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Home , Normed vector space, Norm (mathematics), Triangle inequality

Chapter II Elements of

Functional Analysis is generally understood a “ for infinite di- mensional vector spaces.” Most of the vector spaces that are used are spaces of (various types of) functions, therfeore the name “functional.” This chapter in- troduces the reader to some very basic results in Functional Analysis. The more motivated reader is suggested to continue with more in depth texts.

1. Normed vector spaces

Definition. Let K be one of the fields R or C, and let X be a K-vector . A on X is a map X 3 x 7−→ kxk ∈ [0, ∞) with the following properties (i) kx + yk ≤ kxk + kyk, ∀ x, y ∈ X; (ii) kλxk = |λ| · kxk, ∀ x ∈ X, λ ∈ K; (iii) kxk = 0 =⇒ x = 0. (Note that conditions (i) and (ii) state that k . k is a .)

Examples 1.1. Let K be either R or C, and let J be some non-empty set. A. Define `∞(J) = α : J → : sup |α(j)| < ∞ . K K j∈J We equip the space `∞(J) with the - structure defined by point-wise K K addition and point-wise multiplication. This vector space carries a natural norm k . k∞, defined by kαk = sup |α(j)|, α ∈ `∞(I). ∞ K j∈J When = , the space `∞(J) is simply denoted by `∞(J). When J = - the set K C C N of natural numbers - instead of `∞( ) we simply write `∞, and instead of `∞( ) R N R N we simply write `∞. B. Consider the space   K   c0 (J) = α : J → K : inf sup |α(j)| = 0 . F ⊂J j∈J F finite r

K Remark that, if we equip J with the discrete , then c0 (J) is the space of those (automatically continuous) functions : J → K which have limit 0 at ∞. In particular, cK(J) is a linear subspace of `∞(J). We also equip cK(J) with the norm 0 K 0 k . k∞. As above, when K = C, we ommit it from the notation, and when J = N, we also ommit it.

1 2 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Definition. Suppose X is a , with norm k . k. Then there is a natural d on X, defined by d(x, y) = kx − yk, x, y ∈ X. The toplogy on X, defined by this metric, is called the norm topology.

Exercise 1 ♦. Let X be a normed vector space, over K(= R, C). Prove that, when equipped with the norm toplogy, X becomes a . That is, the maps X × X 3 (x, y) 7−→ x + y ∈ X K × X 3 (λ, x) 7−→ λx ∈ X are continuous.

Exercise 2 ♦. Let K be one of the fields R or C, and let J be a non-empty set. A. Prove that cK(J) is a closed linear subspace in `∞(J). 0 K B. Recall (see I.2) that for a α : J → K we denoted by [[α]] the of α, defined as [[α]] = {j ∈ J : α(j) 6= 0}. Consider the space fin (J) = α : J → : [[α]] finite . K K Prove that fin (J) is a linear subspace in cK(J), which is dense in the K 0 norm topology. The next example is based on the notion of summability discussed in I.2.

Example 1.2. Let K be either R or C, let J be a non-empty set, and let p ∈ [1, ∞) be a . We define `p (J) = α : J → : |α|p : J → [0, ∞) summable . K K Remark that fin (I) ⊂ `q (J). K K For α ∈ `p (J) we define K 1   p X p kαkp = |α(j)| . j∈J When = , the space `p (J) is simply denoted by `p(J). When J = - K C C N the set of natural numbers - instead of `p ( ) we simply write `p , and instead of R N R `p(N) we simply write `p. The proof of the fact that the `p spaces (1 ≤ p < ∞) are normed vector spaces is included in the following exercises. To make the formulation a little easier, we use the following terminology. Definition. Two “numbers” p, q ∈ [1, ∞] are said to be H¨olderconjugate, if 1 1 1 p + q = 1. Here we use the convention ∞ = 0.

Exercise 3 ♦. Let K be one of the fields R or C, let J be a non-empty set, and let p, q ∈ [1, ∞] be two H¨older conjugate “numbers.” Consider the set q  B (J) = β ∈ fin (J): kβkq ≤ 1 , K K §1. Normed vector spaces 3 and define, for functions α : J → and β ∈ fin (J), the quantity K K X hα, βi = α(j)β(j). j∈[[β]] Prove that, for α : J → K, the following are equivalent: (i) α ∈ `p (J); K (ii) sup |hα, βi| < ∞. β∈Bq (J) K Moreover, one has the equality p sup |hα, βi| = kαkp, ∀ α ∈ ` (J). K β∈Bq (J) K

Hint: Use H¨older (Appendix D).

