R n and C n
Formally, the vector spaceR n is defined as the set consisting of all sequences ofnreal numbers:
O Christensen, Functions, Spaces, and Expansions: Mathematical Tools 1 in Physics and Engineering, Applied and Numerical Harmonic Analysis,
DOI 10.1007/978-0-8176-4980-7 1, c Springer Science+Business Media, LLC 2010
Usually a vector belonging toR n orC n is written as x= (x 1 , x 2 , , x n ).
Whenever convenient we will also write the vectorxas a column, x ⎛
The article assumes that readers are familiar with the definition of inner products in vector spaces; for those who are not, it is recommended to refer to Definition 4.1.1 In \( \mathbb{R}^n \), the canonical inner product is defined as \( \langle x, y \rangle = \sum_{k=1}^{n} x_k y_k \) for \( x, y \in \mathbb{R}^n \) Similarly, in \( \mathbb{C}^n \), the canonical inner product is given by \( \langle x, y \rangle = \sum_{k=1}^{n} x_k y_k \) for \( x, y \in \mathbb{C}^n \).
The spaces \( \mathbb{R}^n \) and \( \mathbb{C}^n \) can be defined with various inner products; however, we will consistently utilize the inner products specified in equations (1.1) and (1.2) In these spaces, the length of a vector \( x \), referred to as the norm, is a fundamental concept.
We will focus on the vector spaceC n ; the theory forR n is parallel, except that we do not have the complex conjugation in the inner product.
In the theory forC n , the concept of a basis plays a crucial role We state the formal definition:
Definition 1.1.1 (Basis inC n ) Consider a collection of vectors{ e k } m k=1 inC n
(i) { e k } m k=1 is a basis for C n if span{ e k } m k=1 = C n and the vectors { e k } m k=1 are linearly independent.
(ii) { e k } m k=1 is an orthonormal basis for C n if { e k } m k=1 is a basis and e k , e :=δ k,
It is well-known that any basis forC n contains exactlym=nelements.
If{ e k } n k=1 is a basis forC n , any vectorv ∈C n has a unique representation v n k=1 c k e k (1.4) for some scalarsc k , k= 1, , n; in case{ e k } n k=1 is an orthonormal basis this representation takes the form (Exercise1.1) v n k=1 v , e k e k (1.5)
Example 1.1.2 (Canonical orthonormal basis forC n )Let the vectors
Then{ e k } n k=1 forms an orthonormal basis forC n with respect to the canon- ical inner product This basis is usually called thecanonical orthonormal basis for C n
Let us present another orthonormal basis for C n , the discrete Fourier transform basis An understanding of this basis will motivate our later study of Fourier series and the Fourier transform:
Example 1.1.3 (Discrete Fourier transform basis)Fork= 1, , n, define the vectorse k ∈C n by e k = 1
That is, the th coordinate ofe k is
The vectors { e_k } from k=1 to n, defined by (1.6), form an orthonormal basis for C^n Since these vectors exist in an n-dimensional vector space, demonstrating that they constitute an orthonormal system is sufficient to establish their linear independence and ability to span C^n A direct calculation confirms that || e_k || = 1 for all k Furthermore, when k equals j, the definition of e_k and a change of summation index indicate that e_k and e_j are equal to 1/n.
Using the formula (1−x)(1 +x+ã ã ã+x n−1 ) = 1−x n withx=e 2πi(k−j)/n , we get e k , e j = 1 n
The basis{ e k } n k=1 is called the discrete Fourier transform basis Using this basis, every sequencev ∈C n , v ⎛
=1 v e −2πi(−1)(k−1)/n e k (1.8) Written out in coordinates, this means that thejth coordinate is v j = 1 n n k=1 n
The vector v := { v , e k } n k=1 consisting of the coefficients in (1.7) is called thediscrete Fourier transform(DFT) ofv We return to the discrete
From linear algebra we know many equivalent conditions for a set of vectors to constitute a basis for C n Let us list the most important characterizations:
Theorem 1.1.4 (Characterization of basis forC n ) Considernvectors inC n , e 1 ⎛
⎠, and write them as columns in ann×n matrix,
Then the following are equivalent:
(i) The columns in E (i.e., the given vectors) constitute a basis forC n (ii) The rows in E constitute a basis forC n
(iii) The determinant ofE is nonzero.
