Nicholas Lawrance | ICRA 20111Nicholas Lawrance | Thesis Defence1 1 Functional Analysis I Presented...
-
date post
20-Dec-2015 -
Category
Documents
-
view
218 -
download
0
Transcript of Nicholas Lawrance | ICRA 20111Nicholas Lawrance | Thesis Defence1 1 Functional Analysis I Presented...
Nicholas Lawrance | ICRA 2011 1Nicholas Lawrance | Thesis Defence 11
Functional Analysis I
Presented by Nick Lawrance
Nicholas Lawrance | ICRA 2011 2
What we want to take from this...
• My hope is that a proper understanding of the fundamentals will provide a good basis for future work
• Clearly, not all of the maths will be directly useful. We should try to focus on areas that seem like they might provide utility
• The topic areas are not fixed yet
Nicholas Lawrance | ICRA 2011 8
Sequences
• N = {1, 2, 3, ...} is countably infinite• The rational numbers Q are countable, the real
numbers R are not
• Examples
• Can also have a finite index set, and a subset of the index results in a family of elements
1 : 1, 2,3,...X n n
{ 1,0,1}
I
X I
x
R; ; ;
Nicholas Lawrance | ICRA 2011 9
Supremum and Infimum
• Easy to think of as maximum and minimum, but not strictly correct. They are the bounds but do not have to exist in the set
A = {-1, 0, 1} sup(A) = 1 inf(A) = -1
B = {n-1: n = [1, 2, 3, ...]} sup(B) = 1 inf(B) = 0
0 0.2 0.4 0.6 0.8 1-1
-0.5
0
0.5
1
Nicholas Lawrance | ICRA 2011 10
lp - norms
• For an n-dimensional space
• 2-dimensional Euclidean space unit spheres for a range of p values
1 2
1 2
1
1
, , ,
, , ,
( , )
n
n
n pp
j jj
x
y
d x y
1
1 1 2 2( , ) 1p p pd x y
Nicholas Lawrance | ICRA 2011 11
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
dimension 1
dim
ensi
on 2
lp norm based unit circles for 2-dimensional space
p = 0p = 0.5
p = 1
p = 2
p = 5
p = 20p = 1000000
Nicholas Lawrance | ICRA 2011 13
Examples
• Euclidean R, R2, R3, Rn.• Complex plane C• Sequence space l∞
– Remember a sequence is an ordered list of elements where each element can be associated with the natural numbers N
• Discrete metric space
such that
Nicholas Lawrance | ICRA 2011 14
• Function space C[a,b]
• X is the set of continuous functions of independent variable t є J, J = [a,b]
tx
y
d(x, y)
ba
Nicholas Lawrance | ICRA 2011 15
lp-space
• Note that this basically implies that each point is a finite distance from the ‘origin’
• Sequence can be finite or not
Nicholas Lawrance | ICRA 2011 17
• Balls cannot be empty (they must contain the centre which is a member of X)
• In a discrete metric space, sphere of radius 1 contains all members except x0, S(x0, 1) = X- x0
Nicholas Lawrance | ICRA 2011 18
Open and closed sets
ε > 0
ε > 0
x0
x
xx0
B(x0, ε)
B(x0, ε)
Neighbourhood
Nicholas Lawrance | ICRA 2011 21
0 1 2 3 4 5 6-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
t
x(t)
y(t)
Nicholas Lawrance | ICRA 2011 22
• We need
• Let f(t) = |x(t) – y(t)|• Find the stationary points
[0,2 ]
max ,t
d x t y t
sin cos
cos sin 0
cos sin [0,2 ]
3 7,
4 4
2
f t t t
dft t
dtt t t
t
f t
Nicholas Lawrance | ICRA 2011 26
Accumulation points and closure
B(x0, ε)
M X
x0
• Accumulation point if every neighbourhood of x0 contains a y є M distinct from x0
Nicholas Lawrance | ICRA 2011 27
a) Closure of the integers is the integers
b) Closure of Q is R
c) Closure of rational C is C
d) Closure of both disks is {z | |z| ≤ 1}