I'm working on a thesis concerning the Littlewood-Paley decomposition of distributions and the use of paraproducts. I'm referring to the book of Bahouri, Chemin, Danchin and during a proof (page 88 for who is interested) they have to make an estimate of $L^p$ norm of the $j'$-th block of a smooth function, and they write the following passage (I change a bit the notation to make it more understandable): $$ \sum_{j>j'} \|\Delta_{j'}f_j\|_{L^p}\leq C \sum_{j>j'}\|f_j\|_{L^p}$$ where $f_j$ are smooth functions defined over $\mathbb{R}^d$ and $\Delta_{j'}f_j$ is the $j'$-th Littlewood-Paley block of $f_j.$ That C seems to be independent of $j$ and $j'$, and looks like something which comes from a Holder inequality.
What I tried to do was to write the block as a convolution of $f_j$ with some $\eta_{j'}$, and Holder inequality leads to: $$ \|\Delta_{j'}f_j\|_{L^p} =(\int |f_j\star \eta_j(x)|^pdx)^{(1/p)}\leq\\ \leq (\int(\int|f_j(x)\eta_j(y-x)|dy)^pdx)^{1/p}\leq \\ \leq\|f_j\|_{L^p}(\int(\int |\eta_j(y-x)|^{p'}dy)^{p/p'}dx)^{1/p},$$ where the $\eta_j(\cdot)$ is the Fourier antitrasform of a function on $\mathbb{R}^d$ with support on an annulus of radii of order $2^j$ and values on $[0,1]$, and $p'$ is the conjugate exponent of $p,$ and the last passage follows from Holder Inequality.
My doubt is: is the last factor independent of $j$? It seems not to be, as the area of the annulus increases with $j$. What do you suggest me? Thanks a lot.
Why don't you just use Young's convolution inequality since $\eta_{j'}$ is scaled like an approximate identity so that $\|\eta_{j'}\|{L^1}= = \|\eta\|{L^1}?