Exercise 4 ♦. Let K be either R or C, let J be a non-empty set, and let p ∈ [1, ∞). Prove that, when equipped with pointwise addition and scalar mul- p tiplication, then (` (J), k .kp) is a normed vector space. K Hint: Use the preceding exercise.

Exercise 5 ♦. Let K be one of the fields R or C, let J be a non-empty set, and let p, q ∈ [1, ∞] be two H¨older conjugate “numbers.” A. If α ∈ `p (J) and β ∈ `q (J), then αβ ∈ `1 (J), and K K K kαβk1 ≤ kαkp · kβkq. B. Using part A, extend the pairing constructed in Exercise 3 to a h .,. i : `p (J) × `q (J) → , defined by K K K X hα, βi = α(j)β(j), ∀ α ∈ `p (J), β ∈ `q (J). K K j∈J C. Prove that, for every α ∈ `p (J), one has the equality K q kαkp = sup {|hα, βi| : β ∈ ` (J), kβkq ≤ 1} . K Exercise 6 ♦. Let K be one of the fields R or C, and let J be a non-empty set. A. Prove that fin (J) is a dense linear subspace in `p (J), for every p ∈ [1, ∞). K K B. Why was the case p = ∞ ommitted? We now examine linear continuous maps between normed spaces.

Proposition 1.1. Let K be either R or C, let X and Y be normed K-vector spaces, and let T : X → Y be a K-. The following are equivalent: (i) T is continuous; (ii) sup kT xk : x ∈ X, kxk ≤ 1 < ∞; (iii) sup kT xk : x ∈ X, kxk = 1 < ∞; (iv) T is continuous at 0. Proof. (i) ⇒ (ii). Assume T is continuous, but sup kT xk : x ∈ X, kxk ≤ 1 < ∞, which means there exists some sequence (xn)n≥1 ⊂ X such that

(a) kxnk ≤ 1, ∀ n ≥ 1; (b) limn→∞ kT xnk = ∞. 4 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Put −1 zn = kT xnk xn, ∀ n ≥ 1. On the one hand, we have

kxnk 1 kznk = ≤ , ∀ n ≥ 1, kT xnk kT xnk which gives limn→∞ kznk = 0, i.e. limn→∞ zn = 0. Since T is assumed to be continuous, we will get

(1) lim T zn = T 0 = 0. n→∞ −1 On the other hand, since T is linear, we have T zn = kT xnk T xn, so in particular we get kT znk = 1, ∀ n ≥ 1, which clearly contradicts (1). (ii) ⇒ (iii). This is obvious, since the supremum in (iii) is taken over a subset of the set used in (ii). (iii) ⇒ (iv). Let (xn)n≥1 ⊂ X be a with limn→∞ xn = 0. For each n ≥ 1, define  −1  kxnk xn, if xn 6= 0 un =  any vector of norm 1, if xn = 0 so that we have kunk = 1 and xn = kxnkun, ∀ n ≥ 1. Since T is linear, we have

(2) T xn = kxnkT un, ∀ n ≥ 1.  If we define M = sup kT xk : x ∈ X, kxk = 1 , then kT unk ≤ M, ∀ n ≥ 1, so (2) will give kT xnk ≤ M · kxnk, ∀ n ≥ 1, and the condition limn→∞ xn = 0 will force limn→∞ T xn = 0. (iv) ⇒ (i). Assume T is continuous at 0, and let us prove that T is continuous at any point. Start with some arbitrary x ∈ X and an arbitrary sequence (xn)n≥1 ⊂ X with limn→∞ xn = x. Put zn = xn − x, so that limn→∞ zn = 0. Then we will have limn→∞ T zn = 0, which (use the linearity of T ) means that

