An additive group
is said to be a module over
, or R-module for short, if the scalars, the members of a ring
, satisfy the following properties: if
and
- (i) Both
and
are in ![{\displaystyle G}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5f3c8921a3b352de45446a6789b104458c9f90b)
- (ii)
(associativity)
- (iii)
and
(distribution law)
- (iv)
![{\displaystyle 1_{R}x=x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95ab100ba0f80ae5277ed7c954b53f15cb1940b1)
By definition, every abelian group itself is a module over
, since
and
is a scalar. Finally, a linear space is a module over a field. Defining the notion of dimension is a bit tricky. However, we can safely say a
-vector space is finite-dimensional if it has a finite basis; that is, we can find linear independent vectors
so that
. Such a basis need not be unique.
3 Theorem Let
be a finite-dimensional
-vector space. Then
has the same dimension as
does; that is, every basis for
has the same cardinality as every basis for
does.
It can be shown that the map
cannot be defined constructively.[1] (TODO: need to detail this matter)
1 Theorem If
is a TVS and every finite subset of
is closed, it then follows that
is a Hausdorff space.
Proof: Let
with
be given. Moreover, let
be the complement of the singleton
, which is open by hypothesis. Since the function
is continuous at
and
is in
, we can find an
open and such that
. Here, we used, and would do so henceforward, the notation
the union of
taken all over
and
. Furthermore, since the function
is continuous and so is its inverse, namely
, we may assume that
by replacing
by the intersection of
and
. By repeating the same construction for each
where
, we find
so that
. It then follows that
and
are disjoint. Indeed, if we write
for some
, then
, a contradiction.
A vector space is said to be normed if it is a metric space and its metric
has the form:
![{\displaystyle d(x,y)=\|x-y\|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94df456fedac6b08c33cb4dffa5345a0ce0891f0)
Here, the function
, called a norm, has the property (in addition to that it induces the metric) that
for any scalar
. We note that:
![{\displaystyle \|x+y\|=d(x,-y)\leq d(x,0)+d(0,-y)=\|x\|+\|y\|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/178a8017c5f5e84c349b7d7126e63f45bfd80437)
and
for any
.
It may go without saying but a vector space is infinite-dimensional if it is not finite-dimensional.
3 Theorem Let
,
be normed spaces. If
is an infinite-dimensional and if
is nonzero, there exists a linear operator
that is not continuous.
A normed space is said to be complete when every Cauchy sequence in it converges in it.
3 Theorem Let
is a subspace of a Banach space
carrying the same norm. Then the following are equivalent:
- (a)
is complete.
- (b)
is closed in
.
- (c)
implies
.
Proof: (i) Show (a)
(b). If
is complete, then every Cauchy sequence in
has the limit in
; thus,
is closed. Conversely, if
is closed, then every Cauchy sequence converges in
since
is complete. Hence,
is complete. (ii) Show (a)
(c). Let
be a Cauchy sequence. Then
as
.
Thus,
is Cauchy, and converges in
since the completeness. Conversely, since a Cauchy sequence is convergent, we can find its subsequence
such that
. Then
.
If the summation condition holds, then it follows that
converges in
. Hence,
converges in
as well.
3 Corollary
is incomplete but dense in
.
Proof:
is not closed in
. Since
has empty interior,
.
We say a set has dense complement if its closure has empty interior.
The next is the theorem whose importance is not what it says literally but that of consequences. Though the theorem can be proved more generally for a pseudometric space; e.g., F-space, this classical formulation suffices for the remainder of the book.
3 Theorem A complete normed space
which is nonempty is never the union of a sequence of subsets of
with dense complement.
Proof: Let
be a sequence of subsets of
with dense complement. Since
has empty interior and
has nonempty interior, there exists an nonempty open ball
with the radius
. Since
has empty interior and
has empty interior, again there exists an nonempty open ball
with the radius
. Iterating the construction ad infinitum we get the decreasing sequence
. Now let
be the sequence of the centers of
. Then
is Cauchy since: for some
as
.
