Chapter II Elements of Functional Analysis Functional Analysis is generally understood a “linear algebra for infinite dimensional vector spaces.” Most of the vector spaces that are used are spaces of (various types of) functions, therfeore the name “functional.” This chapter introduces 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 space. A norm 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 seminorm.) Examples 1.1. Let K be either R or C, and let J be some non-empty set. A. Define  `∞ K (J) = α : J → K : sup |α(j)| < ∞ . j∈J

`∞ K (J)

We equip the space with the K-vector space structure defined by point-wise addition and point-wise scalar multiplication. This vector space carries a natural norm k . k∞ , defined by kαk∞ = sup |α(j)|, α ∈ `∞ K (I). j∈J

`∞ C (J)

When K = C, the space is simply denoted by `∞ (J). When J = N - the set ∞ ∞ of natural numbers - instead of `∞ R (N) we simply write `R , and instead of ` (N) ∞ we simply write ` . B. Consider the space     cK (J) = α : J → K : inf sup |α(j)| = 0 . 0 F ⊂J finite

j∈JrF

Remark that, if we equip J with the discrete topology, then cK 0 (J) is the space of those (automatically continuous) functions : J → K which have limit 0 at ∞. In ∞ K particular, cK 0 (J) is a linear subspace of `K (J). We also equip c0 (J) with the norm k . k∞ . As above, when K = C, we ommit it from the notation, and when J = N, we also ommit it. 1

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Definition. Suppose X is a normed vector space, with norm k . k. Then there is a natural metric 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 topological vector space. 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 0 (J) is a closed linear subspace in `K (J). B. Recall (see I.2) that for a function α : J → K we denoted by [[α]] the support of α, defined as

[[α]] = {j ∈ J : α(j) 6= 0}. Consider the space  fin K (J) = α : J → K : [[α]] finite . Prove that fin K (J) is a linear subspace in cK 0 (J), which is dense in the 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 real number. We define  `pK (J) = α : J → K : |α|p : J → [0, ∞) summable . Remark that finK (I) ⊂ `qK (J). For α ∈ `pK (J) we define kαkp =

X

p

|α(j)|

 p1 .

j∈J

When K = C, the space `pC (J) is simply denoted by `p (J). When J = N the set of natural numbers - instead of `pR (N) we simply write `pR , and instead of `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. 1 p

+

Definition. Two “numbers” p, q ∈ [1, ∞] are said to be H¨ older conjugate, if 1 1 = 1. Here we use the convention = 0. q ∞

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  BqK (J) = β ∈ finK (J) : kβkq ≤ 1 ,

§1. Normed vector spaces

3

and define, for functions α : J → K and β ∈ finK (J), the quantity X hα, βi = α(j)β(j). j∈[[β]]

Prove that, for α : J → K, the following are equivalent: (i) α ∈ `pK (J); (ii) sup |hα, βi| < ∞. β∈BqK (J)

Moreover, one has the equality sup β∈BqK (J)

|hα, βi| = kαkp , ∀ α ∈ `pK (J).

Hint: Use H¨ older inequality (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 multiplication, then (`pK (J), k .kp ) is a normed vector space. 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.” p A. If α ∈ `K (J) and β ∈ `qK (J), then αβ ∈ `1K (J), and kαβk1 ≤ kαkp · kβkq . B. Using part A, extend the pairing constructed in Exercise 3 to a bilinear map h . , . i : `pK (J) × `qK (J) → K, defined by X hα, βi = α(j)β(j), ∀ α ∈ `pK (J), β ∈ `qK (J). j∈J

C. Prove that, for every α ∈ `pK (J), one has the equality kαkp = sup {|hα, βi| : β ∈ `qK (J), kβkq ≤ 1} . Exercise 6 ♦ . Let K be one of the fields R or C, and let J be a non-empty set. A. Prove that finK (J) is a dense linear subspace in `pK (J), for every p ∈ [1, ∞). 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-linear map. The following are equivalent: (i) T iscontinuous; (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) kxn k ≤ 1, ∀ n ≥ 1; (b) limn→∞ kT xn k = ∞.

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

kxn k 1 ≤ , ∀ n ≥ 1, kT xn k kT xn k

which gives limn→∞ kzn k = 0, i.e. limn→∞ zn = 0. Since T is assumed to be continuous, we will get (1)

lim T zn = T 0 = 0.

n→∞

On the other hand, since T is linear, we have T zn = kT xn k−1 T xn , so in particular we get kT zn k = 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 sequence with limn→∞ xn = 0. For each n ≥ 1, define   kxn k−1 xn , if xn 6= 0 un =  any vector of norm 1, if xn = 0 so that we have kun k = 1 and xn = kxn kun , ∀ n ≥ 1. Since T is linear, we have T xn = kxn kT un , ∀ n ≥ 1. If we define M = sup kT xk : x ∈ X, kxk = 1 , then kT un k ≤ M , ∀ n ≥ 1, so (2) will give kT xn k ≤ M · kxn k, ∀ n ≥ 1, (2)



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 zn k = 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 absolute value 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 kT xk ≤ kT k · kxk, ∀ x ∈ X.

