# Version of Littlewood-Paley inequality

For any $p\in(1,\infty)$ there holds the Littlewood-Paley inequality $$\Vert f\Vert_{L^p}\sim_{n,p}\left\Vert \left(\sum_{k\in\mathbb{Z}}\left\vert \dot P_kf(x)\right\vert^2\right)^\frac{1}{2}\right\Vert_{L^p}$$ with the Littlewood-Paley projections $\dot P_kf$ defined by $\widehat{\dot P_kf}=\psi_k\widehat{f}$, where $\mathrm{Supp}(\psi_k)\subset\{2^{k-1}\leq\vert\xi\vert \leq2^{k+1}\}$ and $\sum_{k\in\mathbb{Z}}\psi_k(x)=1$ for $x\neq 0$. If I alternatively define the projections $P_kf$ by $P_kf=\dot P_kf$ for $k\geq1$ and $\widehat{P_0f}=\phi\widehat{f}$ with a bump function $\phi$ with support around the origin and $\phi+\sum_{k\in\mathbb{N}}\psi_k=1$. Does there also hold the similar inequality $$\Vert f\Vert_{L^p}\sim_{n,p}\left\Vert \left(\sum_{k\in\mathbb{N}_0}\left\vert P_kf(x)\right\vert^2\right)^\frac{1}{2}\right\Vert_{L^p}?$$

You seem to think that the implied constants will be completely independent of our choice of Littlewood-Paley partition of unity. If you look at the proof of the standard Littlewood-Paley inequality, you'll see that the estimates for the vector-valued kernel depend on the LP partition of unity. For the upper LP estimate, we can say the following: for all mean zero, $C^{1}$ functions $\phi$ such that $$|\psi(x)|\leq B(1+|x|)^{-n-1},\quad |\nabla\psi(x)|\leq B(1+|x|)^{-n-1}$$ we have that $$\left\|\left(\sum_{j\in\mathbb{Z}}|P_{j}f|^{2}\right)^{1/2}\right\|_{L^{p}}\leq C_{n,p}B\|f\|_{L^{p}},\quad 1<p<\infty$$ Conversely, if $\psi$ is a Schwartz function such that $\sum_{j}\widehat{\psi}(\xi)=1$ for $\xi\neq 0$, then there is a constant $C_{n,\psi}$ depending only on the dimension $n$ and the function $\psi$, such that $$\|f\|_{L^{p}}\lesssim_{p}C_{n,\psi}\left\|\left(\sum_{j}|P_{j}f|^{2}\right)^{1/2}\right\|_{L^{p}}$$
Having said all that, let's turn to what you want to prove. Let $\phi$ be a Schwartz function such that $\widehat{\phi}$ is supported in the annulus $\left\{1/2\leq|\xi|\leq 2\right\}$ and $\sum_{j}\widehat{\phi}(2^{-j}xi)=1$ for all $\xi\neq 0$. Define a Schwartz function $\psi$ by $$\widehat{\psi}(\xi)=\begin{cases}\sum_{j\leq 0}\widehat{\phi}(2^{-j}\xi) & {\xi\neq 0}\\ 1 & {\xi=0}\end{cases}$$ Technically, you need to check that $\psi$ is a well-defined Schwartz function, but this is easy. Let $P_{\leq 0}$ be the operator given by convolution with $\psi$. We then have the following result, $$\|f\|_{L^{p}}\sim_{n,p,\phi}\|P_{\leq 0}f\|_{L^{p}}+\left\|\left(\sum_{j=1}^{\infty}|P_{j}f|^{2}\right)^{2}\right\|_{L^{p}},\quad f\in L^{p}(\mathbb{R}^{n})$$ for all $1<p<\infty$.
