A “really” intro level text on functional analysis, at the master/advanced undergrad level.
It is very well written – clear & smooth, and a good portion of the exercises are elementary so that one can reinforce her understanding of definitions and theorems without getting stuck too often.
Highly recommended for anyone trying to learn functional analysis.
CH1 Preliminaries
A bunch of inequalities
Let be conjugate exponents, i.e., . Here are complex numbers.
Young’s inequality: .
Holder’s inequality:
(can let
if the resulting series are convergent; also, this follows
from Young)
Holder’s inequality, integral version:
( are Lebesgue measurable functions defined on
and the integrals are all finite by assumption)
Cauchy Schwarz inequality: setting in Holder’s inequality.
Minkowski’s inequality: if , then
(can let if the resulting series are convergent; proof mostly relies on triangle inequality. Note that if , sign of Minkowski reversed)
Minkowski also has integral version:
if , then we have .
Zorn’s lemma
poset: , is a partial order (consistent ranking, but may be incomplete)
chain: , is a linear order (complete: any two elements can be ranked)
least element of : an element s.t. (so must be able to be compared with all elements of )
lower bound of : an element s.t. ( must be able to be compared with all elements of )
minimal element of : an element s.t. ( doesn’t have to be able to be compared with every element of , just make sure that if it can for some )
Zorn’s lemma: If is a poset s.t. every chain in has an upper bound, then contains a maximal element.
As an application, can show every abstract vector space has a Hamel basis.
for a vector space over a field , A subset of vectors in is a Hamel basis for if is linearly independent and spans
proof is actually short. But seems Schauder basis introduced later is the focus.
CH2 Metric Spaces
separable metric space
having a countable dense subset.
some standard topology concepts in baby Rudin Ch2. Then some general topology definitions and results. I remind myself those that I’m starting to forget.
- neighborhood: given metric space , a neighborhood of is a subset of s.t. it contains an open ball , for some .
Cantor intersection property: if is complete, then for a sequence of contracting nonempty closed subsets of , there exists s.t.
First, perhaps the book should mention that the sequence of sets has the property that their diameters tend to 0:
Then, according to Wiki, this is actually a variant of “Cantor’s intersection theorem”, which is stated in general topology (and another version which is stated for subsets of .
This can be used to show that is uncountable. If it is, we write it as a sequence . Then we find a interval with length , such that it doesn’t contain . Then, divide this interval into 3 closed subintervals of equal length. Then at least one of these subintervals doesn’t contain . Keep doing this, we construct a sequence of closed intervals, contracting in diameter, but it contains an element in the limit which is not a term of the sequence .
Continuous mapping between metric spaces
1. For a mapping
between metric spaces, continuity at a point defined in
the
way is equivalent to continuity defined via sequences.
- Continuity of the mapping on the whole domain is equivalent to the notion that inverse image of any open set is open.
isometry (mapping between metric spaces):
a mapping between metric spaces that preserves distance:
is isometry if
. (by def, this is a continuous mapping, and injective)
isometric metric spaces
is isometric to
if there exists a bijective isometry from
onto
. (note bijectivity implies existence of inverse isometric
map, so if can just say
are isometric to each other)
extension of isometry to the whole space
For two complete metric spaces, if they each have a dense subset, and
there is an isometry (mapping) between them, then there exists an
isometry between the two metric spaces themselves, and the isometric
mappings are consistent with the one associated with dense subsets.
thus, perhaps we can say that if two metric spaces are isometric, then their Banach completions are isometric.
completing metric spaces first, completion of
means
such that:
1.
is complete 2. $(X,d) is isometric to a dense subset of
for example, is a complete of .
Can show that any metric space has a completion (long proof), and it is somewhat unique in the sense that if we have two completions of the same metric space, then the two completions are isometric to each other.
Baire Category Theorem, v1
Let
be a sequence of dense open sets in a
complete metric space
. Then,
.
Baire Category Theorem, v2
Let
be a complete metric space. If
, where
are closed sets, then at least one of these sets contain
an open ball.
