)|dy \quad (1)$$
Aplying Holder's Inequality to $(1)$, we get that
$$(1) \displaystyle \leq \left[ \int_{\Omega} |K(x,y)|^{\alpha a} dy\right]^{1/a} \cdot \left[ \int_{\Omega} |K(x,y)|^{(1-\alpha)c} |v(y)|^{(1-\beta)c}dy \right]^{1/c} \cdot \left[\int_{\Omega}|v(y)|^{\beta b}dy \right]^{1/b}, $$
where $\displaystyle 1 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}. \quad (1^{\prime})$
Since we want the $L^{p}$ norm of $v$ to appear in the RHS, choose $\beta b = p$.
Thus, $\displaystyle b = \frac{p}{\beta}. \quad (2)$
Also, let $(1-\beta)c = p$, which implies that $\displaystyle \beta = 1 - \frac{p}{c}$.
Raising (1) to the power of $q$ and taking the integral with respect to x, we obtain that
$$ \int |u(x)|^{q} dx \leq \int \left( \left[ \int|K(x,y)|^{\alpha a}dy \right]^{q/a}\left[ \int|K(x,y)|^{(1-\alpha)c}|v(y)|^{(1-\beta)c}dy\right]^{q/c}||v||_{L^{p}}^{\beta q}\right) dx. $$
Now, in order for us to be able to switch the order of integration, we need $\displaystyle \frac{q}{c} = 1$, so $q = c. \quad (4)$
Therefore, $(3)$ becomes $\displaystyle \beta = 1 - \frac{p}{q} \quad (3^{\prime})$,
and $(2)$ becomes $\displaystyle b = \frac{p}{\displaystyle \left( 1 - \frac{p}{q}\right)} = \frac{pq}{q-p} \quad (2^{\prime})$.
Substituting $(4)$ and $(2^{\prime})$ into $(1^{\prime})$, we obtain an expression for $a$:
$\displaystyle a = \frac{p}{p-1} \quad (5)$
Bounding everything, we obtain
$$||u||_{L^{q}}^{q} \leq \sup_{x}\left( \int|K(x,y)|^{\alpha a}dy\right)^{q/a} \cdot \left[\sup_{y}\int |K(x,y)|^{(1-\alpha)c}dx \right] ||v||_{L^{p}}^{p+\beta q}. \quad (6)$$
Notice also that $\displaystyle \frac{q}{a} = 1$, so
$$ (6) = \left[ \sup_{x}\int|K(x,y)^{\alpha a}dy\right] \cdot \left[ \sup_{y} \int |K(x,y)|^{(1-\alpha)c}dx \right] \cdot ||v||_{L^{p}}^{p+\beta q}. \quad (6^{\prime})$$
Now, we need $\sup_{x}$, $\sup_{y}$, and each of these kernels to be finite, so choose $\alpha$ so that $\alpha a = (1-\alpha)c$, which implies then that $\displaystyle \alpha = \frac{c}{a+c}$.
Then, substituting our values for $a$ and $c$ [$(5)$ and $(4)$], respectively, we obtain that
$\alpha = \frac{q}{\frac{p}{p-1}+q} = \frac{q(p-1)}{p+pq-q} \quad (7)$
and that
$(1-\alpha) = \frac{p}{p+pq-q}. \quad (7^{\prime})$
So, $\alpha a = \frac{pq}{p+pq-q}$, and $(1-\alpha)c = \frac{pq}{p+pq-q}. \quad (7^{\prime\prime})$
Thus, $(6^{\prime})$ becomes
$$ = \left[ \sup_{x}\int |K(x,y)|^{\frac{pq}{p+pq-q}}dy\right] \cdot \left[ \sup_{y} \int |K(x,y)|^{\frac{pq}{p+pq-q}}dx\right]||v||_{L^{p}}^{p+(1-\frac{p}{q})\cdot q}. \quad (8)$$
But, since $\displaystyle p + \left( 1-\frac{p}{q}\right)q = q$, $(8)$ becomes
$$ = \left[\sup_{x}\int|K(x,y)|^{\frac{pq}{p+pq-q}}dy\right]\left[ \sup_{y} \int |K(x,y)|^{\frac{pq}{p+pq-q}} dx \right] ||v||_{L^{p}}^{q}$$.
Which brings me to my question.
Taking $q$th roots, you see that we have the following:
$$ ||u||_{L^{q}(\Omega)} \leq \left[\sup_{x}\int|K(x,y)|^{\frac{pq}{p+pq-q}}dy\right]^{1/q}\left[\sup_{y}\int |K(x,y)|^{\frac{pq}{p+pq-q}}dx\right]^{1/q}||v||_{L^{p}}$$,
but what I need is:
$$ ||u||_{L^{q}(\Omega)} \leq \left[\sup_{x}\int|K(x,y)|^{\frac{pq}{p+pq-q}}dy\right]^{\frac{p+pq-q}{pq}}\left[\sup_{y}\int |K(x,y)|^{\frac{pq}{p+pq-q}}dx\right]^{\frac{p+pq-q}{pq}}||v||_{L^{p}}$$,
so that I will have the $||K(x,y)||_{L^{\frac{pq}{pq+p-q}}}$ norms on the RHS that I need.
