# How to use Hölder inequality to prove this integral inequality?

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 $$$$\| Tf\|_{r} \leq (\int_{\mathbf{R}^n} \int_{\mathbf{R}^n} |K(x,y)|^rdx)^{s/r}dy) ^{1/s} \| f \|_{s'}$$$$ (Assume all the integral are not infinity, and $$\| 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{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} Another attempt is to use Hölder plus Minkowski's inequality \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} I am stuck at it.

Can anybody help me?

## 1 Answer

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.

• Thank you for your reply. Yes, we indeed have $s \geq r$(I forgot to type it). I will try your advice. Commented Mar 3, 2023 at 3:16
• I am still stuck, could you explain this to me more explicitly? (I have renewed some details in the question) Commented Mar 3, 2023 at 3:41
• Sorry, forget about Fubini. You should have proven Minkowski Integral Inequality, see en.m.wikipedia.org/wiki/Minkowski_inequality. Write out \|Tf\|_r, apply Minkowski, then the result follows immediately from Hölder with exponents s,s’.
– Nick
Commented Mar 3, 2023 at 4:27