SUMMARY Flashcards
YOUNGS INEQUALITY
let 1/p + 1/q = 1 with 1<p< ∞.
if a,b>=0,
then ab<= (a^p)/p + (b^q)/q
HOLDERS INEQUALITY
let x,y∈K^n and 1/p + 1/q = 1 with 1<p< ∞
Σ|xiyi|<= (Σ|xi|^p)^1/p (Σ|yi|^q)^1/q
*sums from i=1 to n
MINKOWSKIS INEQUALITY
let x,y∈K^n and 1/p + 1/q = 1 with 1<=p< ∞
(Σ|xi+yi|^p)^1/p <= (Σ|xi|^p)^1/p + (Σ|yi|^p)^1/p
*sums from i=1 to n
DISTANCE BETWEEN X∈X AND V
(V subset of NLS X)
d(x,V) := inf{||x-v||: v ∈ V}
CLOSURE of V
(V subset of NLS X)
V(-) := {x∈X: d(x,V)=0}
V IS OPEN
(V subset of NLS X)
∀x∈V, ∃ε>0: B(x;ε) := {y∈X :||x-y||< ε} ⊆V
DENSE
the subset E ⊆X when E(-)=X
where X is a metric space
SEPARABLE
metric space X if it contains a countable dense subset
CAUCHY-SCHWARZ INEQUALITY
if X is an IPS then:
1. |<x,y>|^2 <= <x,x><y,y>
2. ||x|| = √<x,x> is a norm
3. xn -> x, yn -> y ⟹ <xn,yn> -> <x,y>
PARALLELOGRAM LAW
||x+y||^2 + ||x-y||^2 = 2(||x||^2 + ||y||^2)
POLARISATION IDENTITY (K=R)
4<x,y> = ||x+y||^2 - ||x-y||^2
POLARISATION IDENTITY (K=C)
4<x,y> = ||x+y||^2 - ||x-y||^2 + i||x+iy||^2 - i||x-iy||^2
x⊥y
if <x,y>=0
PYTHAGOREAM THEOREM
x⊥y ⟹ ||x+y||^2 = ||x||^2 + ||y||^2
ORTHOGONAL COMPLEMENT OF V
V^⊥ = {x∈X: <x,v>=0 for all v∈V} is a closed linear subspace
where V subset of IPS X