Derive Hausdorff-Young inequality from Paley's inequality Given a sequence $(c_j)_{j\in\mathbb{Z}}$ of complex numbers with $\lim_{|j|\to\infty}c_j=0$. Define the rearrangement $c_j^*$ as follows: for $j\geq 0$, $c_j^*$ is the $j+1-$th largest element of the set $\{|c_0|,|c_1|,\dots\}$; for $j<0$, $c_j^*$ is the $|j|-$th largest element of $\{|c_{-1}|, |c_{-2}|,\dots\}$.
Let $f\in L^2(\mathbb{T})$ and $1<p\leq 2$. Set $c_j=\hat{f}(j)$ for all $j\in\mathbb{Z}$.
(a) Prove Paley's inequality:
$$\left(\sum_{j\in\mathbb{Z}}(1+|j|)^{p-2}|c_j^*|^p\right)^{\frac{1}{p}}\leq C_p\Vert f\Vert_p$$
where $C_p>0$ only depends on $P$.
(b) By splitting the sum in (a) dyadically, prove the Hausdorff-Young inequality for $1<p\leq 2$:
$$\left(\sum_{j\in\mathbb{Z}}|\hat{f}(j)|^q\right)^{\frac{1}{q}}\leq C'_p\Vert f\Vert_p$$
where $q>0$ and $\frac{1}{p}+\frac{1}{q}=1$, $C'_p>0$ only depends on $p$.
Part (a) is straight forward which can be proved by Marcinkiewicz interpolation theorem. I am stuck on part (b), I didn't see how it helps by splitting the sum in (a), and I have no idea how to proceed. Any hints are appreciated.
 A: Following the hint, we write
\begin{align*}
\sum_{j\in\mathbb{Z}}(1+\left|j\right|)^{p-2}\left|c_{j}^{*}\right|^{p}&=\left|c_{0}^{*}\right|^{p}+\sum_{k=0}^{\infty}\sum_{2^{k}\leq\left|j\right|<2^{k+1}}(1+\left|j\right|)^{p-2}\left|c_{j}^{*}\right|^{p}\\
&\gtrsim\left|c_{0}^{*}\right|^{p}+\sum_{k=0}^{\infty}2^{k}(1+2^{k+1})^{p-2}\left(\left|c_{2^{k+1}}^{*}\right|^{p}+\left|c_{-2^{k+1}}^{*}\right|^{p}\right), \tag{1}
\end{align*}
where we use that $\left\{\left|c_{j}\right|^{*}\right\}$ are a decreasing rearrangement to obtain the last equality. Since
$$2^{k}(1+2^{k+1})^{p-2}\sim_{p}2^{k(p-1)},$$
we see that (1) is
$$\sim_{p}\left|c_{0}^{*}\right|^{p}+\sum_{k=0}^{\infty}(2^{k\frac{p-1}{p}}\left|c_{2^{k+1}}^{*}\right|)^{p}+\sum_{k=0}^{\infty}(2^{k\frac{p-1}{p}}\left|c_{-2^{k+1}}^{*}\right|)^{p} \tag{2}$$
Since $q\geq p$, we have by the nesting property of $\ell^{r}$ spaces that (2) is
\begin{align*}
&\geq\left(\left|c_{0}^{*}\right|^{q}+\sum_{k=0}^{\infty}(2^{k\frac{p-1}{p}}\left|c_{2^{k+1}}^{*}\right|)^{q}+\sum_{k=0}^{\infty}(2^{k\frac{p-1}{p}}\left|c_{-2^{k+1}}^{*}\right|)^{q}\right)^{p/q}\\
&=\left(\left|c_{0}^{*}\right|^{q}+\sum_{k=0}^{\infty}2^{k}\left(\left|c_{2^{k+1}}^{*}\right|^{q}+\left|c_{-2^{k+1}}\right|^{q}\right)\right)^{p/q}, \tag{3}
\end{align*}
where we use that $q=p/p-1$. Since all the terms in the series are nonnegative, the order of summation is irrelevant, so we obtain that
\begin{align*}
\sum_{j\in\mathbb{Z}}\left|c_{j}\right|^{q}=\sum_{j\in\mathbb{Z}}\left|c_{j}^{*}\right|^{q}&=\left|c_{0}^{*}\right|^{q}+\sum_{k=0}^{\infty}\sum_{2^{k}\leq\left|j\right|<2^{k+1}}\left|c_{j}^{*}\right|^{q}\\
&\leq 3\left|c_{0}^{*}\right|^{q}+\sum_{k=1}^{\infty}\sum_{2^{k}\leq\left|j\right|<2^{k+1}}\left|c_{j}^{*}\right|^{q}\\
&\lesssim 3\left|c_{0}^{*}\right|^{q}+\sum_{k=0}^{\infty}2^{k}(\left|c_{2^{k+1}}^{*}\right|^{q}+\left|c_{-2^{k+1}}^{*}\right|^{q}), \tag{4}\\
\end{align*}
from which together with Paley's inequality, we conclude that
$$\left(\sum_{j\in\mathbb{Z}}\left|\widehat{f}(j)\right|^{q}\right)^{1/q}\lesssim_{p}\left\|f\right\|_{p} \tag{5}$$