The reason this is called “category thm” is that we call a subset of a topological space “nowhere dense” if its closure contains no interior points. (the interior of of a topological space is defined as the union of all subsets of that are open in )
A set is 1st category if it is a countable union of nowhere dense sets. If not, the set is called 2nd category. Then, Baire v2 says that every complete metric space is of 2nd category.
This can actually be used to “prove” is uncountable by writing , and using the fact that is a closed set.
Similarly, this can be used to show that is not a complete metric space.
Compactness
define a metric space to be “compact” if every sequence in it has a convergent subsequence.
(later, a topological space is defined “compact” if every open cover has finite subcover.)
A compact set in a metric space is closed and bounded, but the converse is only true for . For general metric spaces, The Borel-Lebesgue theorem says that it is compact iff every open cover of contains a finite subcover.
continuous mapping & compact domain (metric
space results)
1. If a mapping is continuous, then take any compact set, its image
under this mapping is compact.
- If a mapping is continuous, and is compact, then is bounded, and attains its maximum value.
later, analogous results for topological spaces.
Topological Spaces
a more general structure than that of a metric space.
This book uses the notation . Given this topology, let . Then is a topology on , and is called the *subspace topology**. Always remember that a topology is just a collection of sets we call “open”.
Hausdorff topological space: if every two distinct points in the space have disjoint neighborhoods.
base/basis of topology a family of open subsets of s.t. every open subset of is the union of sets belonging to this family. Say this family generates the topology. Many topologies are more easily defined in terms of a generating set.
characterizing topological base Given a topology
, can show that
is its base iff:
1.
2.
,
, there exists a set
s.t.
Intuitively, first make sure the base can cover all of ; then, make sure it doesn’t generate slices (intersections) too thin.
As an example, in any metric space, the collection of open balls is a base of the metric topology on .
Continuous mapping between topological spaces
If are topological spaces, a mapping is continuous at if for any neighborhood of the point , there is a neighborhood of s.t. .
A mapping between topological spaces are continuous iff the inverse image under of every open set in is an open set in .
is discrete topology iff every mapping from into any topological space is continuous.
Homeomorphic spaces two topological spaces are homeomorphic if there is a bijection s.t. and are continuous.
Metrizable topological space
if the topology is homeomorphic to a metric space.
note that if two metric spaces are isometric, then they are homeomorphic as topological spaces. But not the converse. For example, consider and , with the usual Euclidean metric. It is easy to construct a linear monotone bijective mapping from one to another, hence homeomorphic; yet, you can never get in the space , so the two spaces can not be isometric.
compactness, topological spaces is defined to be compact if every open cover of contains a finite sub-cover. is compact if is compact in the subspace topology.
2 results for continuous mapping & compact domain (topological space results) Let be a continuous mapping. Then for any compact , is a compact set in .
If is compact, then a continuous function is bounded and attains its maximum.
product topology
Given
and
, a topology on
is defined by taking as a base the collection of sets of
the form
, where
for
.
(note that you actually need to show the collection of sets so
constructed indeed is a base)
CH3 Special Spaces (Topological/Metric/Normed vector spaces)
Topological, metric, and normed vector spaces will be introduced. together with operators on them, they are the main focus of Functional Analysis.
This chapter introduces some important special cases, and establish their topological and metric properties.
First we should clarify the difference between metric space and vector space.
metric space in general have no notion of addition and scalar multiplication (note we are not looking at the special case of normed space, which induces a metric, but also is by definition a vector space)
vector space in general have no notion of distance.
topological vector space defined as a vector space over field , endowed with a topology such that vector addition and scalar multiplication are continuous functions. That is:
for any vector , for any neighborhood of this vector, there exists neighborhood of and neighborhood of s.t.
for any vector , for any neighborhood of it, there exists a neighborhood of the scalar and neighborhood of , s.t.
metric vector space a topological space whose topology is the topology of a metric/pseudometric space.
