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I'm reading Fourier Analysis and Nonlinear Partial Differential Equations by Rapha\"el Danchin et al. There are lines on page 9 reads:

Using Young's inequality for $\mathbb{Z}$ equipped with the counting measure, we may now deduce that $$ I(f,g,h)\leqslant C\sum_{j,k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |jq+k|-\varepsilon |jq+\ell|-\varepsilon |k-\ell|} $$ $$\leqslant \frac{C}{\varepsilon}\sum_{k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |k-\ell|} $$ $$ \leqslant\frac{C}{\varepsilon^2}\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}} $$ where: $$ \frac{1}{p}+\frac{1}{q}=1+\frac{1}{r},\,\,\varepsilon\triangleq\frac{1}{4}\left(\frac{1}{p}-\frac{1}{r}\right),\,\,(p,q,r)\in (1,\infty)^3, c_k,d_{\ell}>0. $$

I really don't know how to obtain the last two "$\leqslant$". For the last "$\leqslant$" I can only get $\frac{C}{\varepsilon}$ but $\frac{C}{\varepsilon^2}$ because $2^{-\varepsilon |k-\ell|}$ are always no bigger than 1.

Anyone could help me? Any advice will be appreciated.


Actually, this is to prove the Young's inequality for weak $L^q$ space, i.e., $$ \lVert f*g\rVert_{L^r(G, \mu)}\leqslant C\lVert f\rVert_{L^p(G, \mu)}\lVert g\rVert_{L_w^q(G, \mu)} $$ by homogeneity, he assume $\lVert f\rVert_{L^p(G,\mu)}=\lVert g\rVert_{L_w^q(G,\mu)}=\lVert h\rVert_{L^{r'}(G,\mu)}=1$, and then prove $$ I(f,g,h)=\int_{G^2}f(y)g(y^{-1}\cdot x)h(x)\,d\mu(x)\,d\mu(y)\leqslant C $$ instead.

So I think it's enough once we get (Juli\'{a}n's answer) $$ I(f,g,h)\leqslant C\sum_{j,k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |jq+k|-\varepsilon |jq+\ell|-\varepsilon |k-\ell|} $$ $$ \leqslant 2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}\sum_{k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |k-\ell|} $$ $$ \leqslant 2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}} $$ $$ \leqslant 2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{r'}} $$ By the definition of $\varepsilon$ we can denotes $2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}$ by $C(q,r)$. And by the assumption of norm 1, the coefficients of atomic decompositions of $f$ and $h$ we know $\lVert(c_k)\rVert_{\ell^{p}}\leqslant 2, \lVert(d_{\ell})\rVert_{\ell^{r'}}\leqslant 2$. So we get $I\leqslant C(q,r)$. I don't know why the $\frac{1}{\varepsilon}$ and $\frac{1}{\varepsilon^2}$ appear. Or am I miss understand something?

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What are $c_k,d_l$ and $I$? – Ilya Feb 28 '12 at 12:21
@Ilya: $I=\int_{G^2}f(y)g(y^{-1}\cdot x)h(x)\,d\mu(x)\,d\mu(y)$, $G$ is a locally compact topological group and $\mu$ a left-invariant Haar measure. $f\in L^p$ and $g$ in weak $L^q$, $h\in L^{r'}$. $f=\sum_k c_kf_k,h=\sum_{\ell}d_{\ell}h_{\ell}$ are atomic decompositions. But I don't think these matters. – Y.Z Feb 28 '12 at 12:44
There is no $j$ index in the last summation, right? – Ilya Feb 28 '12 at 12:58
@Ilya: I'm so sorry! No $j$ index in the last summation! Corrected. – Y.Z Feb 28 '12 at 13:09
up vote 3 down vote accepted

For the first inequality, since $|2\,q\,j+k+\ell|\le|j\,q+k|+|j\,q+\ell|$, we have $$ \sum_{j\in\mathbb{Z}}2^{-\varepsilon |jq+k|-\varepsilon |jq+\ell|}\le\sum_{j\in\mathbb{Z}}2^{-\varepsilon|2qj+k+\ell|}\quad\forall k,\ell\in\mathbb{Z}. $$ For fixed $k$ and $\ell$ choose $j_0\in\mathbb{Z}$ such that $\Bigl|\dfrac{k+\ell}{2\,q}-j_0\Bigr|<1$. Then $$ |2\,q\,j+k+\ell|=|2\,q(j+j_0)+k+\ell-2\,q\,j_0|\ge|2\,q(j+j_0)|-2\,q. $$ Then $$\begin{align*} \sum_{j\in\mathbb{Z}}2^{-\varepsilon|2qj+k+\ell|}&\le2^{\varepsilon2q}\sum_{j\in\mathbb{Z}}2^{-\varepsilon|2q(j+j_0)|}\\ &=2^{\varepsilon2q}\sum_{j\in\mathbb{Z}}2^{-\varepsilon|2qj|}\\ &\le2^{\varepsilon2q+1}\sum_{j=0}^\infty2^{-\varepsilon2qj}\\ &=2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}\\ &\le\frac{C_q}{\varepsilon}, \end{align*}$$ where $C_q$ is independent of $k$ and $\ell$.

The second inequality holds because $\varepsilon\le1$.

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Thanks for your help! I have a question about the $\frac{1}{\varepsilon}$ and I edit the post for that. Another question is where did we use the Young's inequality? I can just see I use H\"older's inequality after the last summation. Thanks! – Y.Z Feb 28 '12 at 16:27
Observe that $\bigl(1-2^{\varepsilon2q}\bigr)^{-1}\sim C/\epsilon$ as $\epsilon\to0$. As for the use of Young's inequality, I assumed it was used to bound $I$ im terms of the sum. – Julián Aguirre Feb 28 '12 at 18:36
Here may be a Young's inequality: $$\sum_{k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |k-\ell|}\leqslant \lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}}\lVert(2^{-\vareps‌​ilon |\cdot|})\rVert_{\ell^{\infty}} =\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}} .$$ But still I don't know why $\frac{1}{\varepsilon}$ becomes $\frac{1}{\varepsilon^2}$. – Y.Z Feb 29 '12 at 2:33

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