Minkowski's Inequality

Solution 1:

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, $$ \begin{align} \left\|\int f(\cdot,y)\;\mathrm{d}y\right\|_p&=\sup_{\|h\|_{L^q}=1}\int\int h(x)f(x,y)\;\mathrm{d}y\;\mathrm{d}x\\ &\le\int\sup_{\|h\|_{L^q}=1}\int h(x)f(x,y)\;\mathrm{d}x\;\mathrm{d}y\\ &=\int\|f(.,y)\|_p\;\mathrm{d}y \end{align} $$