(v) E defines an injective mapping from C n intoC n
(vi) E defines a surjective mapping from C n ontoC n
(vii) The columns in E are linearly independent.
(viii) E has rank equal to n.
Example 1.1.5 (Basis forC 2 )Consider the matrix
The operator associated withE acts onR 2 by matrix multiplication That is, forx= (x 1 , x 2 )∈R 2 ,
Since det(E) =−2, the equivalent conditions stated in Theorem1.1.4are satisfied Thus, the columns in the matrixE form a basis forC 2
Abstract vector spaces
Vector spaces are fundamental in linear algebra, extending its central concepts to abstract settings A vector space, denoted as V, is defined as a nonempty set equipped with two operations: addition and scalar multiplication The addition operation combines any two elements v and w from V to produce another element in V, represented as v + w Similarly, scalar multiplication takes an element v from V and a scalar α from the set of complex numbers, resulting in another element in V, denoted as αv or vα.
These operations have to satisfy the following requirements:
Definition 1.2.1 (Vector space) Consider a nonempty setV, equipped with operations of addition and scalar multiplication Assume that the following rules are satisfied:
(i) For all v , w ∈V, we have that v+w=w+v ;
(ii) For all v , w , u ∈V, we have that(v+w) +u=v+ (w+u);
(iii) There exists an element, called 0 , inV, such that for all v ∈V, v+0=v;
(iv) For each v ∈V there exists an element, called − v , in V, with the property that v+ (− v) =0;
(vi) For allα, β∈Cand all v ∈V,
(vii) For allα∈Cand all v , w ∈V, α(v+w) =α v+α w;
In that case we say that V, equipped with the operations of addition and scalar multiplication, forms a vector space.
It is immediate to check that the setsR n andC n ,equipped with the usual operations of addition and scalar multiplication, satisfy the conditions in Definition1.2.1.
A complex vector space is defined by allowing scalar multiplication with complex numbers, while a real vector space is similar but uses real numbers as scalars.
In this book, all vector spaces, except for R^n, are considered over the complex numbers rather than the real numbers R Consequently, all definitions will be presented specifically for the complex case.
Example 1.2.2 (Functions on a set A ) Let A denote an arbitrary nonempty set, and let V denote the collection of all functions f :A→C.
Given functionsf, g∈V,we define the functionf+g∈V by
Also, givenf ∈V andα∈C,define the functionαf ∈V by
Direct verification shows thatV equipped with these definitions of addition and scalar multiplication satisfies all the conditions in Definition 1.2.1. Thus, the setV forms a (complex) vector space
Example 1.2.3 (Polynomials on R) LetW denote the set of polyno- mials onRof degree at mostN for someN ∈N That is, the elements in
P(x) =a N x N +a N −1 x N−1 +ã ã ã+a 0 , x∈R, for some scalar coefficients a 0 , a 1 , , a N ∈ C Given a polynomial P of that form and another polynomialQof the same form,
Q(x) =b N x N +b N−1 x N−1 +ã ã ã+b 0 , we define the polynomialP+Qby
(P+Q)(x) = (a N +b N )x N + (a N−1 +b N −1 )x N−1 +ã ã ã+ (a 0 +b 0 ). Also, forα∈C, we define the polynomialαP by
We see thatP+Q∈W and thatαP ∈W; furthermore, direct verification shows that all the conditions in Definition 1.2.1are satisfied Thus, with our definitions of addition and scalar multiplication, the set W forms a
In a complex vector space V, a collection of vectors {v_k} from k=1 to N can form a linear combination represented as α_1 v_1 + α_2 v_2 + + α_N v_N, where α_1, α_2, , α_N are coefficients in the complex numbers C If all coefficients are set to zero, the resulting linear combination produces the zero vector The vectors {v_k} are considered linearly independent if no other combination of these vectors can result in the zero vector.
Definition 1.2.4 (Linear independence) Let{ v k } N k=1 be a collection of vectors inV If α 1 v 1 +α 2 v 2 +ã ã ã+α N v N =0 ⇒α 1 =α 2 =ã ã ã=α N = 0, then { v k } N k=1 are linearly independent; if not, the vectors are linearly dependent.