0 = lim kT znk = lim kT xn − T xk, n→∞ n→∞ thus proving that limn→∞ T xn = T x.  Remark 1.1. Using the notations above, the quantities in (ii) and (iii) are in fact equal. Indeed, if we define  M1 = sup kT xk : x ∈ X, kxk ≤ 1 ,  M2 = sup kT xk : x ∈ X, kxk = 1 , then as observed during the proof, we have M2 ≤ M1. Conversely, if we start with some arbitrary x ∈ X with kxk ≤ 1, then we can always write x = kxku, for some u ∈ X with kuk = 1. In particular we will get

kT xk = kxk · kT uk ≤ kxk · M2 ≤ M2. §1. Normed vector spaces 5

Taking supremum in the above inequality, over all x ∈ X with kxk ≤ 1, will then give the inequality M1 ≤ M2. Notations. Let K be either R or C, and let X and Y be normed K-vector spaces. We define  L(X, Y) = T : X → Y : T K-linear and continuous . For T ∈ L(X, Y) we define (see the above remark) kT k = sup kT xk : x ∈ X, kxk ≤ 1 = sup kT xk : x ∈ X, kxk = 1 When Y = K (equipped with the as the norm), the space L(X, K) will be denoted simply by X∗, and will be called the topological dual of X.

Proposition 1.2. Let K be either R or C, and let X and Y be normed K-vector spaces. (i) The space L(X, Y) is a K-vector space. (ii) For T ∈ L(X, Y) we have (3) kT k = min C ≥ 0 : kT xk ≤ Ckxk, ∀ x ∈ X . In particular one has (4) kT xk ≤ kT k · kxk, ∀ x ∈ X. (iii) The map L(X, Y) 3 T 7−→ kT k ∈ [0, ∞) is a norm. Proof. The fact that L(X, Y) is a vector space is clear. (ii). Assume T ∈ L(X, Y). We begin by proving (4). Start with some arbitrary x ∈ X, and write it as x = kxku, for some u ∈ X with kuk = 1. Then by definition we have kT uk ≤ kT k, and by linearity we have kT xk = kxk · kT uk ≤ kxk · kT k. To prove the equality (3) let us define the set  CT = C ≥ 0 : kT xk ≤ Ckxk, ∀ x ∈ X .

On the one hand, by (4) we know that kT k ∈ CT . On the other hand, if we take an arbitrary C ∈ CT , then for every u ∈ X with kuk = 1, we will have kT uk ≤ Ckuk = C, so taking supremum, over all u with kuk = 1, will immediately give kT k ≤ C. Since we now have kT k ≤ C, ∀ C ∈ CT , we clearly get kT k = min CT . (iii). Let T,S ∈ L(X, Y). Using (4), we have k(T + S)xk = kT x + Sxk ≤ kT xk + kSxk ≤ (kT k + kSk) · kxk, ∀ x ∈ X. Then using (3) we get kT + Sk ≤ kT k + kSk. If T ∈ L(X, Y) and λ ∈ K, then the equality k(λT )xk = |λ| · kT xk, x ∈ X will immediately give kλT k = |λ| · kT k. Finally if T ∈ L(X, Y) has kT k = 0, then using (4) one immediately gets T = 0.  6 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

The inequality (4) is referred to as the norm inequality. Exercise 7 ♦. Let J be some non-empty set, let K be either R or C. Prove that for 1 ≤ p < q < ∞ one has the inclusions `p (J) ⊂ `q (J) ⊂ cK(J). Moreover, prove K K 0 that the inclusion maps are linear and continuous. Below is an interesting application of Proposition 1.2. Proposition 1.3. Let X be a vector space, and let k . k and k . k0 be two norms on X. The following are equivalent: (i) the norm defined by k . k and k . k0 coincide; (ii) there exist constants C, D > 0 such that Ckxk ≤ kxk0 ≤ Dkxk, ∀ x ∈ X. Proof. Consider the identity maps T :(X, k . k) 3 x 7−→ x ∈ (X, k . k0) T 0 :(X, k . k0) 3 x 7−→ x ∈ (X, k . k) Remark that condition (i) is equivalent to the fact that both T and T 0 are contin- uous. In turn, this is equivalent to the existence of two numbers D,D0 > 0, such that kxk0 ≤ Dkxk and kxk ≤ D0kxk0, ∀ x ∈ X. 0 This last condition is obviously equivalent to (ii) by taking C = 1/D .  Definition. Two norms k . k and k . k0 satisfying the above conditions are said to be equivalent. Proposition 1.4. On a finite dimensional vector space, any two norms are equivalent. Proof. We can assume that we work on the vector space Kn. Start with some n (arbitrary) norm k . k on K , and let us compare with the norm k . k∞, defined by  k(α1, . . . , αn)k∞ = max |α1|,..., |αn| . n Let {e1,..., en} be the standard for K , i.e. ek = (δ1k, δ2k, . . . , δnk), k = n 1, . . . , n, so that any vector v = (α1, . . . , α) ∈ K is written as