(4)

(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. 

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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 `pK (J) ⊂ `qK (J) ⊂ cK 0 (J). Moreover, prove 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 topologies 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 continuous. In turn, this is equivalent to the existence of two numbers D, D0 > 0, such that kxk0 ≤ Dkxk and kxk ≤ D0 kxk0 , ∀ x ∈ X. This last condition is obviously equivalent to (ii) by taking C = 1/D0 .  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 (arbitrary) norm k . k on Kn , and let us compare with the norm k . k∞ , defined by  k(α1 , . . . , αn )k∞ = max |α1 |, . . . , |αn | . Let {e1 , . . . , en } be the standard basis for Kn , i.e. ek = (δ1k , δ2k , . . . , δnk ), k = 1, . . . , n, so that any vector v = (α1 , . . . , α) ∈ Kn is written as v = α1 e1 + · · · + αn en . Remark that using the norm inequality, this presentation gives kvk ≤ |α1 | · ke1 k + · · · + |αn | · ken k. Then if we put C = ke1 k + · · · + ken k, this proves that we have the inequality (5)

kvk ≤ Ckvk∞ , ∀ v ∈ Kn .

Among other things, this proves that the identity map (6)

Id : (Kn , k . k∞ ) → (Kn , k . k)

is continuous.  Consider now the set T = v ∈ Kn : kvk∞ = 1 , which is obviously compact in (Kn , 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 kvk ≥ D · kvk∞ , ∀ v ∈ Kn .

(7)

This is trivial if v = 0. If v 6= 0, we can write it as v = αv 0 , with α = kvk∞ , and kv 0 k = 1, by taking v 0 = kvk1 ∞ v. Since v 0 ∈ T , we have kv 0 k ≥ D, so we have kvk = |α| · kv 0 k ≥ 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 ∈ `pK (J) we define q the map θα : `K (J) → K by X θα (β) = hα, βi = α(j)β(j), β ∈ `qK (J). j∈J

We know that θα is linear, and by Exercise 5, we have θα (β) ≤ kαkp · kβkq , ∀ β ∈ `q (J), K so θα is continuous, and in fact (again by Exercise 5) we have the inequality kθα k = kαkp . This gives rise to a linear map
∗ (8) Θ : `pK (J) 3 α 7−→ θα ∈ `qK (J) , which is isometric, in the sense that kΘαk = kαkp , ∀ α ∈ `pK (J).

(9)

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 : `1K (J) → `∞ K (J) is not surjective, unless J is finite. Hints: Assume J is infinite. If we take 1 ∈ `∞ K (J) to be the constant function 1, then show that kλ1 + βk ≥ |λ|, ∀ λ ∈ K, β ∈ finK (J). Consider the subspace f (J) = {λ1 + β : β ∈ fin (J), λ ∈ K}, fin K K and define the map f (J) 3 λ1 + β 7−→ λ ∈ K. φ0 : fin K Then φ0 is linear, continuous, and has the property that (10) φ0 fin (I) = 0, φ0 (1) = 1, K

(11)

f (I). |φ0 (γ)| ≤ kγk, ∀ γ ∈ fin K

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

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CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

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

Prove that σα is linear and continuous. Prove that the map
∗ Σ : `1K (I) 3 α 7−→ σα ∈ cK 0 (I) is an isometric linear isomorphism 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 triangle inequality, start with two elements v 1 , v 2 ∈ X/Y. For every ε > 0, we choose xε1 ∈ v 1 and xε2 ∈ v 2 , such that kxεk k ≤ ε + kv k kX/Y , k = 1, 2. Since xε1 + xε2 ∈ v 1 + v 2 , we get kv 1 + v 2 kX/Y ≤ kxε1 + xε2 k ≤ kxε1 k + kxε2 k ≤ 2ε + kv 1 kX/Y + kv 2 kX/Y . Since the inequality kv 1 + v 2 kX/Y ≤ 2ε + kv 1 kX/Y + kv 2 kX/Y holds for all ε > 0, it will force the inequality kv 1 + v 2 kX/Y ≤ kv 1 kX/Y + kv 2 kX/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 kxε k ≤ ε + kvkX/Y . Since λxε ∈ λv, it follows that
kλvkX/Y ≤ kλxε k = |λ| · kxε k ≤ |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→∞ kxn k = 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 that kxε k ≤ kvkX/Y + ε. Since we have T xε = Tˆ[x] = Tˆv, we get kTˆvk = kT xε k ≤ kT k · kxε k ≤ kT k · (kvkX/Y + ε). Since the inequality kTˆvk ≤ kT k · (kvkX/Y + ε) holds for all ε > 0, it will give kTˆvk ≤ 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.

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 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 αλ xλ = 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 .