Example 1.2.5 (Linear independence of polynomials)Consider the vector spaceV in Example1.2.3 Note that the polynomials
1, x, , x N (1.10) belong toV.Now, assume that for someα 0 , α 1 , , α N ∈C,we have that α 0 +α 1 x+ã ã ã+α N x N = 0 for allx∈R.
Since a nontrivial polynomial of degreeN at most can haveN roots, this implies that α 0 =α 1 =ã ã ã=α N = 0.
Thus, the polynomials in (1.10) are linearly independent
Often, we encounter subsets of vector spaces having themselves the structure of a vector space:
Definition 1.2.6 (Subspace) LetV be a vector space A subset W ⊆V which itself is a vector space (when equipped with the operations of addition and scalar multiplication inV), is called a subspace ofV.
The set just containing the element 0is always a subspace of V Fur- thermore,V is a subspace of itself Any subspaceW for whichW =0and
In practice, one often verifies that a subsetW ofV is a subspace via the following lemma The reader is asked to provide the proof in Exercise1.4.
Lemma 1.2.7 (Characterization of subspace) A nonempty subsetW of a vector spaceV is a subspace ofV if and only if α v+β w ∈W for all v , w ∈W, α, β∈C (1.11)
In the context of vector spaces, let V represent the space of all functions from R to C, as illustrated in Example 1.2.2 Notably, the subset W discussed in Example 1.2.3 is included within V Additionally, it is important to note that any linear combination of two polynomials with a maximum degree of N will also yield a polynomial of the same degree or less, confirming that the property (1.11) holds true.
Finite-dimensional vector spaces
LetV be a (complex) vector space Given a collection of vectors{ v k } N k=1 of vectors in V, we define the span of the vectors as the set of all linear combinations: span{ v k } N k=1 :={α 1 v 1 +α 2 v 2 +ã ã ã+α N v N |α 1 , α 2 , , α N ∈C}.
In general, span{ v k } N k=1 will be a nontrivial subspace of V However, for special choices of the vectors{ v k } N k=1 it might happen that the span of the vectors equalsV This leads to a definition:
Definition 1.3.1 (Dimension) A vector space V has dimension N,
N ∈N, if there exists a collection of linearly independent vectors{ v k } N k=1 such that
A vector space, for which the condition in Definition 1.3.1is satisfied for some numberN ∈N,is said to befinite-dimensional;otherwise, the vector space isinfinite-dimensional.
We now define the concept of abasisin a finite-dimensional vector space.
In Section2.5we treat the infinite-dimensional case.
Definition 1.3.2 (Basis in finite-dimensional vector space) A collec- tion of vectors{ e k } n k=1 in V is a basis for V if span{ e k } n k=1 =V and the vectors{ e k } n k=1 are linearly independent.
Example 1.3.3 (Basis for vector space of polynomials)The vector spaceV in Example1.2.3is finite-dimensional In fact, as we have seen in Example1.2.5, the polynomials
1, x, x 2 , , x N (1.12) are linearly independent, and each polynomialP of degree at mostN is a linear combination of these, i.e.,
The argument shows thatV has dimensionN+ 1, and that the vectors in
In this book we will mainly consider vector spaces consisting of functions
— and most of them will actually be infinite-dimensional.
Topology in R n
As we have seen in (1.3), thelength(later to be called thenorm) of a vector x= (x 1 , , x n )∈R n is
Given a point x = (x 1 , , x n ) ∈ R n , the ball centered at x and with radiusr >0 is defined as the set
Definition 1.4.1 (Open and closed sets in R n ) Consider a subset U ofR n
(i) The subset U is open if for each x ∈U there exists a number δ >0 such that B(x , δ)⊆U.
(ii) The complement of the subsetU is defined as the set
(iii) The subsetU is said to be closed if the complement U c is open.
(iv) The closure of U, to be denoted U , is the smallest closed set in R n that contains U.
Note that it is easy to find sets in R n that are neither open nor closed Intuitively, the closure of a subset of R n is obtained by adding the “boundary”:
Example 1.4.2 (Subsets of Rand R 2 )We consider some subsets ofR andR 2 :
(i) The subset ]−1,1[ ofRis open; the closure of the set is [−1,1].
(ii) The subset [−1,1[ of Ris neither open nor closed; its closure is the set [−1,1].