For metric vector space, continuity of vector + and scalar * can be expressed using sequences: is a metric vector space if:
norm & seminorm Given vector space over field . A function is a seminorm on if:
If in addition, , then it is a norm.
Banach space
a normed space that is a complete metric space.
Note that normed space is easily seen to be Hausdorff.
We can show that normed and seminormed spaces are metric spaces via defining . This requires proof because we are not just showing it is a metric space (this just requires verifying is a metric. We are actually showing it is a “metric vector space”, so we need to verify that + and scalar* are continuous operations.
normed subspace (of normed space ) Take , and restrict the norm on to on .
Abstract/genreal normed spaces will be studied in Ch4. This chapter deals with special cases of normed spaces.
Finite-Dim Spaces
. Equip this with the -norm for .
If , the norm is . If , we just have .
If , then the “norm” function defined above is not a norm, but we can still define a metric
can show that all the spaces so defined, for , are complete & separable metric spaces.
(background info for isomorphic topological spaces) isomorphism (mapping) of vector spaces source Two vector spaces over the same field are called isomorphic if there is a bijection which preserves addition and scalar multiplication:
And this correspondence is called isomorphism of vector spaces.
Isomorphic topological spaces Topological vector spaces are called isomorphic if there exists a bijection s.t.:
- is an isomorphism of vector spaces
- is a homeomorphism of topological spaces
Can show that , the spaces are all isomorphic. As a matter of fact, in Ch4, we will show that every two -dim normed spaces are isomorphic.
Sequence spaces
for , elements in are defined to be sequences s.t. converges. This is a vector space.
for , we use the sup-norm
We can show that for , the space is complete and separable. While for , the space is complete, but not separable.
We can show that if , then . But note we are talking about vector (sub)spaces. is NOT a normed subspace of any .
sequence spaces
space of all convergent sequences
space of all sequences converging to 0
space of all sequences s.t. for all but finitely many .
Can show and are complete (hence Banach), and separable. is not complete. (later will show its Banach completion is )
sequence space
the space is the vector space of all sequences with terms in .
there are many ways of making it into a topological vector space. Begin with a general approach – turning it into a polynormed space.
polynormed space Let be a family of seminorms on vector space . We can show that finite intersections of sets of the form form the basis of a topology on .
Assume the seminorms separate points in : , there exists s.t. .
Then, the space is Hausdorff (every two points have distinct neighborhoods). If is countable, we call a polynormed space.
(natural) metric for polynormed space For polynormed space with seminorms , define a metric via: . Can show this metric satisfies translation invariance: .
We can define two topologies on a polynormed space .
topology defined by the family of seminorms (I think it means the topology generated by all the open balls associated with whatever )
metric topology defined by the metric
Turns out, these two topologies are the same.
Some other properties about polynormed space
1.
converges to
, i.e.,
, iff
The polynormed space is a metric vector space (verify vector + and scalar * are continuous).
for the special sequence space , define a family of seminorms by , and let be the associated metric. Then, the space is complete, separable, but not normable (topology being equivalent to the topology defined by a norm).
Space of Continuous Functions
Usually we equip this vector space of continuous functions with the sup-norm:
We can show that such normed space is complete and separable.
In general, the similarly defined normed space can be shown to be isomorphic to .
The class of is defined as the family of -valued functions on the interval s.t. the derivative exists and is continuous. Can equip this space with a norm:
For , cannot use the sup-norm, since there could be unbounded continuous functions on . But we can introduce polynormed structure on it.
The Spaces
This stands for functions of bounded variation. The total variation is defined as : P is a partition of
We can show that $V^{b}_{a}(x) is a seminorm on the space . (but not a norm since as long as is a constant function on .) To make it a norm, we need to define . And with this norm, we can show that the normed space is Banach (complete). However, it is not separable.
CH4 Normed Spaces
convergence in general normed spaces
Given , recall convergence in this normed space is convergence in the associated metric space, which can be stated in terms of sequences.
