Minkowski's inequality for integrals, including $p=\mathrm{\infty }$

If $p=1$, then $\mathrm{\mid }f+g\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }+\mathrm{\mid }g\mathrm{\mid }$ pointwise, so [L4, L5, given] $\mathrm{\parallel }f+g{\mathrm{\parallel }}_1=\int \mathrm{\mid }f+g\mathrm{\mid }d\mu \le \int \mathrm{\mid }f\mathrm{\mid }d\mu +\int \mathrm{\mid }g\mathrm{\mid }d\mu =\mathrm{\parallel }f{\mathrm{\parallel }}_1+\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{.}$

Assume $1<p<\mathrm{\infty }$ and let $q:=p\mathrm{/}(p-1)$. Then [L1, L2, L4, L5, L6, given, algebra] $\int \mathrm{\mid }f+g{\mathrm{\mid }}^{p}d\mu \le 2^{p-1}(\int \mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu +\int \mathrm{\mid }g{\mathrm{\mid }}^{p}d\mu )<\mathrm{\infty }\mathrm{.}$ Indeed, [L6] applied pointwise to the two-term families $(\mathrm{\mid }f(x)\mathrm{\mid },\mathrm{\mid }g(x)\mathrm{\mid })$ and $(1,1)$ gives $\mathrm{\mid }f(x)\mathrm{\mid }+\mathrm{\mid }g(x)\mathrm{\mid }\le 2^{1\mathrm{/}q}(\mathrm{\mid }f(x){\mathrm{\mid }}^p+\mathrm{\mid }g(x){\mathrm{\mid }}^p{)}^{1\mathrm{/}p},$ so $\mathrm{\mid }f+g{\mathrm{\mid }}^p\le (\mathrm{\mid }f\mathrm{\mid }+\mathrm{\mid }g\mathrm{\mid }{)}^p\le 2^{p-1}(\mathrm{\mid }f{\mathrm{\mid }}^p+\mathrm{\mid }g{\mathrm{\mid }}^p)$ pointwise. Thus $f+g\in {\mathcal{L}}^p(\mu )$. Put $C:=\mathrm{\parallel }f+g{\mathrm{\parallel }}_p$. If $C=0$, the claim is immediate. Otherwise $\mathrm{\mid }f+g{\mathrm{\mid }}^p=\mathrm{\mid }f+g\mathrm{\mid }\mathrm{\mid }f+g{\mathrm{\mid }}^{p-1}\le \mathrm{\mid }f\mathrm{\mid }\mathrm{\mid }f+g{\mathrm{\mid }}^{p-1}+\mathrm{\mid }g\mathrm{\mid }\mathrm{\mid }f+g{\mathrm{\mid }}^{p-1}\mathrm{.}$ Because $(p-1)q=p$, the function $\mathrm{\mid }f+g{\mathrm{\mid }}^{p-1}$ lies in ${\mathcal{L}}^q(\mu )$ and has $q$-norm $C^{p-1}$. Integrating and applying [L1] with conjugate exponents $p$ and $q$ to each term yields $C^p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p{(\int \mathrm{\mid }f+g{\mathrm{\mid }}^{(p-1)q}d\mu )}^{1\mathrm{/}q}+\mathrm{\parallel }g{\mathrm{\parallel }}_p{(\int \mathrm{\mid }f+g{\mathrm{\mid }}^{(p-1)q}d\mu )}^{1\mathrm{/}q}\mathrm{.}$ Since $(p-1)q=p$, this becomes $C^p\le (\mathrm{\parallel }f{\mathrm{\parallel }}_p+\mathrm{\parallel }g{\mathrm{\parallel }}_p)C^{p-1}\mathrm{.}$ If $C>0$, divide by $C^{p-1}$ to obtain the claim.

For $p=\mathrm{\infty }$, let $M:=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $N:=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Then [L2, L3, given] $\mathrm{\mid }f+g\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }+\mathrm{\mid }g\mathrm{\mid }\le M+N\mathrm{.}$ Indeed, outside the union of the two null exceptional sets supplied by [L3], one has $\mathrm{\mid }f\mathrm{\mid }\le M$ and $\mathrm{\mid }g\mathrm{\mid }\le N$. Therefore $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{\mathrm{\infty }}\le M+N$.

Steps 1.1, 1.2, and 1.3 prove the $p=1$, $1<p<\mathrm{\infty }$, and $p=\mathrm{\infty }$ cases.