Observe that the identity $P_{\leq 0}+\sum_{j\geq 1}P_{j}$ holds for all Schwartz functions $f\in\mathcal{S}(\mathbb{R}^{n})$. Indeed, by our hypothesis that $\sum_{j\in\mathbb{Z}}\widehat{\phi}(2^{-j}\xi)=1$, $\xi\neq 0$, we have that for $\xi\neq 0$, $$\widehat{f}(\xi)=\sum_{j\in\mathbb{Z}}\widehat{\phi}(2^{-j}\xi)\widehat{f}(\xi)=\sum_{j\leq 0}\widehat{\phi}(2^{-j}\xi)\widehat{f}(\xi)+\sum_{j\geq 1}\widehat{P_{j}f}(\xi)=\widehat{P_{\leq 0}f}(\xi)+\sum_{j\geq 1}\widehat{P_{j}f}(\xi)$$ By nonnegativity, $$\left\|\left(\sum_{j\geq 1}|P_{j}f|\right)^{1/2}\right\|_{L^{p}}\leq\left\|\left(\sum_{j\in\mathbb{Z}}|P_{j}f|^{2}\right)^{1/2}\right\|_{L^{p}}\lesssim_{n,p,\psi}\|f\|_{L^{p}}$$ By Young's inequality, $$\|P_{\leq 0}f\|_{L^{p}}\leq\|\psi\|_{L^{1}}\|f\|_{L^{p}}$$ These two estimates together with Minkowski's inequality give us one side of the inequality.
We now show the other side of the inequality. We have the trivial estimate $$\left(\sum_{j\geq 1}|P_{j}f|^{2}\right)^{1/2}\geq|P_{1}f|,$$ which implies by Minkowski that $$\left\|\left(\sum_{j\geq 1}|P_{j}f|^{2}\right)^{1/2}\right\|_{L^{p}}+\left\|P_{\leq 0}f\right\|_{L^{p}}\geq\left\|P_{\leq 1}f\right\|_{L^{p}}$$ Now by the support properties of $\widehat{\psi}$, we have that $P_{k}P_{\leq 1}=P_{k}$ if $k\leq 0$. Whence, $$\left(\sum_{k\leq0}|P_{k}f|^{2}\right)^{1/2}\leq\left(\sum_{k\in\mathbb{Z}}|P_{k}P_{\leq 1}f|^{2}\right)^{1/2}$$ Taking $L^{p}$ norms and using the standard two-sided LP inequality, we obtain $$\left\|\left(\sum_{k\leq 0}|P_{k}f|^{2}\right)^{1/2}\right\|_{L^{p}}\lesssim_{n,p,\phi}\left\|P_{\leq 1}f\right\|_{L^{p}}$$ Putting these results together and using Minkowski's inequality and two-sided LP inequality again, we conclude that \begin{align*} \|f\|_{L^{p}}&\lesssim_{n,p,\phi}\left\|\left(\sum_{j\in\mathbb{Z}}|P_{j}f|^{2}\right)^{1/2}\right\|_{L^{p}}\\ &\leq\left\|\left(\sum_{j\leq 0}|P_{j}f|^{2}\right)^{1/2}\right\|_{L^{p}}+\left\|\left(\sum_{j\geq 1}|P_{j}f|^{2}\right)^{1/2}\right\|_{L^{p}}\\ &\lesssim_{n,p,\phi}\|P_{\leq 0}f\|_{L^{p}}+2\left\|\left(\sum_{j\geq 1}|P_{j}f|^{2}\right)^{1/2}\right\|_{L^{p}} \end{align*}
• Thank you for you answer! I know that the implied constants do change if your Littlewood-Paley decomposition is different but I forgot to denote it. I've now proved this version of the LP inequality by applying Calderon Zygmund theorem and showing that the Kernel $K(x,y)=(a__k(x-y))_{k\in\mathbb{N}_0}$ with $a_0(x)=\mathcal{F}^{-1}\phi(x)$ and $a_k(x)=2^{nk}\mathcal{F}^{-1}\psi(2^kx)$ for $k\geq1$ is singular an then going along with the standard proof for the LP-inequlity. You solution is also very neat! thank you. Nov 23, 2015 at 21:23