简明赋范线性空间与泛函分析讲义（英文版）

Course 212: Hilary Term 2001 Part III: Normed Vector Spaces and Functional Analysis D. R. Wilkins Copyright c© David R. Wilkins 1997–2001 Contents 9 Normed Vector Spaces 2 9.1 Bounded Linear Transformations . . . . . . . . . . . . . . . . 7 9.2 The Equivalence of Norms on a Finite-Dimensional Vector Space 11 10 Introduction to Functional Analysis 14 10.1 The Contraction Mapping Theorem and Picard’s Theorem . . 16 1 9 Normed Vector Spaces A set X is a vector space over some field F if • given any x, y ∈ X and λ ∈ F, there are well-defined elements x + y and λx of X, • X is an Abelian group with respect to the operation + of addition, • the identities λ(x+ y) = λx+ λy, (λ+ µ)x = λx+ µx, (λµ)x = λ(µx), 1x = x are satisfied for all x, y ∈ X and λ, µ ∈ F. Elements of the field F are referred to as scalars. We consider here only real vector spaces and complex vector spaces : these are vector spaces over the fields of real numbers and complex numbers respectively. Definition A norm ‖.‖ on a real or complex vector space X is a function, associating to each element x of X a corresponding real number ‖x‖, such that the following conditions are satisfied:— (i) ‖x‖ ≥ 0 for all x ∈ X, (ii) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X, (iii) ‖λx‖ = |λ| ‖x‖ for all x ∈ X and for all scalars λ, (iv) ‖x‖ = 0 if and only if x = 0. A normed vector space (X, ‖.‖) consists of a a real or complex vector space X, together with a norm ‖.‖ on X. Note that any normed complex vector space can also be regarded as a normed real vector space. Example The field R is a one-dimensional normed vector space over itself: the norm |t| of t ∈ R is the absolute value of t. Example The field C is a one-dimensional normed vector space over itself: the norm |z| of z ∈ C is the modulus of z. The field C is also a two- dimensional normed vector space over R. 2 Example Let ‖.‖1, ‖.‖2 and ‖.‖∞ be the real-valued functions on Cn defined by ‖z‖1 = n∑ j=1 |zj|, ‖z‖2 = ( n∑ j=1 |zj|2 ) 1 2 , ‖z‖∞ = max(|z1|, |z2|, . . . , |zn|), for each z ∈ Cn, where z = (z1, z2, . . . , zn). Then ‖.‖1, ‖.‖2 and ‖.‖∞ are norms on Cn. In particular, if we regard Cn as a 2n-dimensional real vector space naturally isomorphic to R2n (via the isomorphism (z1, z2, . . . , zn) 7→ (x1, y1, x2, y2, . . . , xn, yn), where xj and yj are the real and imaginary parts of zj for j = 1, 2, . . . , n) then ‖.‖2 represents the Euclidean norm on this space. The inequality ‖z+w‖2 ≤ ‖z‖2 +‖w‖2 satisfied for all z,w ∈ Cn is therefore just the standard Triangle Inequality for the Euclidean norm. Example The space Rn is also an n-dimensional real normed vector space with respect to the norms ‖.‖1, ‖.‖2 and ‖.‖∞ defined above. Note that ‖.