How to prove triangle inequality for $p$-norm?

Question

If $\mathcal{M}=\{M_i : i\in I_n\}$ is a collection of metric spaces, each with metric $d_i$, we can make $M=\prod_{i\in I_n}M_i$ a metric space using the $p$-norm, we simply set $d : M\times M\to \mathbb{R}$ as:

$$d((p_1,\dots,p_n),(q_1,\dots,q_n))=\left\|(d(p_1,q_1),\dots,d(p_n,q_n))\right\|_p$$

What I want to prove is that the $p$-norm

$$\left\|x\right\|_p=\left(\sum_{i=1}^{n}\left|x_i\right|^p\right)^{1/p}$$

is really a norm. Showing that $\left\|x\right\|_p \geq 0$ being zero if and only if $x = 0$ was easy. Showing that $\left\|kx\right\|_p = \left|k\right|\left\|x\right\|_p$ was also easy. The triangle inequality is the thing that is not being easy to show. Indeed, I want to show that: for every $x,y \in \mathbb{R}^n$ we have:

$$\left(\sum_{i=1}^{n}\left|x_i+y_i\right|^p\right)^{1/p}\leq \left(\sum_{i=1}^{n}\left|x_i\right|^p\right)^{1/p}+\left(\sum_{i=1}^{n}\left|y_i\right|^p\right)^{1/p}.$$

I thought that it might not be as difficult as it seems, but after trying a little without sucess I've searched on the internet and the I found that we need measure theory to prove that. Is there any more elementary proof of this inequality?

Answer 1

(I learned from Terry Tao the following proof, which exploits a symmetry to simplify the task of proving an estimate by normalising one or more inconvenient factors to equal $1$.)

I assume here that $1\leq p<\infty$. We want to show that $$ \|x+y\|_p\leq\|x\|_p+\|y\|_p\tag{*} $$ When the RHS is $0$, the proof is trivial. Suppose it is positive. By homogeneity $\|cx\|_p=|c|\|x\|_p$ we may reduce to the case $\|x\|_p=1-\lambda$ and $\|y\|_p=\lambda$ for some $0\leq\lambda\leq 1$. The cases $\lambda=0,1$ are trivial, so suppose $0<\lambda<1$. Writing $X:=x/(1-\lambda)$ and $Y:=y/\lambda$ we reduce to the convexity estimate: $$ \|(1-\lambda)X+\lambda Y\|_p\leq 1\quad\text{whenever } \|X\|_p=\|Y\|_p=1\ \text{and }0\leq\lambda\leq 1. $$ But since $z\mapsto|z|^p$ is convex for $p\geq 1$, we have the coordinate-wise convexity bound $$ |(1-\lambda)X_i+\lambda Y_i|^p\leq (1-\lambda)|X_i|^p+\lambda |Y_i|^p. $$ Summing $i$ from $1$ to $n$, we obtain $$ \|(1-\lambda)X+\lambda Y\|_p^p\leq 1 $$ and thus the claim follows.

Note that this proof works for the general abstract $L^p$ spaces as well.

Answer 2

If you mean $$ \left\|x\right\|_p=\left(\sum_{i=1}^n\left|x_i\right|^{\color{#C00000}{p}}\right)^{1/p}\tag{1} $$ then Minkowski's Inequality is the triangle inequality for the $p$-norm.

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Duality

Note that by Hölder's Inequality, if $\|y\|_q=1$, where $\frac1p+\frac1q=1$, we have $$ \left|\sum_{i=1}^nx_iy_i\right|\le\|x\|_p\tag{2} $$ Furthermore, if $y_i=\frac{\bar{x}_i|x_i|^{p/q-1}}{\|x\|_p^{p/q}}$, then $\|y\|_q=1$ and $$ \sum_{i=1}^nx_iy_i=\|x\|_p\tag{3} $$ $(2)$ and $(3)$ show that $$ \|x\|_p=\sup_{\|y\|_q=1}\left|\sum_{i=1}^nx_iy_i\right|\tag{4} $$ $(4)$ says that $\ell^p$ is the dual of $\ell^q$.

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Duality Proof of Minkowski's Inequality $$ \|x+y\|_p =\sup_{\substack{\|u\|_q&=1\\\|v\|_q&=1\\u&=v}}\sum_{i=1}^nx_iu_i+y_iv_i \le\sup_{\substack{\|u\|_q&=1\\\|v\|_q&=1}}\sum_{i=1}^nx_iu_i+y_iv_i =\|x\|_p+\|y\|_p\tag{5} $$ The inequality is because the $\sup$ on the left is being taken over a subset of the pairs $(u,v)$ over which the $\sup$ on the right is being taken.