It then follows
converges in
from the compleness of
.
3 Corollary (open mapping theorem) If
and
are Banach spaces, then a continuous linear surjection
maps an open set in
to an open set in
.
Proof: Left as an exercise.
The following gives an nice example of the consequences of Baire's theorem.
3 Corollary (Lipschitz continuity) Let
= the set of functions
such that there exists some
such that:
for all
.
Then (i)
is complete, (ii)
is closed and has dense complement, and (iii) there exists a
that is not in any
; i.e., one that is differentiable nowhere.
Proof: (i)
is complete; thus,
is a Banach space by some early theorem. (ii) Let
be a sequence, and suppose
. Then we have:
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as
|
Thus,
; i.e.,
is closed. Stone-Weierstrass theorem says that every continuous function can be uniformly approximated by some infinitely differentiable function; thus, we find a
such that:
.
If we let
, then
![{\displaystyle v\in {\mathcal {C}}^{0}([0,1])\backslash S_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5431c7dea9b069c18476bd7227251e300982b6a)
Hence,
has dense complement. Finally, (iii) follows from Baire's theorem since (i) and (ii).
More concisely, the theorem says that not every continuity is Lipschitz because of Baire's theorem.
3 Lemma In a topological space
, the following are equivalent:
- (i) Every countable union of closed sets with empty interior has empty interior.
- (ii) Every countable intersection of open dense sets is dense.
Proof: The lemma holds since an open set is dense if and only if its complement has empty interior.
When the above equivalent conditions are true, we say
is a Baire space.
3 Theorem If a Banach space
has a Schauder basis, a unique sequence of scalars such that
as
,
then
is separable.
Proof:
The validity of the converse had been known as a Basis Problem for long time. It was, however, proven to be false in 19-something by someone.
The kernel of a linear operator
, denoted by
, is the set of all zero divisors for
. A kernel of a linear operator is a linear space since
and
implies
. Moreover, a linear operator has zero kernel if and only it is injective.
3 Theorem Let
be a linear functional. Then
is continuous if and only if
is closed.
Proof: If
is continuous, then
is closed since a finite set is closed. Conversely, suppose
is not continuous. Then there exists a sequence
such that
![{\displaystyle \lim _{j\to \infty }f(x_{j})-f(x)=\lim _{j\to \infty }f(x_{j}-x)\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a6c38684652b7c4d4e515ccb7c115a65afa6e89)
In other words,
is not closed.
3 Theorem If
is a linear functional on
, then
![{\displaystyle f(x_{1},x_{2},x_{3},...)=\sum _{1}^{\infty }x_{k}y_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b70d092c0768d9723449fa5a21df16b0fa05b7ef)
Proof: Let
where
if
else
.
The dual of a linear space
, denoted by
, is the set of all of linear operators from
to
(i.e., either
or
). Every dual of a linear space becomes again a linear space over the same field as the original one since the set of linear spaces forms an additive group.
Theorem Let G be a normed linear space. Then
and
.
The duality between a Banach space and its dual gives rise to.
Example: For
finite, the dual of
is
where
.
3 Theorem (Krein-Milman) The unit ball of the dual of a real normed linear space has an extreme point.
Proof: (TODO: to be written)
The theorem is equivalent to the AC. [2]
3 Theorem (Hahn-Banach) Let
be normed vector spaces over real numbers. Then the following are equivalent.
- (i) Every collection of mutually intersecting closed balls of
has nonempty intersection. (binary intersection property)
- (ii) If
is a subspace and
is a continuous linear operator, then
can be extend to a
on
such that
. (dominated version)
- (iii) If the linear variety
does not meet a non-empty open convex subset
of
, then there exists a closed hyper-plane
containing
that does not meet
either. (geometric form)
3 Corollary If the equivalent conditions hold in the theorem,
is complete.