‖2 is the standard Euclidean norm on Rn. Example Let `1 = {(z1, z2, z3, . . .) ∈ C∞ : |z1|+ |z2|+ |z3|+ · · · converges}, `2 = {(z1, z2, z3, . . .) ∈ C∞ : |z1|2 + |z2|2 + |z3|2 + · · · converges}, `∞ = {(z1, z2, z3, . . .) ∈ C∞ : the sequence |z1|, |z2|, |z3|, . . . is bounded}. where C∞ denotes the set of all sequences (z1, z2, z3, . . .) of complex numbers. Then `1, `2 and `∞ are infinite-dimensional normed vector spaces, with norms ‖.‖1, ‖.‖2 and ‖.‖∞ respectively, where ‖z‖1 = +∞∑ j=1 |zj|, ‖z‖2 = ( +∞∑ j=1 |zj|2 ) 1 2 , ‖z‖∞ = sup{|z1|, |z2|, |z3|, . . .}. 3 (For example, to show that ‖z + w‖2 ≤ ‖z‖2 + ‖w‖2 for all z,w ∈ `2, we note that( n∑ j=1 |zj + wj|2 ) 1 2 ≤ ( n∑ j=1 |zj|2 ) 1 2 + ( n∑ j=1 |wj|2 ) 1 2 ≤ ‖z‖2 + ‖w‖2 for all natural numbers n, by the Triangle Inequality in Cn. Taking limits as n→ +∞, we deduce that ‖z + w‖2 ≤ ‖z‖2 + ‖w‖2, as required.) If x1, x2, . . . , xm are elements of a normed vector space X then∥∥∥∥∥ m∑ k=1 xk ∥∥∥∥∥ ≤ m∑ k=1 ‖xk‖, where ‖.‖ denotes the norm on X. (This can be verified by induction on m, using the inequality ‖x+ y‖ ≤ ‖x‖+ ‖y‖.) A norm ‖.‖ on a vector space X induces a corresponding distance function on X: the distance d(x, y) between elements x and y of X is defined by d(x, y) = ‖x − y‖. This distance function satisfies the metric space axioms. Thus any vector space with a given norm can be regarded as a metric space. A norm on a vector space X therefore generates a topology on X: a subset U of X is an open set if and only if, given any point u of U , there exists some δ > 0 such that {x ∈ X : ‖x− u‖ < δ} ⊂ U. The function x 7→ ‖x‖ is a continuous function from X to R, since ‖x‖ − ‖y‖ = ‖(x− y) + y‖ − ‖y‖ ≤ (‖x− y‖+ ‖y‖)− ‖y‖ = ‖x− y‖, and ‖y‖ − ‖x‖ ≤ ‖x− y‖, and therefore |‖x‖ − ‖y‖| ≤ ‖x− y‖. The Cartesian product X1×X2×· · ·×Xn of vector spaces X1, X2, . . . , Xn can itself be regarded as a vector space: if (x1, x2, . . . , xn) and (y1, y2, . . . , yn) are points of X1 ×X2 × · · · ×Xn, and if λ is any scalar, then (x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1 + y1, x2 + y2, . . . , xn + yn), λ(x1, x2, . . . , xn) = (λx1, λx2, . . . , λxn). Lemma 9.1 Let X1, X2, . . . , Xn be normed vector spaces, and let ‖.‖max be the norm on X1 ×X2 × · · · ×Xn defined by ‖(x1, x2, . . . , xn)‖max = max(‖x1‖1, ‖x2‖2, . . . , ‖xn‖n), where ‖.‖i is the norm on Xi for i = 1, 2, . . . , n. Then the topology on X1 ×X2 × · · · ×Xn generated by the norm ‖.‖max is the product topology on X1 ×X2 × · · · ×Xn. 4 Proof It is a straightforward exercise to verify that ‖.‖max is indeed a norm on X, where X = X1 ×X2 × · · · ×Xn. Let U be a subset of X. Suppose that U is open with respect to the product topology. Let u be any point of U , given by u = (u1, u2, . . . , un). We must show that there exists some δ > 0 such that {x ∈ X : ‖x− u‖max < δ} ⊂ U. Now it follows from the definition of the product topology that there exist open sets V1, V2, . . . , Vn in X1, X2, . . . , Xn such that ui ∈ Vi for all i and V1×V2×· · ·×Vn ⊂ U . We can then take δ to be the minimum of δ1, δ2, . . . , δn, where δ1, δ2, . . . , δn are chosen such that {xi ∈ Xi : ‖xi − ui‖ < δi} ⊂ Vi for i = 1, 2, . . . , n. Conversely suppose that U is open with respect to the topology generated by the norm ‖.