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I am wondering how to prove inequality

$\left\| \int f(.,y)dy \right\|_{p}\leq \int \left\| f(.,y) \right\|_{p}dy~~~~?$

Here, $f$ is an integrable function on $\mathbb{R}^n$ and $\displaystyle \left\|f\right\|_{p}=\left( \int|f|^pdx \right)^{1/p}$ for $p\geq 1.$

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    @ Willie Wong thanks for lecture notes.2011-08-18

1 Answers 1

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Let's start with this version of Minkowski's inequality: $ \|f+g\|_p\le\|f\|_p+\|g\|_p $ We can repeatedly apply this inequality to get $ \left\|\sum_k\;f_k\right\|_p\le\sum_k\|f_k\|_p $ Then using the scaling property of $\|\cdot\|_p$, we can write $ \left\|\sum_k\;f_k\Delta_k\right\|_p\le\sum_k\|f_k\|_p\Delta_k $ So we can write a Riemann sum for the integral inequality you cite and pass to the limit to get the integral inequality you cite.

Another way to show this inequality is to compute $\|\cdot\|_p$ using duality. That is, $ \begin{align} \left\|\int f(\cdot,y)\;\mathrm{d}y\right\|_p&=\sup_{\|h\|_{L^q}=1}\int\int h(x)f(x,y)\;\mathrm{d}y\;\mathrm{d}x\tag{1}\\ \int\|f(.,y)\|_p\;\mathrm{d}y&=\int\sup_{\|h\|_{L^q}=1}\int h(x)f(x,y)\;\mathrm{d}x\;\mathrm{d}y\tag{2} \end{align} $ where $\frac{1}{p}+\frac{1}{q}=1$. Since any $h$ used in $(1)$ can be used for all $y$ in $(2)$, it is clear that $(2)$ is at least as big as $(1)$. However, we might be able to do better in $(2)$ by choosing a different $h$ for each $y$. Therefore, $ \begin{align} \left\|\int f(\cdot,y)\;\mathrm{d}y\right\|_p&=\sup_{\|h\|_{L^q}=1}\int\int h(x)f(x,y)\;\mathrm{d}y\;\mathrm{d}x\\ &\le\int\sup_{\|h\|_{L^q}=1}\int h(x)f(x,y)\;\mathrm{d}x\;\mathrm{d}y\\ &=\int\|f(.,y)\|_p\;\mathrm{d}y \end{align} $