Proof: Consider the identity map extended to the completion of
.
3 Corollary Let
be a linear operator from a Banach space
to a Banach space
. If there exists a set
and operators
and
such that
and
, then
can be extended to a Banach space containing
without increase in norm.
A linear space
is called a pre-Hilbert space if for each ordered pair of
there is a unique complex number called an inner product of
and
and denoted by
satisfying the following properties:
- (i)
is a linear operator of
when
is fixed.
- (ii)
(where the bar means the complex conjugation).
- (iii)
with equality only when
.
When only one pre-Hilbert space is being considered we usually omit the subscript
.
We define
and indeed this is a norm. Indeed, it is clear that
and (iii) is the reason that
implies that
. Finally, the triangular inequality follows from the next lemma.
3 Lemma (Schwarz's inequality)
where the equality holds if and only if we can write
for some scalar
.
If we assume the lemma, then since
for any complex number
it follows:
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Proof of Lemma: The lemma is just a special case of the next theorem:
3 Theorem Let
be a pre-Hilbert and
be an orthonormal set (i.e., for
iff
iff
is nonzero.)
- (i)
for any
.
- (ii) The equality holds in (i) if and only if
is maximal in the collection of all orthonormal subsets of
ordered by
.
Proof: (TODO)
3 Theorem Let
be a sequence in a pre-Hilbert space with
. If
, then
for any sequence
of scalars.
Proof: Let
be a set of all pairs
such that
,
and
. By Hölder's inequality we get:
.
Since
,
we get the second inequality. Moreover,
![{\displaystyle \sum _{j=m}^{n}|\alpha _{j}|^{2}\leq \sum _{j=m}^{n}\langle \alpha _{j}u_{j},\alpha _{j}u_{j}\rangle +\sum _{(j,k)\in I}\langle \alpha _{j}u_{j},\alpha _{k}u_{k}\rangle +|\langle \alpha _{j}u_{j},\alpha _{k}u_{k}\rangle |}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d15d9d4deaee54dfcd5918ff13efbea2624b8ed2)
and this gives the first inequality.
3 Theorem (Bessel's inequality) Let
be an orthonormal subset of a pre-Hilbert space. Then for each
in the space,
![{\displaystyle \sum _{u\in U}|\langle x,u\rangle |^{2}\leq \|x\|^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0786034f2f1987725b7d18bbb45c14c9fca37a41)
where the sum can be obtained over some countable subset of
and the equality holds if and only if
is maximal; i.e.,
is contained in no other orthogonal sets.
Proof: First suppose
is finite; i.e.,
. Let
. Since for each
,
, by the preceding theorem or by direct computation,
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Now suppose that
is maximal. Let
. Then by the same reasoning above,
is orthogonal to every
. But since the assumed maximality
. Hence,
. Conversely, suppose that
is not maximal. Then there exists some nonzero
such that
for every
. Thus,
.
The general case follows from the application of Egorov's theorem.
3 Corollary In view of Zorn's Lemma, it can be shown that a set satisfying the condition in (ii) exists. (TODO: need elaboration)
3 Lemma The function
is continuous each time
is fixed.
Proof: If
, from Schwarz's inequality it follows:
as
. ![{\displaystyle \square }](https://wikimedia.org/api/rest_v1/media/math/render/svg/455831d58fa08f311b934d324adcff89a868b4e4)
Given a linear subspace
of
, we define:
. In other words,
is the intersection of the kernels of the continuous functionals
, which are closed; hence,
is closed. (TODO: we can also show that
)
3 Lemma Let
be a linear subspace of a pre-Hilbert space. Then
if and only if
.
Proof: The Schwarz inequality says the inequality
![{\displaystyle |\langle z,z+w\rangle |\leq \|z\|\|z+w\|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9235a2cb71ec7dda270a2e8fec7bf6951659a9f)
is actually equality if and only if
and
are linear dependent.
