Download Union Intersection Cartesian Product Pair (nonempty set, metric

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M. Sami Fadali
EBME Dept., UNR
Metric space.
Linear vector space.
Linear independence & bases.
Normed spaces.
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Union
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Intersection
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Cartesian Product
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Pair (nonempty set, metric) =
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Metric (distance measure) axioms
(i)
(ii)
(iii)
Show that :
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Vector space over
with two functions
is a nonempty set
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Multiplicative identity
Distributive multiplication
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Associative multiplication
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Commutative addition
Associative addition
Additive zero
Additive inverse
(real or complex vector space)
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Vector space is closed under addition
Vector space is closed under scalar
multiplication
Can define an inner product for
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=Class of continuous function over the
closed interval
exists
Vector space is closed under addition
linearly independent set
linearly dependent set
Vector space is closed under scalar
multiplication
not all zero.
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Set of linearly independent vectors
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Any vector
, can be written as the
linear combination
Any nonempty subset of closed under
addition and scalar multiplication.
Example: The set of vectors
Dimension of vector space: number of basis
vectors.
= subspace of
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(i)
iff
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(ii)
(iii)
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+
Show norm is nonnegative
A vector space
with a norm
Can be regarded as metric spaces with
the metric
Satisfies zero iff
, symmetry, and
the triangle inequality for metrics
+
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Make
norm
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a normed vector space with 
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Holder’s Inequality
Let
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One-norm
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Two-norm (Euclidean)
Infinity-norm
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Equivalent
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, then
Minkowski Inequality
, then
Let
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Matrix Identities
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Inverse
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Transpose
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Inverse Transpose
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Algebraic multiplicity
= number of
eigenvalue
repetitions of the
Geometric multiplicity
= number of
linearly independent eigenvectors for
the
eigenvalue.
Diagonalizable matrix:
, always
true for distinct eigenvalues.
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Positive Definite
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Positive Semidefinite
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Negative Definite
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Negative Semidefinite
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Assume symmetric matrix w.l.o.g.
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Indefinite Matrix
for nonzero .
has no definite sign
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For a symmetric matrix
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Proof
=
The results follow by considering the
different cases for the signs of the
eigenvalues.
e.g. positive definite for
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Neighborhood of a point
s.t.
: Set
.
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Limit Point of
: every neighvorhood
of contains a point
.Note
need not be in
Interior point
neighborhood
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: There exists a
.
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Open Set: Every point in the set is an
interior point.
Closed Set: Contains all its limit points.
Complement of set
If is closed then
versa.
Bounded Set :
is open and vice
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Neighborhood: open ball in
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Closed Set
Bounded Set :
Compact Set: closed and bounded.
Convex Set :
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Convergent Sequence of vectors
to a limit :
→
Limit of
→
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Cauchy Sequence
in a metric
space
: for every real number
there is an integer s.t.
Not Convex
Convex
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Complete Metric Space
: every
Cauchy sequence converges (limit is in
the space), i.e. for every Cauchy
s.t.
as
sequence
.
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Banach Space: complete normed space.
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Hilbert Space: complete inner product
space.
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Let and be abstract sets.
Function
is a set of ordered pairs
in the Cartesian product
and if
,
then
Only one element in (range) assigned to
(domain)
Injective (one-one)
Surjective (onto)
s.t.
Bijective: injective and surjective.
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and
Function
metric spaces
s.t.
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is continuous on :
points in
For normed spaces:
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Uniform Continuity:
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Linear Operator (transformation)
s.t.
is continuous at all
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Bounded Operator
and
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Operator Norm
independent of
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All linear operators
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Induced Matrix Norms
are of the form
real matrix
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Function
interval
defined in an open
and the derivative
_|Å
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exists.
Function
set
,
→
defined in an open
and the derivative
_|Å
exists.
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→
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A function is continuously differentiable at a
point if at
 It is continuous.
 Its partial derivatives are continuous.
 A function
is said to be of class
if the
, exist and are
derivatives
continuous.
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= continuously differentiable.
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Lipschitz Condition
Lipschitz on an open set
Globally Lipschitz if
Contraction Mapping
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Sufficient Condition for Lipschitz :
is
If a function
continuously differentiable on an
open set
, then it is locally
Lipschitz on .
 Necessary Condition for Lipschitz
:
uniformly continuous function
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Solution has a known form for linear
case.
 No general form for the solution in the
nonlinear case.
 Questions:
◦ Does a solutions exist?
◦ Is the solution unique?
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Nonlinear differential eqn.
Nonlinear differential eqn.
(i) piecewise-continuous in
(ii) Satisfies the Lipschitz Condition
(i) piecewise-continuous in
(ii) Satisfies the Lipschitz Condition
The eqn. has a unique solution over
s.t. the eqn. has a unique
solution in the interval
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