Let $1<p<2$ and $\frac1p+\frac{1}{q}=1.$ Show that for any $r\in(p,q)$ and for any $u\in L^p(\mathbb{R})$, we have $$\int_\mathbb{R}\left||y|^\frac{1}{q}\widehat{u}(y)\right|^r\frac{dy}{|y|}\leq c(r)\lVert u\rVert^r_{L^p(\mathbb{R})}.$$
My idea: To prove such kind of inequality, it's natural to think of interpolation of operators. Denote $dx,dy$ the Lebesgue measure on $\mathbb{R}$, and let $d_\mu=\frac{dx}{|x|}$. Define an operator $T:u(y)\mapsto|y|^{\frac1q}\widehat{u}(y).$
Then the problem is reduced to prove that $T$ is $(p,r)$ type.
Fisrt, $T$ is obviously $(p,q)$ type. This is because by Hausdorff-Young inequality, we have:$$\int_\mathbb{R}\left[\left||y|^{\frac1q}\widehat{u}(y)\right|\frac{dy}{|y|}\right]=\vert|\widehat{u}\vert|_{L^q(\mathbb{R})}\leq\vert|u\vert|_{L^p(\mathbb{R})}$$
Now we have to prove that $T$ is also $(p,p)$ type, which is equivalent to: $$\int_\mathbb{R}||y|\widehat{u}(y)|^p\frac{dy}{|y|^2}\leq C\vert|u\vert|_{L^p(\mathbb{R})}.$$
Then I got lost. A natural idea is to use interpolation again. Define an operator $S:u\mapsto y\widehat{u}(y)$. Then we just need to prove that $T$ is $(p,p)$ type.
By Plancharel theorem, $S$ is automatically $(2,2)$ type. But $S$ is obvious not $(1,1)$ type as $u(x)=e^{-x^2}$ is a counter example.
So in my opinion, in order to proceed, we either have to show $S$ is weak $(1,1)$ type, or directly show $$\int_\mathbb{R}||y|\widehat{u}(y)|^p\frac{dy}{|y|^2}\leq C\vert|u\vert|_{L^p(\mathbb{R})}.$$
Any hint or solution is highly appreciated!
To show $S$ is of weak (1,1) type, for any $a>0$, $$ \int_{\{|y\hat u(y)|>a\}}|y|^{-2}\mathrm{d}y\le\int_{\{|y|>a/\|\hat u\|_\infty\}}|y|^{-2}\mathrm{d}y\le\frac{2\|u\|_1}{a}. $$