(iii) The set ]0,1[×]4,7[ is open inR 2 ; its closure is [0,1]×[4,7].
(iv) The set [0,1[×]4,7[ is neither open nor closed in R 2 ; its closure is
We now introduce the concepts supremum and infimum for subsets ofR.
Definition 1.5.1 (Supremum) Consider a subsetE ofR.
(i) E is bounded above if there exists a number β∈Rsuch that x≤β, ∀x∈E (1.13)
If a set E is bounded above, the supremum of E, denoted as supE, is the smallest number β that meets the condition (1.13) This can be expressed in three equivalent ways: supE = sup x∈E x or sup {x | x∈E} Conversely, for a set E that is not bounded above, we define supE as infinity (∞).
Definition 1.5.2 (Infimum) Consider a subsetE of R.
(i) E is bounded below if there exists a number α∈Rsuch that α≤x, ∀x∈E (1.14)
If a set E is bounded below, the largest number α that satisfies the condition is referred to as the infimum of E, denoted by infE, inf x∈E x, or inf{ x | x∈E} In cases where the set E is not bounded below, the infimum is defined as infE = −∞.
Example 1.5.3 (Supremum and infimum)By inspection, we see that sup]0,4] = 4, sup[−2,5[= 5, inf]−2,1] =−2, inf(Q∩[π,7]) =π
In the particular case of a functionf defined on a setAand taking real values,f :A→R,we can consider the set
The set A is called the domain of the function f, and E is the range or image.Now, supE= sup x∈A f(x) = sup{f(x)|x∈A}.
A warning is in order In general, the number sup x∈A f(x) does not need to be afunction valuefor the functionf; that is, there might not exist an x 0 ∈Asuch that f(x 0 ) = sup x∈A f(x) (1.16)
In case anx 0 ∈A satisfying (1.16) exists, we write max x∈A f(x) = sup x∈A f(x).
The expression max x∈A f(x) is only used if the supremum’s value appears as a function value.
Similarly, still under the assumption thatf is real valued and with the setE as in (1.15), we have that infE= inf x∈A f(x) = inf{ f(x)|x∈A}.
In case there exists anx 0 ∈Asuch that f(x 0 ) = inf x∈A f(x), we write min x∈A f(x) = inf x∈A f(x).
Example 1.5.4 (Supremum and infimum for functions)
The supremum value is attained, so sup x∈[0,2] f(x) = max x∈[0,2] f(x).
The supremum value is not attained.
The supremum value is attained, so sup x∈]0,1[ f(x) = max x∈]0,1[ f(x).
The concept of supremum is also used for sequences, e.g., indexed byN; in fact, they appear as special cases of the above by lettingA=N.Let us consider such a case:
Example 1.5.5 (Supremum for a sequence)Direct inspection shows that sup n∈N n 2 + 4 n 2 + 121= 1; the supremum is not attained
For any sequence of real numbers {c_k} from k=1 to infinity, the derived sequence {d_k}, defined as d_k := inf(n≥k) c_n = inf{c_k, c_k+1, }, is an increasing sequence If the original sequence {c_k} is bounded above, it follows that the limit as k approaches infinity of d_k exists Conversely, if the sequence {c_k} is unbounded, we define the limit as k approaches infinity of d_k to be infinity.
In any case, lim k→∞ d k is called limes inferior, or lim inf of the given sequence:
Definition 1.5.6 (lim inf and lim sup) Given a sequence {c k } ∞ k=1 of real numbers, define lim inf k→∞ c k := lim k→∞ n≥k inf c n
Similarly, define lim sup k→∞ c k := lim k→∞ sup n≥k c n
If the limit superior and limit inferior of the sequence {c_k} as k approaches infinity are real numbers, then infinitely many elements of the sequence are found within arbitrarily small neighborhoods of these limits To accurately describe this phenomenon, we must introduce the concept of an accumulation point.
Definition 1.5.7 (Accumulation point) Let {c k } ∞ k=1 be a sequence of real numbers A numberc∈Ris an accumulation point for{c k } ∞ k=1 if the set
{k∈Nc k ∈[c−, c+ ]} is infinite for all >0.
