analysis (differential/integral calculus, functional analysis, topology)
metric space, normed vector space
open ball, open subset, neighbourhood
convergence, limit of a sequence
compactness, sequential compactness
continuous metric space valued function on compact metric space is uniformly continuous
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What is commonly known as Minkowski’s inequality is the statement that the p-norm ${\Vert f\Vert_p} \coloneqq \root{p}{\int_X {\vert f\vert^p} d\mu}$ on Lebesgue spaces indeed satisfies the triangle inequality.
Our proof of Minkowski’s inequality is broken down into a few simple lemmas. The plan is to boil it down to two things: the scaling axiom, and convexity of the function $x \mapsto {|x|^p}$ (as a function from real or complex numbers to nonnegative real numbers).
First, some generalities. Let $V$ be a (real or complex) vector space equipped with a function ${\|(-)\|}\colon V \to [0, \infty]$ that satisfies the scaling axiom: ${\|t v\|} = {|t|} \, {\|v\|}$ for all scalars $t$, and the separation axiom: ${\|v\|} = 0$ implies $v = 0$. As usual, we define the unit ball in $V$ to be $\{v \in V \;|\; {\|v\|} \leq 1\}.$
Given that the scaling and separation axioms hold, the following conditions are equivalent:
Since conditions 2. and 3. are pretty obviously equivalent, we just prove 1. and 3. are equivalent. Condition 1. implies condition 3. easily: if $\|u\| = 1 = \|v\|$ and $0 \leq t \leq 1$, we have
Now we prove that 3. implies 1. Suppose ${\|v\|}, {\|v'\|} \in (0, \infty)$. Let $u = \frac{v}{{\|v\|}}$ and $u' = \frac{v'}{{\|v'\|}}$ be the associated unit vectors. Then
where $t = \frac{{\|v\|}}{{\|v\|} + {\|v'\|}}$. If condition 3. holds, then
but by the scaling axiom, this is the same as saying
which is the triangle inequality.
Consider now $L^p$ with its $p$-norm ${\|f\|} = {|f|_p}$. By Lemma , the Minkowski inequality is equivalent to
This allows us to remove the cumbersome exponent $1/p$ in the definition of the $p$-norm.
Define $\phi\colon \mathbb{C} \to \mathbb{R}$ by $\phi(x) = |x|^p$. Then $\phi$ is convex, i.e., for all $x, y$,
for all $t \in [0, 1]$.
The function $g: x \mapsto {|x|}$ is convex, and for $1 \lt p$ the function $f: t \mapsto t^p$ for $t \geq 0$ is monotone increasing and convex, by the first and second derivative tests. Thus $g(t x + (1-t)y) \leq t g(x) + (1-t)g(y)$ and then
so $f \circ g: x \mapsto {|x|^p}$ is convex.
Let $u$ and $v$ be unit vectors in $L^p$. By condition 4, it suffices to show that ${|t u + (1-t)v|_p^p} \leq 1$ for all $t \in [0, 1]$. But
by Lemma . Using $\int {|u|^p} = 1 = \int {|v|^p}$, we are done.
Another commonly seen proof of Minkowski’s inequality derives it with the help of Hölder's inequality; see there for some commentary on this. But this is probably not the first thing one would think of unless one knows the trick, whereas the alternative proof given above seems geometrically motivated and fairly simple.
Wikipedia, Minowski’s inequality
Last revised on December 23, 2022 at 10:52:43. See the history of this page for a list of all contributions to it.