It's been a while but here we go:
for orange to be a metric 4 conditions must be met:
- ๐(๐,๐) = 0
proof
since ๐(x) - ๐(x) will always be 0 for any ๐ and any x in domain
- ๐(๐,๐) > 0 if ๐ != ๐.
proof
|๐(x) - ๐(x)| >= 0 by definition, so ๐(๐,๐) must be >= 0. we only have to prove that:
๐(๐,๐) = 0 -> ๐=๐
Consider the contrapositive: ๐!=๐ -> ๐(๐,๐) != 0
since ๐!=๐ โx s.t ๐(x) != ๐(x)
but then |๐(x) - ๐(x)| > 0
thus ๐(๐,๐) > 0
thus ๐(๐,๐) = 0 -> ๐=๐
- ๐(๐,๐) = ๐(๐,๐)
proof
|๐(x) - ๐(x)| = |-1(-๐(x) + ๐(x))|
|-1(-๐(x) + ๐(x))| = |-1(๐(x) - ๐(x))|
|-1(๐(x) - ๐(x))| = |๐(x) - ๐(x)| since |-q| =|q|
so for any x |๐(x) - ๐(x)| = |๐(x) - ๐(x)|
which means ๐(๐,๐) = ๐(๐,๐)
- The Triangle Inequality:๐(๐,๐) <= ๐(๐,๐) + ๐(๐, ๐)
proof
let x be the element in [a,b] s.t |๐(x) - ๐(x)| is maximized
let y be the element in [a,b] s.t |๐(y) - ๐(y)| is maximized
let z be the element in [a,b] s.t |๐(z) - ๐(z)| is maximized
๐(๐,๐) <=๐(๐,๐) + ๐(๐, ๐) is equivalent to
|๐(y) -๐(y)| +|๐(z) - ๐(z)| >= |๐(x) - ๐(x)|
Let's start with the following (obvious) inequality:
|๐(y) -๐(y)| +|๐(z) - ๐(z)| >= |๐(y) -๐(y)| +|๐(z) - ๐(z)|
|๐(y) -๐(y)| +|๐(z) - ๐(z)| >= |๐(x) -๐(x)| +|๐(z) - ๐(z)| since |๐(y) - ๐(y)| is maximized
|๐(x) -๐(x)| +|๐(z) - ๐(z)| >= |๐(x) -๐(x)| +|๐(x) - ๐(x)| since |๐(z) - ๐(z)| is maximized
|๐(x) -๐(x)| +|๐(z) - ๐(z)| >= ||๐(x) -๐(x)| +|๐(x) - ๐(x)|| since |q| + |p| >= 0 so |q| + |p| = ||q| +|p||
||๐(x) -๐(x)| +|๐(x) - ๐(x)|| >=|๐(x) -๐(x) +๐(x) - ๐(x)| = |๐(x) - ๐(x)| since |q| >= q forall q
therefore |๐(y) -๐(y)| +|๐(z) - ๐(z)| >= |๐(x) - ๐(x)|
since all 4 conditions are satisfied the ๐ is a metric!
It should be ||๐(x) -๐(x)| +|๐(x) - ๐(x)|| >=|๐(x) -๐(x) +๐(x) - ๐(x)| = |๐(x) - ๐(x)| I missed the abs that I added in the previous step.
let me make the variables less annoying:
||x-y|+|y-z|| >= |x-y+y-z| = |x-z| we are getting rid of the abs around |x-y| and |y-z| so the 2 y's can cancel out. We can do this because |x-y| >= x-y because |q| >= q