‖max. Let u be any point of U . Then there exists δ > 0 such that {x ∈ X : ‖x− u‖max < δ} ⊂ U. Let Vi = {xi ∈ Xi : ‖xi − ui‖ < δ} for i = 1, 2, . . . , n. Then, for each i, Vi is an open set in Xi, ui ∈ Vi, and V1 × V2 × · · · × Vn ⊂ U . We deduce that U is also open with respect to the product topology, as required. Proposition 9.2 Let X be a normed vector space over the field F, where F = R or C. Then the function from X ×X to X sending (x, y) ∈ X ×X to x+y is continuous. Also the function from F×X to X sending (λ, x) ∈ F×X to λx is continuous. Proof Let (u, v) ∈ X ×X, and let ε > 0 be given. Let δ = 1 2 ε. If (x, y) ∈ X ×X satisfies ‖(x, y)− (u, v)‖max < δ, then ‖x− u‖ < δ and ‖y − v‖ < δ, and hence ‖(x+ y)− (u+ v)‖ ≤ ‖x− u‖+ ‖y − v‖ < ε. This shows that the function (x, y) 7→ x+ y is continuous at (u, v) ∈ X ×X. Next let (µ, u) ∈ F×X, and let ε > 0 be given. Let δ = minimum ( ε 2(‖u‖+ 1) , ε 2(|µ|+ 1) , 1 ) . Now λx − µu = λ(x − u) + (λ − µ)u for all λ ∈ F and x ∈ X. Thus if (λ, x) ∈ F×X satisfies ‖(λ, x)− (µ, u)‖max < δ, then |λ− µ| < ε 2(‖u‖+ 1) , ‖x− u‖ < ε 2(|µ|+ 1) , |λ| < |µ|+ 1, 5 and hence ‖λx− µu‖ ≤ |λ| ‖x− u‖+ |λ− µ| ‖u‖ < ε 2 + ε 2 = ε. This shows that the function (λ, x) 7→ λx is continuous at (µ, u) ∈ F×X, as required. Corollary 9.3 Let X be a normed vector space over the field F, where F = R or C. Let (xn) and (yn) be convergent sequences in X, and let (λn) be a convergent sequence in F. Then the sequences (xn + yn) and (λnxn) are convergent in X, and lim n→+∞ (xn + yn) = lim n→+∞ xn + lim n→+∞ yn, lim n→+∞ (λnxn) = ( lim n→+∞ λn )( lim n→+∞ xn ) . Proof Let x = lim n→+∞ xn, y = lim n→+∞ yn and λ = lim n→+∞ λn. Using Lemma 9.1, together with the definition of convergence in metric spaces, it follows easily that the sequences (xn, yn) and (λn, xn) converge to (x, y) and (λ, x) in X×X and F×X respectively. The convergence of (xn + yn) and λnxn to x+ y and λx respectively now follows from Proposition 9.2. Let X be a normed vector space, and let x1, x2, x3, . . . be elements of X. The infinite series +∞∑ n=1 xn is said to converge to some element s of X if, given any ε > 0, there exists some natural number N such that ‖s− m∑ n=1 xn‖ < ε for all m ≥ N (where ‖.