Definition 1.5.8 (Bounded set, compact set) Consider a subsetE of
(i) E is bounded if E is bounded above and below, i.e., if there exists a numberα >0 such that
(ii) E is compact if E is bounded and closed.
In particular, a sequence{c k } ∞ k=1 consisting of real or complex numbers is bounded if there exists a constantC >0 such that
Any bounded sequence of real numbers has at least one accumulation point:
Lemma 1.5.9 states that for a bounded sequence of real numbers {c_k} from k=1 to infinity, the limit inferior (lim inf) and limit superior (lim sup) as k approaches infinity are accumulation points of the sequence Furthermore, if the limit of the sequence as k approaches infinity exists, then the limit inferior, limit superior, and the limit itself are all equal.
We ask the reader to prove Lemma1.5.9in Exercise1.11.
Given a bounded sequence{c k } ∞ k=1 of real numbers, put c:= lim sup k→∞ c k
Given any > 0, Lemma 1.5.9 implies that for all N ∈ N there exists a k > N such that
In particular, there exists ak 1 ∈Nsuch that
2; and there exists ak 2 > k 1 such that
Continuing this way, we obtain a sequence ã ã ã> k n > k n−1 >ã ã ã> k 2 > k 1 ≥1 such that for anyn∈N,
The sequence{c k n } ∞ n=1 is said to be asubsequenceof{c k } ∞ k=1 By definition, c k n →c asn→ ∞.
This proves the so-calledBolzano–Weierstrass Lemma:
Lemma 1.5.10 (Bolzano–Weierstrass lemma) Every bounded se- quence{c k } ∞ k=1 of real numbers has a convergent subsequence.
The sequence{c k } ∞ k=1 does not converge But the subsequence
The concepts of limit inferior (lim inf) and limit superior (lim sup) extend to real-valued functions For a sequence of functions \( f_k: A \rightarrow \mathbb{R} \) where \( k = 1, 2, \ldots \), the limit inferior and limit superior are defined as follows: \( \liminf_{k \to \infty} f_k(x) = \lim_{k \to \infty} \inf_{n \geq k} f_n(x) \).
, x∈A, and lim sup k→∞ f k (x) := lim k→∞ sup n≥k f n (x)
Note that with our definitions of supremum and infimum, the functions lim inf k→∞ f k and lim sup k→∞ f k might assume the values±∞.
At a few instances we will use the following inequality, which is a discrete version of the so-calledFatou’s Lemma:
Lemma 1.5.12 (Fatou’s lemma) Let f k : N → [0,∞[, k ∈ N, be a sequence of functions Then
We guide the reader through a proof of Lemma1.5.12in Exercise1.12.
This section focuses on functions defined on the real numbers or their subintervals, taking values in either the real or complex numbers We begin by presenting a precise definition of continuity, along with the related concept of uniform continuity.
Definition 1.6.1 (Continuity and uniform continuity) Let I⊆Rbe an interval, and consider a functionf :I→C.
(i) The function f is continuous at the point x 0 ∈ I if for each >0 there exists aδ >0 such that
(ii) The function f is continuous if f is continuous at every point in I.
(iii) The functionf is uniformly continuous if for each >0 there exists aδ >0 such that
Uniform continuity imposes a stricter condition than continuity, as demonstrated by the contrast between fixed points in the interval I and the variability of both x and y in different scenarios.
We state some of the important properties for continuous functions on bounded and closed intervals:
Theorem 1.6.2 (Uniform continuity) A continuous function on a bounded and closed interval[a, b] is uniformly continuous.
A functionf defined on a setA, f :A→C,isboundedif the range
E:={f(x)x∈A} is bounded, i.e., if there exists a constantK >0 such that
If no such constantK exists, the functionf isunbounded.
Under certain conditions on the domain, a continuous function is bounded We ask the reader to provide the proof of the following result in Exercise1.13:
Theorem 1.6.3 (Continuous functions on [a, b]) Consider a continu- ous real-valued function f defined on a bounded and closed interval [a, b].
(ii) f attains its supremum, i.e., there existsx 0 ∈[a, b]such that f(x 0 ) = sup{f(x)| x∈[a, b]}. (iii) f attains its infimum.