Cauchy sequence:
A series in is convergent if the partial sums converges to some element in .
A series in is absolutely convergent if converges in .
In , absolute convergence implies convergence. But not true in general normed spaces.
Shauder basis
a sequence of vectors in is a Schauder basis for if , ! sequence of scalars s.t. the series converges to . That is, .
Recall Hamel basis: for a vector space over a field , A subset of vectors in is a Hamel basis for if is linearly independent and spans . We will show later this chapter that Hamel basis is either finite or uncountable. But Schauder basis apparently can be at most countable.
This series expression for is called the expansion of wrt basis .
Can show that if has a Schauder basis then is separable. Thus, cannot have a basis. But we can show that for have Schauder basis.
Banach completion
Given a normed space which may not be complete. Then there exists a Banach space and an isometry s.t.
- is dense in
- is unique up to isometry of normed spaces
Recall previously we have some extension results of metric spaces. Here is another one for normed spaces. We will natural start with that result, but more work is needed: we need to extend vector-+ and scalar-* into the extended metric space to make it a vector space, and then introduce a norm on this extended space that makes it a Banach space.
As an example, with norm is not complete. Its completion is , the space of Lebesgue measurable functions on such that (Lebesgue integral) is finite.
Linear Operators and Functionals
continuous linear operator A linear operator is **continuous at if .
note that this is consistent with the definition of continuous mapping between metric spaces.
bounded linear operator A linera operator is bounded if there exists s.t. .
Some results about bounded/continuous operators:
a linear operator is continuous at implies that it is bounded
a linear operator is bounded implies that it is uniformly continuous. That is, , as long as , we have .
a linear operator is bounded iff it is continuous
null space of a bounded operator, denoted , is a closed set
in general, the null space of an operator being closed does not imply the operator is bounded. But if the operator is known to be a linear functional (linear operator mapping into ), then it null space being closed does imply boundedness
Example of bounded/unbounded operator:
be the space of polynomials on with sup-norm. differentiation operator defined by , for is unbounded. For example, let , then . Then , but .
be with the sup-norm. Let be a continuous function defined on . Define the integral operator . Can show this is a linear and bounded operator from to itself. We call the kernel of .
operator norm
For a bounded operator , define .
this is equivalent to the def:
, we have
Denote as the set of all bounded linear operators from to . Then the operator norm defined above can serve as a norm for this space.
for the special case of linear functionals, where , then we call the dual space of , denoted by . The norm of a linear functional is then given by .
We can show that for the normed space (, it is Banach if is Banach.
Book calculated some operator norms for simple spaces on p84-85. One example is used a lot in spectral theory, the multiplication operator on :
given function , define
can show that .
Finite-dim Normed Spaces
Equivalence of norms:
Let be a finite-dim vector space. Then every two norms on are equivalent, in the sense that there exists s.t. .
therefore, every two norms on finite-dim vector space define the same topology on
If a normed space is finite-dim, then every linear operator is bounded.
isomorphic normed spaces isomorphic normed spaces is defined as a special case of isomorphic topological vector spaces.
if is an isomorphism of normed spaces, then are bounded opeartors
if are isomorphic normed spaces, then is complete iff is complete.
every two -dim normed spaces are isomorphic
every finite-dim subspace of a normed space is complete, and hence, is closed
a subset of a finite-dim normed space is compact iff it is closed and bounded
As an application, can show that a Hamel basis of a Banach space is either finite or uncountable. (Baire category thm is also used in the proof)
Compactness in Normed Spaces
In , we know the equivalence between closed&bounded and compact. But not true for ANY infinite dim spaces: **the closed unit ball in a normed space is compact iff is finite-dim.
In this chapter the book defined orthogonality, not using inner product, but using norms: if .
And presented a theorem characterizing the “perpendiculars” to the null space of some bounded linear functional.
here means and
the claim is that a vector is perpendicular to the null space of iff
It is possible that we cannot find such , in the sense that can be arbitrarily close to 1 but never equals to 1, for any .