So, what I am hoping someone will tell me is, where did I go wrong in "misplacing" these powers? How do I fix this so that I will get the correct norm in my final answer?
Thank you in advance! I really would appreciate your help!!
Let's introduce a couple of notations for greater clarity. For any $r\in [1,\infty]$, we define
$$X_r := \sup_{x\in\Omega} \lVert K(x,\cdot)\rVert_r, \quad Y_r := \sup_{y\in\Omega} \lVert K(\cdot,y)\rVert_r, \quad A_r = \max \{ X_r,Y_r\},$$
and for brevity we write $\lVert\cdot\rVert_r$ instead of $\lVert\cdot\rVert_{L^r(\Omega)}$. The assertion is
$$\lVert u\rVert_q \leqslant A_r\lVert v\rVert_p\tag{$\ast$}$$
for all $1\leqslant p \leqslant q$, where
$$r = \begin{cases} \frac{pq}{pq+p-q} &, q < \infty\\ \quad 1 &, p=q=\infty\\ \;\;\frac{p}{p-1} &, 1 < p < q=\infty\\ \quad\infty &, 1 = p < q = \infty.\end{cases}$$
Let's first get the simple cases out of the way:
$p=q=\infty$: Then $r=1$ and we have, for every $x\in \Omega$, $$\lvert u(x)\rvert \leqslant \int \lvert K(x,y)\rvert\cdot\lvert v(y)\rvert\,dy \leqslant \int \lvert K(x,y)\rvert\,dy\cdot \lVert v\rVert_\infty \leqslant X_1\lVert v\rVert_\infty \leqslant A_1\lVert v\rVert_\infty,$$ and $(\ast)$ follows.
$p=1, q=\infty$: Then $r=\infty$ and, for every $x\in \Omega$, $$\lvert u(x)\rvert \leqslant \int \lvert K(x,y)\rvert\cdot\lvert v(y)\rvert\,dy \leqslant \lVert K(x,\cdot)\rVert_\infty \int \lvert v(y)\rvert\,dy \leqslant X_\infty \lVert v\rVert_1 \leqslant A_\infty\lVert v\rVert_1,$$ and again $(\ast)$ follows directly.
$1 < p < q = \infty$: Then $r = \frac{p}{p-1} = p'$ is the conjugate exponent of $p$, and $$\lvert u(x)\rvert \leqslant \int \lvert K(x,y)\rvert\cdot\lvert v(y)\rvert\,dy \leqslant \lVert K(x,\cdot)\rVert_r\cdot\lVert v\rVert_p \leqslant X_r\lVert v\rVert_p \leqslant A_r\lVert v\rVert_p.$$ Once more, $(\ast)$ follows directly.
$1 = p = q$: Then $r = 1$ too, and we can directly change the order of integration: $$\lVert u\rVert_1 \leqslant \int \left(\int \lvert K(x,y)\,dx\right)\lvert v(y)\rvert\,dy = \int \lVert K(\cdot,y)\rVert_1\lvert v(y)\rvert\,dy \leqslant Y_1\int\lvert v(y)\rvert\,dy \leqslant A_1\lVert v\rVert_1.$$
$1 = p < q < \infty$: Then $r = q$, and $$\lvert u(x)\rvert \leqslant \int\left(\lvert K(x,y)\rvert\cdot\lvert v(y)\rvert^{1/q}\right)\lvert v(y)\rvert^{1-1/q}\,dy \leqslant \left(\int \lvert K(x,y)\rvert^q\lvert v(y)\rvert\,dy\right)^{1/q}\lVert v\rVert_1^{1-1/q}.$$ Raising to the $q$-th power and integrating gives $$\lVert u\rVert_q^q \leqslant \int \underbrace{\lVert K(\cdot,y)\rVert_q^q}_{\leqslant Y_q^q}\lvert v(y)\rvert\,dy \leqslant A_q^q\lVert v\rVert_1^q,$$ and taking $q$-th roots yields $(\ast)$.
$1 < p = q < \infty$: Then $r = 1$ and $$\lvert u(x)\rvert \leqslant \int \lvert K(x,y)\rvert^{1-1/p}\left(\lvert K(x,y)\rvert^p\lvert v(y)\rvert\right)\,dy \leqslant \underbrace{\lVert K(x,\cdot)\rVert_1^{1-1/p}}_{\leqslant X_1^{1-1/p}}\left(\int \lvert K(x,y)\rvert\cdot\lvert v(y)\rvert^p\,dy\right)^{1/p}.$$ Raising to the $p$-th power and integrating, switching the order of integration, then gives $$\lVert u\rVert_p^p \leqslant X_1^{p-1}Y_1 \lVert v\rVert_p^p \leqslant A_1^p\lVert v\rVert_p^p,$$ and taking the $p$-th root gives $(\ast)$.