In Theorems1.6.2and1.6.3it is assumed that the functionf is defined on a bounded and closed interval The conclusions might fail if this hypothesis is removed:
Example 1.6.4 We illustrate the necessity of the hypotheses in Theorems 1.6.2and1.6.3:
(i) Let f(x) = x −1 , x ∈]0,1[ The interval ]0,1[ is bounded, but not closed The functionf is continuous, but not uniformly continuous. The function is unbounded.
(ii) Let f(x) = x 2 , x ∈ [0,∞[ The interval [0,∞[ is closed, but not bounded The functionfis continuous, but not uniformly continuous.
We will now consider a sequence of functions{f k } ∞ k=1 ,all of them defined on an intervalI Related to such a sequence one can introduce various types of convergence:
Definition 1.6.5 (Pointwise convergence, uniform convergence)
Let {f k } ∞ k=1 be a sequence of functions defined on an interval I and f :I→Ca given function.
(i) If for eachx∈I and each >0 there exists anN ∈Nsuch that
|f(x)−f k (x)|< for allk≥N, then{f k } ∞ k=1 is said to converge pointwise tof. (ii) If for each >0 there exists an N∈Nsuch that sup x∈I |f(x)−f k (x)|< for allk≥N, then{f k } ∞ k=1 is said to converge uniformly tof.
Note that the difference between the two types of convergence is rather subtle:
• In order to check pointwise convergence, wefix x and ask forf k (x) being close tof(x) for large values ofk;
• In order to check uniform convergence, we ask forf k (x) being close tof(x) for large values ofk,simultaneously for allx.
Note also that if (i) in Definition1.6.5holds, then necessarily f(x) = lim k→∞ f k (x), x∈I.
The limit of a sequence of continuous functions might not be continuous itself On the other hand, the limit of a uniformly convergent sequence of continuous functions is continuous:
Theorem 1.6.6 (Continuity of uniform limit) Assume that {f k } ∞ k=1 is a sequence of continuous functions defined on an interval I If {f k } ∞ k=1 converges uniformly to a functionf :I→C, then f is continuous.
We guide the reader through a proof of Theorem1.6.6in Exercise1.15. For later use we state the definition of piecewise continuous functions formally:
Definition 1.6.7 (Piecewise continuous function) Let I ⊆ R be an interval A functionf :I→Cis piecewise continuous if the intervalI can be split into a finite collection of subintervals on whichf is continuous.
We note that the results for continuous functions considered in this sec- tion do not extend to piecewise continuous functions The following example illustrates this.
Example 1.6.8 (Piecewise continuous function)The function f : [−1,3]→R, f(x) :⎧⎪
0 ifx∈[−1,0], x −1 ifx∈]0,2], x ifx∈]2,3], is piecewise continuous However, f is unbounded and not uniformly continuous We ask the reader to prove these claims in Exercise1.14
The Riemann integral enables the integration of scalar-valued piecewise continuous functions across closed and bounded intervals Additionally, it permits integration for specific functions over unbounded intervals, such as when a function is defined on the interval [0,∞[ and involves the expression α.
0 f(x)dx has a limit asα→ ∞.Then theimproper Riemann integral
Similarly, assuming that the limits α→∞ lim α
1.7 Integration and summation 19 exist, we define
In this section we collect some basic inequalities concerning integrals. Three types of integrals will appear in this book:
(i) The (Riemann) integral, concerning integration of bounded and piecewise continuous functions over bounded intervals;
(ii) The improperRiemann integral, concerning integration of piecewise continuous functions over unbounded domains;
(iii) The Lebesgue integral, to be discussed in Section5.2.
The following results hold in all the settings (i)–(iii) We do not specify these assumptions on the functions and the intervals in the statements of the results.
