Consider an integral operator $Tf(x)=\int_{\mathbf{R}^n}K(x,y)f(y)dy.$ And $s,r \in(0,\infty), s \geq r$ are two indices. I would like to prove \begin{equation} \| Tf\|_{r} \leq (\int_{\mathbf{R}^n} \int_{\mathbf{R}^n} |K(x,y)|^rdx)^{s/r}dy) ^{1/s} \| f \|_{s'} \end{equation} (Assume all the integral are not infinity, and $\| f \|_r$ stands for the standard $L^r$ norm, that is $(\int_{\mathbf{R}^n} |f(x)|^r dx)^{1/r}.$)
Since on the right hand side, the term of $K$ first integrate to x then to y, my attempt is to use Fubini theorem and Hölder inequality, \begin{equation} \begin{aligned} | Tf(x) | &\leq \int_{\mathbf{R}^n} | K(x,y)| |f(y)| dy \\ &=\int_{\mathbf{R}^n} | K(x,y)| |f(y)|^{1/r} |f(y)|^{1/r'} dy \\ & \leq (\int_{\mathbf{R}^n} | K(x,y)|^r|f(y)| dy)^{1/r} (\int_{\mathbf{R}^n} |f(y)| dy)^{1/r'} \\ Then \quad \int_{\mathbf{R}^n} | Tf(x) |^r dx &\leq \int_{\mathbf{R}^n}(\int_{\mathbf{R}^n} | K(x,y)|^r|f(y)| dy) (\int_{\mathbf{R}^n} |f(y)| dy)^{r/r'} dx \end{aligned} \end{equation} Another attempt is to use Hölder plus Minkowski's inequality \begin{equation} \begin{aligned} \| Tf \|_r&=(\int_{\mathbf{R}^n}|\int_{\mathbf{R}^n} K(x,y)f(y)dy|^r dx)^{1/r} \\ &\leq (\int_{\mathbf{R}^n}|\int_{\mathbf{R}^n} |K(x,y)|^sdy|^{r/s} dx)^{1/r} (\int_{\mathbf{R}^n} |f(y)|^{s'} dy)^{1/s'} \\ But \ s \geq q, \ Minkowski \ cannot \ get \ the \ desired \ outcome. \end{aligned} \end{equation} I am stuck at it.
Can anybody help me?
Right, you do want to apply Fubini, then Hölder, but after that, to bring the power r inside the x-integral, you’ll need to apply Minkowski’s integral inequality as well.