‖ denotes the norm on X). We say that a normed vector space X is complete if and only if every Cauchy sequence in X is convergent. (A sequence x1, x2, x3, . . . is a Cauchy sequence if and only if, given any ε > 0, there exists some natural number N such that ‖xj − xk‖ < ε for all j and k satisfying j ≥ N and k ≥ N .) A complete normed vector space is referred to as a Banach space. (The basic theory of such spaces was extensively developed by the famous Polish mathematician Stefan Banach and his co-workers.) Lemma 9.4 Let X be a Banach space, and let x1, x2, x3, . . . be elements of X. Suppose that +∞∑ n=1 ‖xn‖ is convergent. Then +∞∑ n=1 xn is convergent, and∥∥∥∥∥ +∞∑ n=1 xn ∥∥∥∥∥ ≤ +∞∑ n=1 ‖xn‖. 6 Proof For each natural number n, let sn = x1 + x2 + · · ·+ xn. Let ε > 0 be given. We can find N such that +∞∑ n=N ‖xn‖ < ε, since +∞∑ n=1 ‖xn‖ is convergent. Let sn = x1 + x2 + · · ·+ xn. If j ≥ N , k ≥ N and j < k then ‖sk − sj‖ = ∥∥∥∥∥ k∑ n=j+1 xn ∥∥∥∥∥ ≤ k∑ n=j+1 ‖xn‖ ≤ +∞∑ n=N ‖xn‖ < ε. Thus s1, s2, s3, . . . is a Cauchy sequence in X, and therefore converges to some element s of X, since X is complete. But then s = +∞∑ j=1 xj. Moreover, on choosing m large enough to ensure that ‖s− sm‖ < ε, we deduce that ‖s‖ ≤ ∥∥∥∥∥ m∑ n=1 xn ∥∥∥∥∥+ ∥∥∥∥∥s− m∑ n=1 xn ∥∥∥∥∥ ≤ m∑ n=1 ‖xn‖+ ∥∥∥∥∥s− m∑ n=1 xn ∥∥∥∥∥ < +∞∑ n=1 ‖xn‖+ ε. Since this inequality holds for all ε > 0, we conclude that ‖s‖ ≤ +∞∑ n=1 ‖xn‖, as required. 9.1 Bounded Linear Transformations Let X and Y be real or complex vector spaces. A function T :X → Y is said to be a linear transformation if T (x + y) = Tx + Ty and T (λx) = λTx for all elements x and y of X and scalars λ. A linear transformation mapping X into itself is referred to as a linear operator on X. Definition Let X and Y be normed vector spaces. A linear transformation T :X → Y is said to be bounded if there exists some non-negative real num- ber C with the property that ‖Tx‖ ≤ C‖x‖ for all x ∈ X. If T is bounded, then the smallest non-negative real number C with this property is referred to as the operator norm of T , and is denoted by ‖T‖. Lemma 9.5 Let X and Y be normed vector spaces, and let S:X → Y and T :X → Y be bounded linear transformations. Then S + T and λS are bounded linear transformations for all scalars λ, and ‖S + T‖ ≤ ‖S‖+ ‖T‖, ‖λS‖ = |λ|‖S‖. 7 Moreover ‖S‖ = 0 if and only if S = 0. Thus the vector space B(X, Y ) of bounded linear transformations from X to Y is a normed vector space (with respect to the operator norm). Proof ‖(S+T )x‖ ≤ ‖Sx‖+‖Tx‖ ≤ (‖S‖+‖T‖)‖x‖ for all x ∈ X. Therefore S+T is bounded, and ‖S+T‖ ≤ ‖S‖+‖T‖. Using the fact that ‖(λS)x‖ = |λ| ‖Sx‖ for all x ∈ X, we see that λS is bounded, and ‖λS‖ = |λ| ‖S‖. If S = 0 then ‖S‖ = 0. Conversely if ‖S‖ = 0 then ‖Sx‖ ≤ ‖S‖ ‖x‖ = 0 for all x ∈ X, and hence S = 0. The result follows. Lemma 9.6 Let X, Y and Z be normed vector spaces, and let S:X → Y and T :Y → Z be bounded linear transformations. Then the composition TS of S and T is also bounded, and ‖TS‖ ≤ ‖T‖ ‖S‖. Proof ‖TSx‖ ≤ ‖T‖ ‖Sx‖ ≤ ‖T‖ ‖S‖ ‖x‖ for all x ∈ X. The result fol- lows. Proposition 