Chapter 5. Limits and Continuity in Metric Spaces
Limits of Sequences in Metric Spaces
5.1 Definition: Let (xn)n≥p be a sequence in a metric space X. We say that the sequence (xn)n≥p is bounded when the set {xn}n≥p is bounded, that is when there exists a ∈ X andr>0suchthatxn ∈B(a,r)forallindicesn≥p.
For a ∈ X, we say that the sequence (xn)n≥p converges to a (or that the limit of xn isequaltoa)andwewrite lim xn =a(orwewritexn →a)when
n→∞
∀ε>0 ∃m∈Z≥p ∀n∈Z≥pn ≥ m =⇒ d(xn,a) < ε.
We say that the sequence (xn)n≥p converges (in X) when it converges to some point a ∈ X, and otherwise we say that (xn)n≥p diverges (in X).
5.2 Theorem: (Basic Properties of Limits of Sequences) Let (xn)n≥p be a sequence in a metric space X, and let a ∈ X.
(1) If (xn)n≥p converges then its limit is unique.
(2)Ifq≥pandyn =xn foralln≥q,then(xn)n≥p convergesifandonlyif(yn)n≥q
converges and, in this case, lim yn = lim xn. n→∞ n→∞
(3) If (xnk )k≥q is a subsequence of (xn)n≥p, and lim xn = a, then lim xnk = a. n→∞ k→∞
(4) If (xn)n≥p converges then it is bounded.
(5)Wehave lim xn =ainX ifandonlyif lim d(xn,a)=0inR.
n→∞ n→∞
Proof: We prove Parts 1, 4 and 5 and leave the proofs of the other parts as an exercise.
ToprovePart1,supposethatxn →ainXandxn →binX. Weneedtoshowthat
a = b. Suppose, for a contradiction, that a ̸= b, and note that d(a, b) > 0. Since xn → a
andxn →b,wecanchoosem∈Z≥p suchthatwhenn≥mwehaved(xn,a)< d(a,b),and 2
d(xn, b) < d(a,b) . Then we have d(a, b) ≤ d(a, xm) + d(xm, b) < d(a,b) + d(a,b) = d(a, b), 222
giving the desired contradiction.
To prove Part 4, suppose that (xn)n≥p converges, say xn → a in X. Choose m ∈ Z≥p
suchthatwhenn≥mwehaved(xn,a)<1. Thenforalln∈Z≥p wehaved(xn,a)≤r where r = max d(xp, a), d(xp+1, a), · · · , d(xm−1, a), 1 so that xn ∈ B(a, r+1).
To prove Part 5, note that since d(xn, a) ≥ 0 we have d(xn, a) = d(xn, a) − 0 and so
lim xn = a in X ⇐⇒ n→∞
⇐⇒ ⇐⇒
∀ε>0 ∃m∈Z≥p ∀n∈Z≥p d(xn,a) < ε ∀ε>0∃m∈Z≥p ∀n∈Z≥p d(xn,a)−0<ε
limd(xn,a)=0inR. n→∞
5.3 Note: Because of Part 2 of
(xn)n≥p does not effect whether or not the sequence converges and it does not effect the limit. For this reason, we often omit the initial index p from our notation and write (xn) for the sequence (xn)n≥p. Also, we often choose a specific value of p, usually p = 1, in the statements or the proofs of various theorems with the understanding that any other initial value would work just as well.
the above theorem, the initial index p of a sequence
1
5.4 Theorem: (Components of Sequences in Rm) Let (xn)n≥1 be a sequence in Rm, say
xn = xn,1,xn,2,···,xn,m ∈ Rm, and let a = (a1,a2,···,am) ∈ Rm. Then using any of
themetricsd1,d2 ord∞ inRm,wehave lim xn =ainRm ifandonlyif lim xn,k =ak n→∞ n→∞
in R for all indices k with 1 ≤ k ≤ m.
Proof: Letp=1,2or∞. Supposethat lim xn =ainRm. Let1≤k≤mandletε>0.
n→∞
Choosem∈Z+ suchthatwhenn≥mwehavedp(xn−a)<ε,thatis∥xn−p∥p <ε.
