# Missing a necessary power in this proof - please help.

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!!

• If you work it out, assuming that $\frac{1}{p}+\frac{1}{q} = 1$, we obtain that $\frac{p+pq-q}{pq} = \frac{2}{q}$. What I'm thinking is that since we need the kernel to be integrable, its absolute value needs to be <1. So, if we take the sup of something that is <1, the sup is going to be 1, so then whether you have it raised to $\frac{1}{q}$ or $\frac{2}{q}$ doesn't matter. But, this doesn't seem right... – ALannister May 4 '14 at 20:16

## 1 Answer

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

• $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$,
• $b\beta = p$
• $a\alpha = r$
• $(1-\alpha)c = r$
• $(1-\beta)c = p$
• $c = q$

and these are exactly correct. Your mistake is

Notice also that $\dfrac{q}{a}=1$,

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

$$X_r^{r/q} Y_r^{r/q} = X_r^{1/2}Y_r^{1/2} \leqslant A_r^{1/2}A_r^{1/2} = A_r.$$

• Doesn't the argument you used for the $1<p<q<\infty$ case apply to all the possibilities for r? Once you have the formula in terms of $r=\frac{pq}{pq+p-q}$, can't you then go back and plug in what $p$ and $q$ are in all the different cases to get the desired results in each case? I.e., isn't that case general enough to cover them all? – ALannister May 5 '14 at 16:34
• In the other cases, you have $\alpha \in \{0,1\}$ or $\beta \in \{0,1\}$, which requires a bit of care to handle. One can do that, but I found it simpler to avoid it first. – Daniel Fischer May 5 '14 at 20:51