I was reading Fourier analysis by J. Duoandikoetxea, and checking out the proof of the $(L^p,L^p)$ inequality $$ \left| \left| \left( \sum_j|T_jf_j|^2 \right)^{\frac{1}{2}} \right| \right|_p\leq C\left| \left| \left( \sum_j|f_j|^2 \right)^{\frac{1}{2}} \right| \right|_p,~1<p<\infty $$ which is an $L^p$ extension of the sequence $\{T_j\}$ of linear operators bounded on $L^2(\omega)$ ($\omega$ is an $A_2$ weight).
The position of the inequality towards the end is $$ \left| \left| \left( \sum_j|T_jf_j|^2 \right)^{\frac{1}{2}} \right| \right|^{2}_{p}\leq C_{\delta}\int_{\mathbb{R}^n} \left( \sum_j|f_j|^2 \right)M(u^{1/\delta})^{\delta},~0<\delta<1 $$ where $M(u^{1/\delta})^{\delta}$ is an $A_1-$weight and $C_{\delta}$ is a constant that depends on $\delta$.
Here is my calculation based on the suggestion of using Hölder's inequality to finish the proof. Is my calculation correct?:
The integral on the right side can be rewritten as: $$ C_{\delta}\left| \left| \left( \sum_j|f_j|^2 \right)M(u^{1/\delta})^{\delta} \right| \right|^2_1\leq C_{\delta}\left( \int_{\mathbb{R}^n}\left( \sum_j|f_j|^2 \right)^{\frac{p}{2}} \right)^{\frac{2}{p}} \left(\left( M(u^{1/\delta})^{\delta} \right)^{\frac{p'}{2}}\right)^{\frac{2}{p'}}. $$ via Hölders's inequality where $p$ and $p'$ are conjugate pairs. Take the square roots. Is my calculation correct? If not, what have I done wrong?
This result is for $p>2$. For $p<2$, it states
$$ \text{"Finally, if $p<2$ then, since the adjoint operators $T^{\ast}_{j}$ are also bounded on $L^2(\omega),~\omega\in A_2,$ we get the $(L^p,L^p)$ inequality by duality"} $$ I am not sure how to write this. Can someone help me out?
Thanks in advance for your help and comments.
I took a look at the text, and if I'm reading the statement of the theorem correctly, we want to show that the linear map $T$ defined on $L^{p}(\mathbb{R}^{n},\ell^{2})$ (or a dense subspace) by $$T\left(\left\{f_{j}\right\}_{j}\right):=\left\{T_{j}f_{j}\right\}_{j}$$ And we are supposing that the operators $T_{j}$ are bounded on $L^{p}(\mathbb{R}^{n},w)$, for any $A_{2}$ weight $w$, uniformly in $j$ and with uniform constant depending only on the weight $w$.
In the case $p<2$, unless I'm missing something, you've stopped short of showing the desired inequality. You need to use $L^{p}$ boundedness of the Hardy-Littlewood maximal operator $M$ for $1<p<\infty$ and you need to make a choice of $0<\delta<1$ at some point. Fix $\delta$ so that $\delta(p/2)'=p_{0}>1$. By Holder's inequality,
\begin{align*} C_{\delta}\left\|\left(\sum_{j}\left|f_{j}\right|^{2}\right)M(u^{1/\delta})^{\delta}\right\|_{L^{1}}^{2}&\leq C_{\delta}\left\|\sum_{j}\left|f_{j}\right|^{2}\right\|_{L^{p/2}}\left\|M(u^{1/\delta})^{\delta}\right\|_{L^{(p/2)'}}\\ &\leq C_{\delta}\left\|\sum_{j}\left|f_{j}\right|^{2}\right\|_{L^{p/2}}\left(\left\|M\right\|_{L^{p_{0}}\rightarrow L^{p_{0}}}\left\|u^{1/\delta}\right\|_{L^{p_{0}}}\right)^{\frac{p_{0}}{(p/2)'}}\\ &=C_{\delta}\left\|\sum_{j}\left|f_{j}\right|^{2}\right\|_{L^{p/2}}\left\|M\right\|_{L^{p_{0}}\rightarrow L^{p_{0}}}^{\delta}\underbrace{\left\|u\right\|_{L^{(p/2)'}}}_{=1}\\ &\leq C_{\delta}\left\|M\right\|_{L^{p_{0}}\rightarrow L^{p_{0}}}\left\|\sum_{j}\left|f_{j}\right|^{2}\right\|_{L^{p/2}} \end{align*} I'm a little confused about the parameter dependence of the constant $C$; my choice of $\delta\in(0,1)$ is controlled by $1<p<\infty$. A priori, I don't see why $C$ should be independent of $p$.
We've shown that $T$ is bounded for $p\geq 2$. Now consider the adjoint operator $T^{*}$ which is given on $L^{p}(\mathbb{R}^{n},\ell^{2})\cap L^{2}(\mathbb{R}^{n},\ell^{2})$, for $p\geq 2$, by $$T^{*}\left(\left\{f_{j}\right\}_{j}\right)=\left\{T_{j}^{*}f_{j}\right\}_{j}$$ Since $L^{2}(\mathbb{R}^{n},w)$ is a Hilbert space and therefore the $T_{j}^{*}:L^{2}(\mathbb{R}^{n},w)\rightarrow L^{2}(\mathbb{R}^{n},w)$ are bounded uniformly in $j$ with $\left\|T_{j}^{*}\right\|=\left\|T_{j}\right\|$, we see that the adjoint operators satisfy the hypotheses of the theorem we're trying to prove. So we can we use our first result to obtain that $T^{*}$ is bounded on $L^{p}(\mathbb{R}^{n},\ell^{2})$ for $p\geq 2$. But $(T^{*})^{*}=T$ by reflexivity and by a standard result of functional analysis $\left\|(T^{*})^{*}\right\|=\left\|T\right\|$. Whence $T$ is bounded on $L^{p}$ for $p<2$.