v = α1e1 + ··· + αnen. Remark that using the norm inequality, this presentation gives

kvk ≤ |α1| · ke1k + ··· + |αn| · kenk.

Then if we put C = ke1k + ··· + kenk, this proves that we have the inequality n (5) kvk ≤ Ckvk∞, ∀ v ∈ K . Among other things, this proves that the identity map n n (6) Id : (K , k . k∞) → (K , k . k) is continuous.  n Consider now the set T = v ∈ K : kvk∞ = 1 , which is obviously compact n in (K , k . k∞). Using the continuity of (6), it follows that T is also compact in (Kn, k . k). In particular, since 0 6∈ T , it follows that when we consider the metric defined by k . k on Kn, the number D = dist(T, 0) = inf{kvk : v ∈ T } §1. Normed vector spaces 7 is strictly positive. We claim that we have the inequality n (7) kvk ≥ D · kvk∞, ∀ v ∈ K .

This is trivial if v = 0. If v 6= 0, we can write it as v = αv0, with α = kvk∞, and 1 kv0k = 1, by taking v0 = v. Since v0 ∈ T , we have kv0k ≥ D, so we have kvk∞

kvk = |α| · kv0k ≥ D · |α| = D · kvk∞.

Now (7) and (5) show that k . k and k . k∞ are indeed equivalent norms.  The following discussion is based on Exercise 5.

Example 1.3. Let J be a non-empty set, let K be one of the fields R or C, and let p, q ∈ [1, ∞] be the H¨older conjugate. For every element α ∈ `p (J) we define q K the map θα : ` (J) → by K K X q θα(β) = hα, βi = α(j)β(j), β ∈ ` (J). K j∈J

We know that θα is linear, and by Exercise 5, we have q θα(β) ≤ kαkp · kβkq, ∀ β ∈ ` (J), K so θα is continuous, and in fact (again by Exercise 5) we have the inequality kθαk = kαkp. This gives rise to a linear map p q ∗ (8) Θ : ` (J) 3 α 7−→ θα ∈ ` (J) , K K which is isometric, in the sense that p (9) kΘαk = kαkp, ∀ α ∈ ` (J). K In particular, Θ is injective. Exercise 8 ♦. Using the above notations, but assuming p ∈ (1, ∞], the map (8) is surjective. Exercise 9*. Prove that, in the case p = 1, the map ∗ Θ: `1 (J) → `∞(J) K K is not surjective, unless J is finite. Hints: Assume J is infinite. If we take 1 ∈ `∞(J) to be the constant function 1, then show that K kλ1 + βk ≥ |λ|, ∀ λ ∈ , β ∈ fin (J). K K Consider the subspace fin (J) = {λ1 + β : β ∈ fin (J), λ ∈ }, fK K K and define the map φ : fin (J) 3 λ1 + β 7−→ λ ∈ . 0 fK K Then φ0 is linear, continuous, and has the property that

(10) φ0 fin (I) = 0, φ0(1) = 1, K (11) |φ (γ)| ≤ kγk, ∀ γ ∈ fin (I). 0 fK

Using the Hahn-Banach Theorem (Appendix E), we can then extend φ0 to a linear map φ : `∞(J) → which will still satisfy (10) and (11), in particular we have φ ∈ `∞(J)∗. Notice K K K 1 however that if we had φ = θα, for some α ∈ ` (J), then this would force hα, βi = 0, ∀ β ∈ fin (J), K K which would force α = 0, i.e. φ = 0. This is impossible, since φ(1) = 1. 8 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Exercise 10 ♦. Use the notations above. For every α ∈ `1 (I), define K σ = θ : cK(I) → . α α K 0 K c0 (I)