9.7 Let X and Y be normed vector spaces, and let T :X → Y be a linear transformation from X to Y . Then the following conditions are equivalent:— (i) T :X → Y is continuous, (ii) T :X → Y is continuous at 0, (iii) T :X → Y is bounded. Proof Obviously (i) implies (ii). We show that (ii) implies (iii) and (iii) implies (i). The equivalence of the three conditions then follows immediately. Suppose that T :X → Y is continuous at 0. Then there exists δ > 0 such that ‖Tx‖ < 1 for all x ∈ X satisfying ‖x‖ < δ. Let C be any positive real number satisfying C > 1/δ. If x is any non-zero element of X then ‖λx‖ < δ, where λ = 1/(C‖x‖), and hence ‖Tx‖ = C‖x‖ ‖λTx‖ = C‖x‖ ‖T (λx)‖ < C‖x‖. Thus ‖Tx‖ ≤ C‖x‖ for all x ∈ X, and hence T :X → Y is bounded. Thus (ii) implies (iii). Finally suppose that T :X → Y is bounded. Let x be a point of X, and let ε > 0 be given. Choose δ > 0 satisfying ‖T‖δ < ε. If x′ ∈ X satisfies ‖x′ − x‖ < δ then ‖Tx′ − Tx‖ = ‖T (x′ − x)‖ ≤ ‖T‖ ‖x′ − x‖ < ‖T‖δ < ε. Thus T :X → Y is continuous. Thus (iii) implies (i), as required. 8 Proposition 9.8 Let X be a normed vector space and let Y be a Banach space. Then the space B(X, Y ) of bounded linear transformations from X to Y is also a Banach space. Proof We have already shown that B(X,Y ) is a normed vector space (see Lemma 9.5). Thus it only remains to show that B(X, Y ) is complete. Let S1, S2, S3, . . . be a Cauchy sequence in B(X,Y ). Let x ∈ X. We claim that S1x, S2x, S3x, . . . is a Cauchy sequence in Y . This result is trivial if x = 0. If x 6= 0, and if ε > 0 is given then there exists some natural number N such that ‖Sj − Sk‖ < ε/‖x‖ whenever j ≥ N and k ≥ N . But then ‖Sjx − Skx‖ ≤ ‖Sj − Sk‖ ‖x‖ < ε whenever j ≥ N and k ≥ N . This shows that S1x, S2x, S3x, . . . is indeed a Cauchy sequence. It therefore converges to some element of Y , since Y is a Banach space. Let the function S:X → Y be defined by Sx = lim n→+∞ Snx. Then S(x+ y) = lim n→+∞ (Snx+ Sny) = lim n→+∞ Snx+ lim n→+∞ Sny = Sx+ Sy, (see Corollary 9.3), and S(λx) = lim n→+∞ Sn(λx) = λ lim n→+∞ Snx = λSx, Thus S:X → Y is a linear transformation. Next we show that Sn → S in B(X, Y ) as n→ +∞. Let ε > 0 be given. Then there exists some natural number N such that ‖Sj−Sn‖ < 12ε whenever j ≥ N and n ≥ N , since the sequence S1, S2, S3, . . . is a Cauchy sequence in B(X, Y ). But then ‖Sjx−Snx‖ ≤ 12ε‖x‖ for all j ≥ N and n ≥ N , and thus ‖Sx− Snx‖ = ∥∥∥∥ limj→+∞(Sjx− Snx) ∥∥∥∥ ≤ limj→+∞ ‖Sjx− Snx‖ ≤ lim j→+∞ ‖Sj − Sn‖ ‖x‖ ≤ 12ε‖x‖ for all n ≥ N (since the norm is a continuous function on Y ). But then ‖Sx‖ ≤ ‖Snx‖+ ‖Sx− Snx‖ ≤ (‖Sn‖+ 12ε) ‖x‖ for any n ≥ N , showing that S:X → Y is a bounded linear transformation, and ‖S − Sn‖ ≤ 12ε < ε for all n ≥ N , showing that Sn → S in B(X, Y ) as n→ +∞. Thus the Cauchy sequence S1, S2, S3, . . . is convergent in B(X, Y ), as required. 