# User:TakuyaMurata/Calculus

## Module and linear space

[edit | edit source]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
- (ii) (associativity)
- (iii) and (distribution law)
- (iv)

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.

## Normed spaces

[edit | edit source]A vector space is said to be *normed* if it is a metric space and its metric has the form:

Here, the function , called a *norm*, has the property (in addition to that it induces the metric) that for any scalar . We note that:

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.*

## Baire's theorem

[edit | edit source]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:

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

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.

## Duality

[edit | edit source]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

In other words, is not closed.

**3 Theorem** *If is a linear functional on , then*

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]

## The Hahn-Banach theorem

[edit | edit source]**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. *

## Hilbert spaces

[edit | edit source]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:

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,

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,*

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,

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 .

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

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.

## Integration

[edit | edit source]**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 ,

- .

## Differentiation

[edit | edit source]*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:*

*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:

.

That is closed puts the limit of in . Finally, the uniqueness follows since if and , then

- or unless .

**3. Corollary (mean value inequality)** *Let be differentiable. Then there exists some for some such that*

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 ,

.

It thus follows: since ,

**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 .

Since and , the Chain Rule gives:

.

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 ,

where by assumption. Finally it now follows:

**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 .

## Addendum

[edit | edit source]**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**The set**has dimension .*