= Minkowski inequality for finite vectors
{c}
{title2=$\lVert u+v\rVert_p\leq\lVert u\rVert_p+\lVert v\rVert_p$}
For <vectors> $u,v\in\mathbb R^n$ or $\mathbb C^n$, <counting measure> turns the <Minkowski inequality> into
$$
\left(\sum_{i=1}^n|u_i+v_i|^p\right)^{1/p}
\leq\left(\sum_{i=1}^n|u_i|^p\right)^{1/p}
+\left(\sum_{i=1}^n|v_i|^p\right)^{1/p},\qquad 1\leq p<\infty.
$$
For $p=\infty$, replace each displayed <norm> by the maximum of its coordinate <absolute values>. Take $u=(1,0)$ and $v=(0,1)$. Both <vectors> have <Lp norm> one for every $p\geq1$, but their sum has <L1 norm> $2$, <Euclidean norm> $\sqrt2$, and <L-infinity norm> $1$. Thus the same <vector addition> gives equality $2=1+1$ at $p=1$, strict inequality $\sqrt2<1+1$ at $p=2$, and strict inequality $1<1+1$ at $p=\infty$.
By contrast, $u=(1,1)$ and $v=2u=(2,2)$ give equality for every $p\geq1$: <absolute homogeneity of a norm> yields $\lVert u+v\rVert_p=3\lVert u\rVert_p=\lVert u\rVert_p+\lVert v\rVert_p$. At $p=2$, the three sizes are $\sqrt2$, $2\sqrt2$, and $3\sqrt2$. Opposite <vectors> behave very differently: $u+(-u)=0$, so the inequality is strict whenever $u$ is nonzero. Cancellation is allowed; the bound describes the largest possible total size.
\Image[/minkowski-vector-equality-and-quasinorm-failure.png]
{title=Strict <Minkowski inequality>, equality for positive multiples, and <quasi-norm> failure}
{description=Arrows show $u$, the translated step $v$, and their sum. Left: perpendicular unit <vectors> give $\sqrt2<2$ for the <Euclidean norm>. Middle: $u=(1,1)$ and $v=(2,2)$ give $3\sqrt2=\sqrt2+2\sqrt2$. Right: $u=(1,0)$ and $v=(0,1)$ give $4>2$ for the <Lq quasi-norm> at $q=1/2$. Pale curves show the corresponding unit level sets.}
Back to article page