Now comes the harder case, $1 < p < q < \infty$. We start by splitting the integrand in three factors on the right hand side,
$$\begin{align} \lvert u(x)\rvert &\leqslant \int \lvert K(x,y)\rvert^\alpha \left(\lvert K(x,y)\rvert^{1-\alpha}\lvert v(y)\rvert^{1-\beta}\right)\lvert v(y)\rvert^\beta\,dy\\ &\leqslant \lVert K(x,\cdot)\rVert_r^{\alpha/r}\left(\int \lvert K(x,y)\rvert^r\lvert v(y)\rvert^p\,dy\right)^{(1-\alpha)/r}\lVert v\rVert_p^{\beta}\\ &\leqslant X_r^\alpha \lVert v\rVert_p^\beta \left(\int \lvert K(x,y)\rvert^r\lvert v(y)\rvert^p\,dy\right)^{(1-\alpha)/r} \tag{1} \end{align}$$
by Hölder's inequality for three factors, provided we have
$$\frac{\alpha}{r} + \frac{1-\alpha}{r} + \frac{\beta}{p} = 1 \quad\text{and}\quad \frac{1-\alpha}{r} = \frac{1-\beta}{p}.$$
The second of these conditions achieves the integrand $\lvert K(x,y)\rvert^r\cdot\lvert v(y)\rvert^p$ in the remaining integral. Determining $\alpha$ and $\beta$ from these conditions yields
$$\alpha = r\left(1-\frac{1}{p}\right),\qquad \beta = p\left(1-\frac{1}{r}\right),$$
and we verify
$$\frac{1-\alpha}{r} = \frac{1}{r} - \left(1-\frac{1}{p}\right) = \frac{1}{r}+\frac{1}{p}-1 = \frac{1}{p}-\left(1-\frac{1}{r}\right) = \frac{1-\beta}{p},$$
and
$$\frac{\alpha}{r} + \frac{1-\alpha}{r} + \frac{\beta}{p} = \frac{1}{r} + \frac{\beta}{p} = \frac{1}{r} + \left(1 - \frac{1}{r}\right) = 1.$$
Also
$$\frac{1-\alpha}{r} = \frac{1}{r} + \frac{1}{p} - 1 = \frac{pq+p-1}{pq} + \frac{q}{pq} - \frac{pq}{pq} = \frac{p}{pq} = \frac{1}{q},$$
so raising $(1)$ to the $q$-th power and integrating gives
$$\begin{align} \lVert u\rVert_q^q &\leqslant X_r^{q\alpha} \lVert v\rVert_p^{q\beta} \int \underbrace{\left(\int \lvert K(x,y)\rvert^r\,dx\right)}_{\leqslant Y_r^r} \lvert v(y)\rvert^p\,dy\\ &\leqslant X_r^{q\alpha} Y_r^r \lVert v\rVert_p^{q\beta+p}\\ &\leqslant A_r^{q\alpha+r}\lVert v\rVert_p^{q\beta+p}. \tag{2} \end{align}$$
Now
$$\begin{gather} q\alpha + r = qr\left(1-\frac{1}{p}\right)+r = r\left(q+1 - \frac{q}{p}\right) = \frac{pq}{pq+p-q}\frac{pq+p-q}{p} = q,\\ q\beta + p = qp\left(1-\frac{1}{r}\right) + p = pq + p - \frac{pq}{r} = pq+p - (pq+p-q) = q, \end{gather}$$
so taking $q$-th roots in $(2)$ yields $(\ast)$.
So, where was your mistake? The splitting of the integrand into three factors is the same, your conditions are
and these are exactly correct. Your mistake is
that is in general not the case. But when it is the case, your
$$||u||_{L^{q}(\Omega)} \leq \underbrace{\left[\sup_{x}\int|K(x,y)|^{\frac{pq}{p+pq-q}}dy\right]^{1/q}}_{X_r^{r/q}}\underbrace{\left[\sup_{y}\int |K(x,y)|^{\frac{pq}{p+pq-q}}dx\right]^{1/q}}_{Y_r^{r/q}}||v||_{L^{p}}$$
happens to give you what you need, since then $pq = p+q$ and hence
$$r = \frac{pq}{pq+p-q} = \frac{pq}{(p+q)+p-q} = \frac{pq}{2p} = \frac{q}{2},$$
so
$$X_r^{r/q} Y_r^{r/q} = X_r^{1/2}Y_r^{1/2} \leqslant A_r^{1/2}A_r^{1/2} = A_r.$$