Theorem 1.7.1 (H¨older’s inequality and Minkowski’s inequality)
Let I ⊆ R be an interval and consider functions f, g : I → C Then the following inequalities hold:
(i) (H¨older’s inequality) For any numbersp, q∈]1,∞[with1/p+1/q= 1,
(ii) (Minkowski’s inequality) For anyp∈[1,∞[,
Proofs of these inequalities are outlined in Exercises1.16and 1.17 We state another important inequality that is valid under the same type of assumptions on the functionf and the intervalI:
Lemma 1.7.2 (Absolute integrability implies integrability) Let I be an interval, and consider a function f :I→C.Assume that
The results in Theorem 1.7.1 and Lemma 1.7.2have discrete versions,reading as follows (Exercises1.18and1.19):
Theorem 1.7.3 (H¨older’s inequality, Minkowski’s inequality)
Consider any scalar sequences {x k } ∞ k=1 ,{y k } ∞ k=1 Then the following inequalities hold:
(i) (H¨older’s inequality) For any numbersp, q∈]1,∞[with1/p+1/q= 1,
(ii) (Minkowski’s inequality) For anyp∈[1,∞[,
Lemma 1.7.4 (Absolute convergence implies convergence) Let {x k } ∞ k=1 be a scalar sequence If ∞ k=1 |x k | is convergent, then ∞ k=1 x k is convergent, and
In this section we define some special functions that are used in the book.
Definition 1.8.1 (Characteristic function) Given a subsetE⊆R, let χ E (x)
The functionχ E is called the characteristic function for the set E.
Characteristic functions for intervals are essential in describing signals that occur within a specific time frame, such as in physical experiments with defined start and end times For instance, an electrical current can be represented by these functions, highlighting its behavior during a limited duration.
1.8 Some special functions 21 shape of a sine-function but only runs over a time interval of length 2π, might be described by a function of the type f(x) = sinx χ [0,2π] (x).
Definition 1.8.2 (Trigonometric polynomial) A trigonometric poly- nomial is a finite linear combination of complex exponential functions having period 1, i.e., an expression
H(x) N 2 k=N 1 c k e 2πikx (1.21) for someN 1 ≤N 2 and somec k ∈C.
Note that a trigonometric polynomial can be rewritten as
Trigonometric polynomials, represented by the expression (a k cos(2πkx) + b k sin(2πkx)), correspond to partial sums of Fourier series and can be analyzed with various periods beyond 1 For a given N, which is defined as N = max(|N1|, |N2|), the coefficients a k and b k belong to the complex numbers, as detailed in Exercise 1.20.
(a k cos(kx) +b k sin(kx)) defines a trigonometric polynomial with period 2π.
Example 1.8.3 (Trigonometric polynomials)The set of trigonometric polynomials with period 2πis a subspace of
Trigonometric polynomials are continuous functions, ensuring that the integral in (1.23) is finite for these functions This characteristic establishes that trigonometric polynomials constitute a subset of L²(−π, π) Furthermore, the sum of two trigonometric polynomials remains a trigonometric polynomial, and multiplying a trigonometric polynomial by a scalar also results in a trigonometric polynomial Consequently, as stated in Lemma 1.2.7, the collection of trigonometric polynomials forms a subspace of L²(−π, π).
1.9 A useful technique: proof by induction
Suppose that we want to prove that a certain statement, involving a number n, holds true for alln∈N This can be done by showing that
(i) The statement holds forn= 1, and
(ii) For an arbitraryn ∈N it holds that if the statement is true for n, then it is also true whennis replaced byn+ 1.
The induction step, known as Part (ii), relies on the induction hypothesis, which assumes that the statement is true for a specific value, n A proof by induction can be visualized as ascending an infinitely tall ladder; if we can successfully take the first step and ensure that we can progress from one level to the next, we can reach any height we desire.
Example 1.9.1 (Induction)We will prove that for all n∈N,
The statement certainly holds forn= 1 Now, assume that (1.24) holds for a certain value ofn, and let us consider the statement with nreplaced by n+ 1: we want to verify that
In order to do so, we use that the hypothesis holds fornto derive that
This completes the proof of the induction step
Let us complete this section with a result that will be used at several instances For, k∈N0 with≥k,define thebinomial coefficient k
Thebinomial formula states the following:
Lemma 1.9.2 (The binomial formula) For all y, z∈Rand any∈N,
A guide to a proof of Lemma1.9.2is given in Exercise1.26.
1.1 Show that if{ e k } n k=1 is an orthonormal basis forC n , then the representation (1.4) takes the form (1.5).
1.3 Consider a bounded interval [a, b]⊂R, and letC[a, b] denote the set of continuous functionsf : [a, b]→C, i.e.,
(i) Show how to define appropriate operations of addition and scalar multiplication such thatC[a, b] equipped with these operations become a vector space.
(ii) IsC[a, b] finite-dimensional? (Hint:for anyn∈N, x n ∈C[a, b].)
