2
$\begingroup$

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.

Edit:

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?

  • 0
    @Ilya: I'm so sorry! No $j$ index in the last summation! Corrected.2012-02-28

1 Answers 1

3

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 $\varepsilon\le1$.

  • 0
    Here may be a Young's inequality: $\sum_{k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |k-\ell|}\leqslant \lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}}\lVert(2^{-\varepsilon |\cdot|})\rVert_{\ell^{\infty}} =\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}} .$ But still$I$don't know why $\frac{1}{\varepsilon}$ becomes $\frac{1}{\varepsilon^2}$.2012-02-29