3 Theorem (Riesz) Let
be a pre-Hilbert space and
be its subspace. The following are equivalent:
- (i)
is a complete.
- (ii)
is dense if and only if
.
- (iii) Every continuous linear functional on
has the form
where y is uniquely determined by
.
Proof: If
and
, then
. (Note: completeness was not needed.) Conversely, if
is not dense, then it can be shown (TODO: using completeness) that there is
such that
.
That is,
. In sum, (i) implies (ii). To show (iii), we may suppose that
is not identically zero, and in view of (ii), there exists a
with
. Since
,
.
The uniqueness holds since
for all
implies that
. Finally, (iii) implies reflexivility which implies (i).
A complete pre-Hilbert space is called a Hilbert space.
3 Corollary Let
be a a closed linear subspace of a Hilbert space</math>
- (i) For any
we can write
where
and
and
are uniquely determined by
.
- (ii) then
.
Proof: (i) Let
be given. Define
for each
. Since
is continuous and linear on
, which is a Hilbert space, there is
such that
. It follows that
for any
; that is,
. The uniqueness holds since if
and
, then
and the representation is unique. (ii) If
, then since
is orthogonal to
. Thus,
and taking closure on both sides we get:
.
Also, if
, then we write:
where
and
and
. Thus,
.
Since
implies that
and
, the corollary follows.
3 Theorem (Fundamental Theorem of Calculus) The following are equivalent.
- (i) The derivative of
at
is
.
- (ii)
is absolutely continuous.
Proof: Suppose (ii). Since we have:
,
for any
,
. ![{\displaystyle \square }](https://wikimedia.org/api/rest_v1/media/math/render/svg/455831d58fa08f311b934d324adcff89a868b4e4)
Differentiation of
at
is to take the limit of the quotient by letting
:
.
When the limit of the quotient indeed exists, we say
is differentiable at
.
The derivative of
, denoted by
, is defined by
= the limit of the quotient at
.
3.8. Theorem The power series:
![{\displaystyle u=\sum _{0}^{\infty }a_{j}z^{j}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ccba1704fa16352572260e156a0ba7554595cd2)
is analytic inside the radius of convergence.
Proof: The normal convergence of
implies the theorem.
To show that every analytic function can be represented by a power series, we will, though not necessarily, wait for Cauchy's integral formula.
We define the norm in
, thereby inducing topology;
.
The topology in this way is often called a natural topology of
, since so to speak we don't artificially induce a topology by defining
.
3. Theorem (Euler's formula)
If
.
Proof:
3. Theorem (Cauchy-Riemann equations) Suppose
. We have:
on
if and only if
on
.
Proof:
3. Corollary Let
are non-constant and analytic in
. If
, then
.
Proof: Let
. Then
. Thus,
, and hence g = 0</math>.
This furnishes examples of functions that are not analytic. For example,
is analytic everywhere and that means
cannot be analytic unless
.
A operator
is bounded if there exists a constant
such that for every
:
.
3.1 Theorem Given a bounded operator
, if
,
and
,
then
.
Proof: Since
can be verified (FIXME) and
is inf,
.
Thus,
.
But if
in the above, then this is absurd since
is sup; hence the theorem is proven.
We denote by
either of the above values, and call it the norm of
3.2 Corollary A operator
is bounded if and only if is continuous.
Proof: If
is bounded, then we find
and since the identity: for every
and
,
is continuous everywhere. Conversely, every continuous operator maps a open ball centered at 0 of radius 1 to some bounded set; thus, we find the norm of
,
, and the theorem follows after the preceding theorem.
3. Theorem If F is a linear space of dimension
, then it has exactly
subspaces including F and excluding {0}.
Proof: F has a basis of n elements.
Theorem If H is complete, then
(i.e., a cartesian space of E) is complete
Proof: Let
be a Cauchy sequence. Then we have:
as
.
Since orthogonality, we have:
,
and both
and
are also Cauchy sequences. Since completeness, the respective limits
and
are in
; thus, the limit
is in E_2.