9 Corollary 9.9 Let X and Y be Banach spaces, and let T1, T2, T3, . . . be bounded linear transformations from X to Y . Suppose that +∞∑ n=0 ‖Tn‖ is con- vergent. Then +∞∑ n=0 Tn is convergent, and∥∥∥∥∥ +∞∑ n=0 Tn ∥∥∥∥∥ ≤ +∞∑ n=0 ‖Tn‖. Proof The space B(X,Y ) of bounded linear maps from X to Y is a Ba- nach space by Proposition 9.8. The result therefore follows immediately on applying Lemma 9.4. Example Let T be a bounded linear operator on a Banach space X (i.e., a bounded linear transformation from X to itself). The infinite series +∞∑ n=0 ‖T‖n n! converges to exp(‖T‖). It follows immediately from Lemma 9.6 (using induc- tion on n) that ‖T n‖ ≤ ‖T‖n for all n ≥ 0 (where T 0 is the identity operator on X). It therefore follows from Corollary 9.9 that there is a well-defined bounded linear operator expT on X, defined by expT = +∞∑ n=0 1 n! T n (where T 0 is the identity operator I on X). Proposition 9.10 Let T be a bounded linear operator on a Banach space X. Suppose that ‖T‖ < 1. Then the operator I − T has a bounded inverse (I − T )−1 (where I denotes the identity operator on X). Moreover (I − T )−1 = I + T + T 2 + T 3 + · · · . Proof ‖T n‖ ≤ ‖T‖n for all n, and the geometric series 1 + ‖T‖+ ‖T‖2 + ‖T‖3 + · · · is convergent (since ‖T‖ < 1). It follows from Corollary 9.9 that the infinite series I + T + T 2 + T 3 + · · · 10 converges to some bounded linear operator S on X. Now (I − T )S = lim n→+∞ (I − T )(I + T + T 2 + · · ·+ T n) = lim n→+∞ (I − T n+1) = I − lim n→+∞ T n+1 = I, since ‖T‖n+1 → 0 and therefore T n+1 → 0 as n→ +∞. Similarly S(I−T ) = I. This shows that I − T is invertible, with inverse S, as required. 9.2 The Equivalence of Norms on a Finite-Dimensional Vector Space Let ‖.‖ and ‖.‖∗ be norms on a real or complex vector space X. The norms ‖.‖ and ‖.‖∗ are said to be equivalent if and only if there exist constants c and C, where 0 < c ≤ C, such that c‖x‖ ≤ ‖x‖∗ ≤ C‖x‖ for all x ∈ X. Lemma 9.11 Two norms ‖.‖ and ‖.‖∗ on a real or complex vector space X are equivalent if and only if they induce the same topology on X. Proof Suppose that the norms ‖.‖ and ‖.‖∗ induce the same topology on X. Then there exists some δ > 0 such that {x ∈ X : ‖x‖ < δ} ⊂ {x ∈ X : ‖x‖∗ < 1}, since the set {x ∈ X : ‖x‖∗ < 1} is open with respect to the topology on X induced by both ‖.‖∗ and ‖.‖. Let C be any positive real number satisfying Cδ > 1. Then ∥∥∥∥ 1C‖x‖x ∥∥∥∥ = 1C < δ, and hence ‖x‖∗ = C‖x‖ ∥∥∥∥ 1C‖x‖x ∥∥∥∥ ∗ < C‖x‖. for all non-zero elements x of X, and thus ‖x‖∗ ≤ C‖x‖ for all x ∈ X. On interchanging the roles of the two norms, we deduce also that there exists a positive real number c such that ‖x‖ ≤ (1/c)‖x‖∗ for all x ∈ X. But then c‖x‖ ≤ ‖x‖∗ ≤ C‖x‖ for all x ∈ X. We conclude that the norms ‖.‖ and ‖.‖∗ are equivalent. 