Prove that σα is linear and continuous. Prove that the map ∗ Σ: `1 (I) 3 α 7−→ σ ∈ cK(I) K α 0 is an isometric linear of K-vector spaces. We conclude with an important construction. Definition. Let X be a normed vector space, and let Y be a closed linear subspace. Consider the quotient vector space X/Y. Recall that X/Y is defined as the space of equivalence classes corresponding to the relation

x1 ∼ x2 ⇐⇒ x1 − x2 ∈ Y. The equivalence class of an element x ∈ X is the set [x] = x + Y = {x + y : y ∈ Y}. The vector space structure on X/Y is then the unique one which makes the quotient map X 3 x 7−→ [x] ∈ X/Y linear. Define, for every equivalence class v ∈ X/Y, the quantity  kvkX/Y = inf kxk : x ∈ v . Proposition 1.5. Let X and Y be as above.

(i) The map k . kX/Y : X/Y → [0, ∞), constructed above, defines a norm. (ii) The linear map Q : X 3 x 7−→ [x] ∈ X/Y is continuous, and has kQk ≤ 1.

Proof. (i). To prove the inequality, start with two elements v1, v2 ∈ ε ε X/Y. For every ε > 0, we choose x1 ∈ v1 and x2 ∈ v2, such that ε kxkk ≤ ε + kvkkX/Y, k = 1, 2. ε ε Since x1 + x2 ∈ v1 + v2, we get ε ε ε ε kv1 + v2kX/Y ≤ kx1 + x2k ≤ kx1k + kx2k ≤ 2ε + kv1kX/Y + kv2kX/Y.

Since the inequality kv1 + v2kX/Y ≤ 2ε + kv1kX/Y + kv2kX/Y holds for all ε > 0, it will force the inequality

kv1 + v2kX/Y ≤ kv1kX/Y + kv2kX/Y. To prove the second condition in the definition of the norm, we start with some arbitrary v ∈ X/Y, and some λ ∈ K. On the one hand, for each ε > 0, there exists some xε ∈ v, such that kxεk ≤ ε + kvkX/Y. Since λxε ∈ λv, it follows that  kλvkX/Y ≤ kλxεk = |λ| · kxεk ≤ |λ| · ε + kvkX/Y .  Since the inequality kλvkX/Y ≤ |λ| · ε + kvkX/Y holds for every ε > 0, it will force

kλvkX/Y ≤ |λ| · kvkX/Y. §1. Normed vector spaces 9

It λ = 0 this already forces the equality kλvkX/Y = |λ| · kvkX/Y. If λ 6= 0, we use the above inequality for 1/λ to get

|λ| · kvkX/Y = |λ| · k(1/λ)(λv)kX/Y ≤ |λ| · |1/λ| · kλvkX/Y = kλvkX/Y. Finally, let us check the third condition. Start with some v ∈ X/Y, with kvkX/Y = 0, and let us show that v = [0]. Fix some element x ∈ v. By assumption, ∞ it follows that there exists a sequence (xn)n=1 in v such that limn→∞ kxnk = 0. On the one hand, this gives the equality

x = lim (x − xn). n→∞

On the other hand, since xn ∈ v = [x], we get x ∼ xn, i.e. x − xn ∈ Y, ∀ n ≥ 1. Since Y is closed, the above equality will force x ∈ Y, which in turn will force x ∼ 0, hence v = [0]. (ii). This part is trivial, since the relation x ∈ [x] immediately gives k[x]kX/Y ≤ kxk, ∀ x ∈ X. 

Definition. The norm k . kX/Y, defined above, is called the quotient norm. Exercise 11*. With the notations above, prove the following. A. For a subset D ⊂ X/Y, the following are equivalent: (i) D is open in X/Y; (ii) D is open in X. B. The quotient map Q : X → X/Y is an open map, i.e. it has the property: A open in X =⇒ Q(A) open in X/Y.