1.5 Give a geometric description (e.g., via a figure) of the set inR 2 given by
1.6 Letm∈Nbe given, and assume that we for each∈ {0,1, , m} have chosen a polynomialP of degree Show that the collection of polynomials{P } m =0 is a linearly independent set, and that span{P } m =0 = span{x } m =0
Can “infimum” be replaced by “minimum”?
1.8 Determine the following numbers, and decide in each case whether
“supremum” can be replaced by “maximum”:
1.9 Determine the following numbers, and decide in each case whether
“infimum” can be replaced by “minimum”:
1.10 Determine the following numbers, and decide in each case whether
“lim inf” resp “lim sup” can be replaced by “limes”:
1.11 This exercise is related to Lemma1.5.9and its hypotheses. (i) Prove Lemma1.5.9.
(ii) Find a sequence{c k } ∞ k=1 of real numbers that does not have an accumulation point.
1.12 The purpose of this exercise is to prove Lemma1.5.12 Under the conditions in Lemma1.5.12, prove the following:
(i) For eachK∈Nthere exists a numberm(K) such that lim inf n→∞ f n (k)≤f m (k) + 1
(ii) Withm(K) chosen as in (i), show that for allk= 1, , K and allm≥m(K),
(iii) Conclude that for allk= 1, , K,
(iv) Conclude the proof by lettingK→ ∞.
1.13 Prove Theorem1.6.3.Hint:take a sequence of numbers{x k } ∞ k=1 ⊂
Use Lemma1.5.10to select a convergent subsequence{x k n } ∞ n=1 of
1.14 Make a draft of the functionf in Example1.6.8, and argue that it is piecewise continuous Argue further thatf is unbounded and not uniformly continuous.
1.15 The purpose of this exercise is to prove Theorem1.6.6 Assume that the hypotheses are satisfied, and letx 0 ∈I and >0 be given.
(i) Argue that there exists anN ∈Nsuch that
(ii) Argue that we can chooseδ >0 such that
(iii) Use the triangle inequality to show that forx∈Iwith
1.16 The purpose of this exercise is to prove Theorem1.7.1(i).
(i) ProveYoung’s inequality:for anya, b >0 and anyp, q >1 with p −1 +q −1 = 1, ab≤ 1 pa p +1 qb q
Hint:consider thexy-plane, the graph of the functiony=x p−1 , and the linesx=a, y=b.
(ii) Prove Theorem1.7.1(i).Hint:put
1.17 The purpose of this exercise is to prove Theorem1.7.1(ii). (i) Prove Theorem1.7.1(ii) forp= 1.
We now assume thatp >1 Chooseq >1 such thatp −1 +q −1 = 1. (ii) Show thatq(p−1) =p.
, using the result in (ii) and H¨older’s inequality applied on each of the two terms.
(v) Complete the proof of Theorem1.7.1(ii) via (iv).
1.18 Prove Theorem1.7.3(i), e.g., by appropriate modifications of the proof of Theorem1.7.1(i) outlined in Exercise1.16.
1.19 Prove Theorem1.7.3(ii), e.g., by appropriate modifications of the proofs of Theorem1.7.1(ii) outlined in Exercise1.17.
1.20 Show that a trigonometric polynomial on the form (1.21) can be rewritten on the form (1.22).
1.21 Letx∈R\ {1}.Show by induction that for anyN ∈N,
1.23 Show that 7 n −4 n is a multiple of 3 for alln∈N.
1.25 Show that if 1 +x >0,then the inequality
1.26 The purpose of this exercise is to prove Lemma1.9.2.
(ii) Prove Lemma1.9.2by induction.
To effectively analyze vector spaces, it is essential to establish additional structure beyond the basic algebraic conditions outlined in Definition 1.2.1 This chapter focuses on the concept of norms in vector spaces and explores their various properties, with a detailed introduction to norms provided in Section 2.1.
In Section 2.2, the topological concepts from Section 1.4 are expanded to encompass general normed spaces Section 2.3 connects these concepts to dense subsets, illustrated by Weierstrass' theorem, which demonstrates the approximation of continuous functions using polynomials Section 2.4 provides a brief overview of operators in normed vector spaces, while Section 2.5 focuses on expansions in normed spaces through the use of bases.