11 Conversely suppose that the norms ‖.‖ and ‖.‖∗ are equivalent. Then there exist constants c and C, where 0 < c ≤ C, such that c‖x‖ ≤ ‖x‖∗ ≤ C‖x‖ for all x ∈ X. Let U be a subset of X that is open with respect to the topology on X induced by the norm ‖.‖∗, and let u ∈ U . Then there exists some δ > 0 such that {x ∈ X : ‖x− u‖∗ < Cδ} ⊂ U. But then {x ∈ X : ‖x− u‖ < δ} ⊂ {x ∈ X : ‖x− u‖∗ < Cδ} ⊂ U, showing that U is open with respect to the topology induced by the norm ‖.‖. Similarly any subset of X that is open with respect to the topology induced by the norm ‖.‖ must also be open with respect to the topology induced by ‖.‖∗. Thus equivalent norms induce the same topology on X. It follows immediately from Lemma 9.11 that if ‖.‖, ‖.‖∗ and ‖.‖] are norms on a real (or complex) vector space X, if the norms ‖.‖ and ‖.‖∗ are equivalent, and if the norms ‖.‖∗ and ‖.‖] are equivalent, then the norms ‖.‖ and ‖.‖] are also equivalent. This fact can easily be verified directly from the definition of equivalence of norms. We recall that the usual topology on Rn is that generated by the Euclidean norm on Rn. Lemma 9.12 Let ‖.‖ be a norm on Rn. Then the function x 7→ ‖x‖ is continuous with respect to the usual topology on on Rn. Proof Let e1, e2, . . . , en denote the basis of Rn given by e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), · · · , en = (0, 0, 0, . . . , 1). Let x and y be points of Rn, given by x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn). Using Schwarz’ Inequality, we see that ‖x− y‖ = ∥∥∥∥∥ n∑ j=1 (xj − yj)ej ∥∥∥∥∥ ≤ n∑ j=1 |xj − yj| ‖ej‖ ≤ ( n∑ j=1 (xj − yj)2 ) 1 2 ( n∑ j=1 ‖ej‖2 ) 1 2 = C‖x− y‖2, 12 where C2 = ‖e1‖2 + ‖e2‖2 + · · ·+ ‖en‖2 and ‖x− y‖2 denotes the Euclidean norm of x− y, defined by ‖x− y‖2 = ( n∑ j=1 (xj − yj)2 ) 1 2 . Also |‖x‖ − ‖y‖| ≤ ‖x− y‖, since ‖x‖ ≤ ‖x− y‖+ ‖y‖, ‖y‖ ≤ ‖x− y‖+ ‖x‖. We conclude therefore that |‖x‖ − ‖y‖| ≤ C‖x− y‖2, for all x,y ∈ Rn, and thus the function x 7→ ‖x‖ is continuous on Rn (with respect to the usual topology on Rn). Theorem 9.13 Any two norms on Rn are equivalent, and induce the usual topology on Rn. Proof Let ‖.‖ be any norm on Rn. We show that ‖.‖ is equivalent to the Euclidean norm ‖.‖2. Let Sn−1 denote the unit sphere in Rn, defined by Sn−1 = {x ∈ Rn : ‖x‖2 = 1}. Now Sn−1 is a compact subset of Rn, since it is both closed and bounded. Also the function x 7→ ‖x‖ is continuous (Lemma 9.12). But any continuous real- valued function on a compact topological space attains both its maximum and minimum values on that space. Therefore there exist points u and v of Sn−1 such that ‖u‖ ≤ ‖x‖ ≤ ‖v‖ for all x ∈ Sn−1. Set c = ‖u‖ and C = ‖v‖. Then 0 < c ≤ C (since it follows from the definition of norms that the norm of any non-zero element of Rn is necessarily non-zero). If x is any non-zero element of Rn then λx ∈ Sn−1, wher