Hint: For part B, use part A, as well as the implication A open =⇒ A + Y open. Proposition 1.6 (Factorization Theorem for linear continuous maps). Let X and Y be normed vector spaces, let T : X → Y be a linear continuous map. (i) The linear subspace N = Ker T is closed. (ii) There exists a unique linear continuous map Tˆ : X/N → Y such that T = Tˆ ◦ Q, where Q : X → X/N is the quotient map. (iii) One has kTˆk = kT k. Proof. (i). This property is trivial, since N = T −1({0}), and {0} is closed. (ii). The existence and uniqueness of a linear map Tˆ : X/N → Y, with T = Tˆ◦Q is clear (this is the usual Factorization Theorem for linear maps), so the only issue here is continuity. Start with some v ∈ X/Y. For every ε > 0, choose xε ∈ v such ˆ ˆ that kxεk ≤ kvkX/Y + ε. Since we have T xε = T [x] = T v, we get ˆ kT vk = kT xεk ≤ kT k · kxεk ≤ kT k · (kvkX/Y + ε). ˆ Since the inequality kT vk ≤ kT k · (kvkX/Y + ε) holds for all ε > 0, it will give ˆ kT vk ≤ kT k · kvkX/Y, which among other things proves that Tˆ is indeed continuous, and we also have the inequality kTˆk ≤ kT k. (iii). By the above estimate we only need to show the inequality kT k ≤ kTˆk. But this is clear, since the inequality kP k ≤ 1 gives kT xk = kTˆ(Qx)k ≤ kTˆk · kQxk ≤ kTˆk · kxk, ∀ x ∈ X. 10 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

 The following is an interesting application of quotient norms.

Proposition 1.7. Let K be either R or C, and let X be a normed K-vector space. For a linear map φ : X → K, the following are equivalent. (i) φ is continuous; (ii) Ker φ is a closed linear subspace of X. Proof. The implication (i) ⇒ (ii) is trivial. To prove the converse, denote for simplicity the kernel of φ simply by N, we assume N is closed, and we prove that φ is continuous. Of course, if N = X, we have φ = 0, and there is nothing to prove, so for the remainder of the proof we can assume that N ( X. Consider the quotient space X/N and the quotient map Q : X 3 x 7−→ [x] ∈ X/N

We know that, when we equip X/N with the quotient norm k . kX/N, the linear map P is continuous. By the Factorization Theorem (algebraic) version, there exists a linear map ψ : X/N → K, such that φ = ψ ◦ Q, so that all we must show is the fact that the linear map

ψ :(X/N, k . kX/N) → K is continuous. On the one hand, ψ is a linear isomorphism. In particular, the map X/N 3 v 7−→ |ψ(v)| ∈ [0, ∞) defines a norm on X/N. On the other hand, X/N is one dimensional, which means that any two norms on X/N are proportional. In our case this means that there exists a constant C > 0, such that

|ψ(v)| = CkvkX/N, ∀ v ∈ X/N, so ψ is indeed continuous.  The following is one generalization of the above result. Exercise 12. Let X and Y be normed vector spaces, with Y finite dimensional, and let T : X → Y be a linear map. Prove that the following are equivalent: (i) T is continuous; (ii) Ker T is a closed linear subspace of X.

Hint: Follow the proof above, but at one point use Exercise 8. The following is a generalization of Proposition 1.6 to the most general setting.

Exercise 13*. Let X be a topological vector space, and let φ : X → K be a linear map. Prove that the following are equivalent: (i) φ is continuous; (ii) Ker φ is a closed linear subspace of X. Hint: Prove the following intermediary steps. (Assume φ is not identically 0.) Step I: If (xλ)λ∈Λ ⊂ X is a net with limλ∈Λ xλ = 0, and if (αλ)λ∈Λ ⊂ K is bounded, then

lim αλxλ = 0. λ∈Λ §1. Normed vector spaces 11

Step II: If φ is not continuous, there exists a net (xλ)λ∈Λ ⊂ X, and some ε > 0, such that

lim xλ = 0 and |φ(xλ)| ≥ ε, ∀ λ ∈ Λ. λ∈Λ Use Step I to improve Step II to the following Step III: If φ is not continuous, there exists a net (xλ)λ∈Λ ⊂ X with

lim xλ = 0 and φ(xλ) = 1, ∀ λ ∈ Λ. λ∈Λ Conclude that if φ is not continuous, then Ker φ cannot be closed, arguing as follows. Choose some x ∈ X with φ(x) = 1. Use the net (xλ)λ∈Λ given by Step III, and consider the net (yλ)λ∈Λ given by yλ = x − xλ.