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 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 (as a function from real or complex numbers to nonnegative real numbers).
First, some generalities. Let be a (real or complex) vector space equipped with a function that satisfies the scaling axiom: for all scalars , and the separation axiom: implies . As usual, we define the unit ball in to be
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 and , we have
Now we prove that 3. implies 1. Suppose . Let and be the associated unit vectors. Then
where . If condition 3. holds, then
but by the scaling axiom, this is the same as saying
which is the triangle inequality.
Consider now with its -norm . By Lemma , the Minkowski inequality is equivalent to
This allows us to remove the cumbersome exponent in the definition of the -norm.
Define by . Then is convex, i.e., for all ,
for all .
The function is convex, and for the function for is monotone increasing and convex, by the first and second derivative tests. Thus and then
so is convex.
Let and be unit vectors in . By condition 4, it suffices to show that for all . But
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.