Minkowsky's inequality

Proposition (Minkowsky inequality) Let $u,v:E\to \mathbb{C}$ two measurable functions, and $p\in [1,+\infty ]$, then

$$\Vert u+v{\Vert }_p\le \Vert u{\Vert }_p+\Vert v{\Vert }_p$$

Proof Assume $1\le p<\infty$, since the case $p=\infty$ is trivial. Suppose $h\in L^{p^{\prime }}$ and $\Vert h{\Vert }_{p^{\prime }}\le 1$. Then

$$\left|\int (u+v)hd\mu \right|\le \int |uh|d\mu +\int |vh|d\mu$$ $$\Vert uh{\Vert }_1+\Vert vh{\Vert }_1\le \Vert u{\Vert }_p+\Vert v{\Vert }_p$$

where the second inequality is true by Holder inequality. Taking the supremum and using the lemma below we get the result. $\square$

Lemma (Formula for $\Vert u{\Vert }_p$) Suppose $(X,\mathcal{A},\mu )$ is a measure space, $1\le p<\infty$ and $u\in L^p(\mu )$. Then

$$\Vert u{\Vert }_p=\sup \left\{\left|\int uhd\mu \right|:h\in L^{p^{\prime }}\text{ and }\Vert h{\Vert }_{p^{\prime }}\le 1\right\}$$

Proof using Holder inequality

$$\left|\int uhd\mu \right|\le \Vert uh{\Vert }_1\le \Vert u{\Vert }_p$$

To prove the other way, consider the function

$$h(x)=\frac{\overline{u}(x)|u(x){|}^{p-2}}{\Vert u{\Vert }_p^{p/p^{\prime }}}$$

(set $h(x)=0$ if $f(x)=0$). then

$$\left|\int uhd\mu \right|=\frac{1}{\Vert u{\Vert }_p^{p/p^{\prime }}}\left|\int |u{|}^{p}d\mu \right|=\Vert u{\Vert }_p$$

and $\Vert h{\Vert }_{p^{\prime }}=1$. $\square$

Remark The case $p=\infty$ fails, this is a counter example. Let $X=\{b\}$ a set with one element, and $\mu (\{b\})=\infty$. Then the functions in $L^1(\mu )$ are the functions that are identically zero, $h(b)=0$ (with the usual convention $0\cdot \infty =0$) so that $\Vert h{\Vert }_1=0$. But the functions in $L^{\infty }(\mu )$ are the ones with bounded values on $b$, just take $f(b)=1$, so that $\Vert b{\Vert }_{\infty }=1$. Clearly the above indentity cannot hold.

If $X$ is $\sigma$-finite, then the case $p=\infty$ also holds.