## Hölder, Minkowski, Cauchy-Schwarz and triangle inequalities

Last update: 23 July 2014

## Hölder, Minkowski, Cauchy-Schwarz and triangle inequalities

Let $q\in {ℝ}_{\ge 0}\text{.}$ Let $p\in {ℝ}_{>1}\cup \left\{\infty \right\}$ be given by $1p+1q=1.$ The functions $h:{ℝ}_{\ge 0}\to {ℝ}_{\ge 0}$ given by $h(x)=xq 1 x 1 y y=x y={x}^{2} y={x}^{3} y={x}^{4}$ are increasing with $h\left(1\right)=1\text{.}$

The functions $h:{ℝ}_{\ge 0}\to {ℝ}_{\ge 0}$ given by $h(x)=x1q 1 x 1 y y=x y={x}^{\frac{1}{2}} y={x}^{\frac{1}{3}} y={x}^{\frac{1}{4}}$ are increasing with $h\left(1\right)=1\text{.}$

The functions $h:{ℝ}_{>0}\to {ℝ}_{\ge 0}$ given by $h(x)=x-1q 1 x 1 y y=\frac{1}{x} y={x}^{-\frac{1}{2}} y={x}^{-\frac{1}{3}}$ are decreasing with $h\left(1\right)=1\text{.}$

Thus the functions $g:{ℝ}_{\ge 0}\to ℝ$ given by $g\left(x\right)={x}^{-\frac{1}{q}}-1$ are decreasing with $g\left(1\right)=0\text{.}$ So $x-1q-1≤0 forx∈ℝ≥1.$ If $f:{ℝ}_{>0}\to ℝ$ is given by $f(x)=x1p -1px$ then $\frac{df}{dx}=\frac{1}{p}{x}^{\frac{1}{p}-1}-\frac{1}{p}=\frac{1}{p}\left({x}^{-\frac{1}{q}}-1\right)$ and so $f$ is decreasing for $x\in {ℝ}_{>1}$ and $f\left(1\right)=1-\frac{1}{p}=\frac{1}{q}\text{.}$ So $x1p-1px≤ 1qforx∈ ℝ≥1.$

Let $a,b\in {ℝ}_{>0}$ with $a\ge b$ and let $x=\frac{a}{b}\text{.}$ Then $1q≥ (ab)1p-1p (ab)=1b (a1pb-1p+1-1pa) =1b(a1pb1q-1pa).$ So $1pa+1qb≥ a1pb1q fora,b∈ℝ>0.$

Let $x=\left({x}_{1},\dots ,{x}_{n}\right)\in {ℝ}^{n}$ and $y=\left({y}_{1},\dots ,{y}_{n}\right)\in {ℝ}^{n}\text{.}$ Then $|xiyi| ‖x‖p‖y‖q ≤1p ( |xi| ‖x‖p ) p +1q ( |yi| ‖y‖q ) q .$ So $∑i=1n |xiyi| ‖x‖p‖y‖q ≤ ∑i=1n ( ≤1p ( |xi| ‖x‖p ) p +1q ( |yi| ‖y‖q ) q ) = 1p+1q=1.$ So $∑i=1n |xiyi| ≤ ‖x‖p‖y‖q.$ So $|∑i=1nxiyi|≤ ∑i=1n|xiyi|≤ ‖x‖p‖y‖q.$

Using $|xi+yi|≤ |xi|+|yi| andp-1=p(1-1p) =p1q=pq$ and $‖ |x1+y1|pq ,…, |xn+yn|pq ‖ q = (∑i=1n((xi+yi)pq)q)1q = (∑i=1n|xi+yi|p)1p·pq = (‖x+y‖p)pq,$ gives $|x+y|pp = ∑i=1n |xi+yi|p =∑i=1n |xi+yi| |xi+yi|p-1 ≤ ∑i=1n (|xi|+|yi|) |xi+yi|pq = ∑i=1n|xi| |xi+yi|pq+ ∑i=1n|yi| |xi+yi|pq ≤ ‖x‖p ‖ ( |x1+y1|pq,…, |xn+yn|pq ) ‖ q + ‖y‖p ‖ ( |x1+y1|pq,…, |xn+yn|pq ) ‖ q = ‖x‖p ‖x+y‖ppq+ ‖y‖p ‖x+y‖ppq = ( ‖x‖p+ ‖y‖p ) ‖x+y‖pp-1.$ Dividing both sides by ${‖x+y‖}_{p}^{p-1},$ then $‖x+y‖p≤ ‖x‖p+ ‖y‖p.$

Let $x=\left({x}_{1},\dots ,{x}_{n}\right)\in {ℝ}^{n}\text{.}$ For $p\in {ℝ}_{\ge 1}$ define $‖x‖p= ( |x1|p+⋯+ |xn|p ) 1p .$ For $x=\left({x}_{1},\dots ,{x}_{n}\right)$ in ${ℝ}^{n}$ define $|x|=|x|2 =(|x1|2+⋯+|xn|2)12 =x12+⋯+xn2$ and, for $x=\left({x}_{1},\dots ,{x}_{n}\right)$ and $y=\left({y}_{1},\dots ,{y}_{n}\right)$ in ${ℝ}^{n}$ define $⟨x,y⟩= x1y1+⋯+xnyn.$

Let $x,y\in {ℝ}^{n}$ with $x=\left({x}_{1},\dots ,{x}_{n}\right)$ and $y=\left({y}_{1},\dots ,{y}_{n}\right)\text{.}$ If $p\in {ℝ}_{>1}$ and $q$ is given by $\frac{1}{p}+\frac{1}{q}=1$ then $|∑i=1nxiyi|≤ ‖x‖p‖y‖q and ‖x+y‖p≤ ‖x‖p+‖y‖p.$

Let $,xy\in {ℝ}^{n}$ with $x=\left({x}_{1},\dots ,{x}_{n}\right)$ and $y=\left({y}_{1},\dots ,{y}_{n}\right)\text{.}$ Then $|⟨x,y⟩|= |∑i=1nxiyi| ≤|x||y| and|x+y|≤ |x|+|y|.$

Special case: Let $x,y\in ℝ\text{.}$ Then $|xy|=|x||y| and|x+y|≤ |x|+|y|.$

## Notes and References

These are a typed copy of handwritten notes from the pdf 140721Holderinequalitiesscanned140721.pdf.