"Pythagorean theorem" for projection onto convex set I'm going through the book on online convex op

sweetymoeyz 2022-07-07 Answered
"Pythagorean theorem" for projection onto convex set
I'm going through the book on online convex optimization by Hazan, and in the first chapter I saw this assertion (which Hazan calls the "pythagorean theorem"):
Let K R d be a convex set, y R d , and x = Π K ( y ). Then for any z K we have:
y z x z .
It is presented without proof - what is a proof for this? Also, how does it relate to the pythagorean theorem?
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Answers (1)

potamanixv
Answered 2022-07-08 Author has 15 answers
Suppose x is the closest point to y in the closed convex set K.
If x=y there is nothing to prove, so we can suppose x y.
The we have that x y , z x 0 for all z K (this is essentially the dual problem).
If z K, we can write z y = t ( x y ) + d, where d ( x y ). Then the above gives x y , t ( x y ) + d + y x = ( t 1 ) x y 2 0 from which we get t 1.
Then z y 2 = d 2 + t 2 x y 2 x y 2 which is the desired result.
Addendum: To see the first condition, suppose z y y x for all z K.
We have z y 2 = z x + x y 2 = z x 2 + x y 2 + 2 z x , x y (this is where Pythagoras appears) which gives z x 2 + 2 z x , x y 0 for all z K. Since w ( t ) = x + t ( z x ) K for all t [ 0 , 1 ], we have t 2 z x 2 + 2 t z x , x y 0, dividing across by t and letting t 0 yields the desired result.

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