The theorem shows in particular that
are complete.
3. Theorem (Hamel basis) The Axiom of Choice implies that every linear space has a basis
Proof: We may suppose the space is infinite-dimensional, otherwise the theorem holds trivially.
FIXEME: Adopt [3]. 3. Theorem (Fixed Point Theorem) Suppose a function f maps a closed subset
of a Banach space to itself, and further suppose that there exists some
such that
for any
and
. Then
has a unique fixed point.
Proof: Let
be a sequence:
. For any
for some
. Then we have:
.
By induction it follows:
.
Thus,
is a Cauchy sequence since:
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That
is closed puts the limit of
in
. Finally, the uniqueness follows since if
and
, then
or
unless
. ![{\displaystyle \square }](https://wikimedia.org/api/rest_v1/media/math/render/svg/455831d58fa08f311b934d324adcff89a868b4e4)
3. Corollary (mean value inequality) Let
be differentiable. Then there exists some
for some
such that
![{\displaystyle \|f(x)-f(y)\|\leq \|f'(z)\|\|x-y\|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e4421f689f1c100c97302f964ae795b8ca5a689)
where the equality holds if
(mean value theorem).
Proof:
Theorem Let
where
and is open. If
are bounded in
, then
is continuous.
Proof: Let
and
be given. Using the assumption, we find a constant
so that:
for
.
Let
. Suppose
and
. Let
.
Then by the mean value theorem, we have: for some
,
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It thus follows: since
,
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Theorem (differentiation rules)' Given
differentiable,
- (a) (Chain Rule)
.
- (b) (Product Rule)
.
- (b) (Quotient Rule)
.
Proof: (b) and (c) follows after we apply (a) to them with
,
and the implicit function theorem.
.
Theorem (Cauchy-Riemann equations) Let
and
. Then
is differentiable if and only if
and
are continuous on
and
on
.
Proof: Suppose
is differentiable. Let
and
and
.
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Since
and
, the Chain Rule gives:
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Conversely, let
. It suffices to show that
. Let
be given and
and
. Since the continuity of the partial derivatives and that
is open, we can find a
so that:
and for
it holds:
and
.
Let
be given and
and
. Using the mean value theorem we have: for some
,
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where
by assumption. Finally it now follows:
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3 Corollary Let
and suppose
is connected. Then the following are equivalent:
- (a)
is constant.
- (b)
is constant.
- (c)
is constant.
Proof: That (a)
(b) is obvious. Suppose (b). Since we have some constant
so that for all
,
,
clearly it holds that
. Thus, (b)
(c). Suppose (c). Then
.
Differentiating both sides we get:
.
Since
, it follows that
and
.
If
, then
. If
, then
is constant since
is connected. Thus, (c)
(a).
We say a function has the open mapping property if it maps open sets to open sets. The maximum principle states that equivalently
- if a function has a local maximum, then the function is constant.
3 Theorem Let
. The following are equivalent:
- (a)
is harmonic.
- (b)
has the mean value property.
3 Theorem Let
. If
has the open mapping property, then the maximum principle holds.
Proof: Suppose
and
is open and connected. Let
. If
has a local maximum, then
is nonempty. Also,
is closed in
since
. Let
. Since
is open, we can find a
so that:
. Since
is open by the open mapping property, we can find a
so that
. This is to say that
for some
. This is absurd since
and
for all
. Thus,
identically on
and it thus holds that
and
is open in
. Since
is connected,
. Therefore,
on
.
Exercise Let
. Then
is a polynomial of degree
if and only if there are constants
and
such that
for all
.
Exercise 2 Let
be linear. Further suppose
has dimension
. Then the following are equivalent:
exists
where ![{\displaystyle \det(f)=\sum _{1}^{n}sgn(\sigma )x_{\sigma (i)j}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/baffa40604f988a43f847d2b36e8bb3e2b6c6a0d)
- The set
has dimension
.