Then when n ≥ m we have
p1/p m p1/p
|xn,k −ak|= |xn,k −ak| ≤ |xn,j −aj| =∥xn −a∥p <ε,
j=1
and so xn,k → ak in R, as required.
Suppose, conversely, that for all k with 1 ≤ k ≤ m we have lim xn,k = ak in R. Let
n→∞
ε>0. Choosem∈Z+ suchthatforalln≥mwehave|xn,k −ak|< ε for1≤k≤m.
m
Then, letting ek denote the kth standard basis vector in Rm, when n ≥ m we have
m m ∥xn − a∥p = (xn,k − ak) ek ≤ (xn,k − ak)ekp
k=1 p k=1
= m xn,k −ak∥ek∥p = m xn,k −ak< m ε =ε
m k=1 k=1 k=1
so that xn → a in Rm, as required.
5.5 Note: When (xn)n≥1 is a sequence in R∞, l1, l2 or l∞, each term xn is itself a sequence (so that (xn) is a sequence of sequences) and we can write xn = (xn,k)k≥1. We have sequences x1 = (x1,1, x1,2, · · ·), x2 = (x2,1, x2,2, x2,3, · · ·), and x3 = (x3,1, x3,2, · · ·) and so on. This is not the same thing as a subsequence (xn), which is a single sequence (xnk)k≥1 = (xn1,xn2,xn3,···).
5.6 Theorem: Components of Sequences in lp. Let p = 1, 2 or ∞. Let (xn)n≥1 be a sequenceinlp,sayxn =(xn,k)k≥1 ∈lp,andleta=(ak)k≥1 ∈lp. If lim xn =ainlp
n→∞
then lim xn,k =ak inRforallk∈Z+. n→∞
Proof: The proof is the same as the first half of the proof of Theorem 5.4. Suppose that lim xn =ainlp. Letk∈Z+ andletε>0. Choosem∈Z+ suchthatwhenn≥mwe
n→∞
have ∥xn − a∥p < ε. Then when n ≥ m we have
p1/p ∞ p1/p |xn,k −ak|= |xn,k −ak| ≤ |xn,j −aj|
j=1
and so xn,k → ak in R, as required.
5.7 Note: Unlike the case in Rm, in the infinite-dimensional spaces lp, when xn,k → ak in R for all indices k, it does not necessarily follow that xn → a in lp. For example, you can verify, as an exercise, that when xn = en (the nth standard basis vector in R∞), we have lim xn,k =0inRforallk∈Z+,but lim xn ̸=0inlp.
n→∞ n→∞ 5.8Exercise:Foreachn∈Z+,letxn∈R∞bethesequencegivenbyxn=1 ek,
=∥xn −a∥p <ε
n n
k=1 that is by xn = (xn,k)k≥1 = 1 , 1 ,···, 1 ,0,0,0,··· with n non-zero terms. Show that
∞nnn∞
(xn) converges in (R , d2) but diverges in (R , d1).
2
5.9 Remark: In the definition of the limit of a sequence in a metric space X (Definition 5.1), we can replace the strict inequality d(xn , a) < ε by the inequality d(xn , a) ≤ ε without changing the meaning. In other words, for a sequence (xn)n≥p in X and an element a ∈ X we have
lim xn = a in X ⇐⇒ ∀ε>0 ∃m∈Z≥p∀n∈Z≥pn ≥ m =⇒ d(xn,a) ≤ ε. n→∞
Hereisaproof. Supposethat lim xn =ainX. Letε>0. Choosem∈Z≥p sothat n→∞
when n ≥ m we have d(xn,a) < ε. Then when n ≥ a we also have d(xn,a) ≤ a. Suppose,
conversely, that for every ε > 0 there exists m ∈ Z≥p such that whenever n ≥ m we have
d(xn,a)≤ε. Letε>0. Choosem∈Z+ suchthatwhenn≥mwehaved(xn,a)≤ ε.
Thenwhenn≥mwehaved(xn,a)≤ ε <ε. 2
5.10 Remark: Note that for h ∈ B[a, b] and r > 0, we have
∥h∥∞ ≤ r ⇐⇒ sup|h(x)|a≤x≤b ≤ r ⇐⇒ |h(x)| ≤ r for all x ∈ [a,b].
2
We also remark that we would not have equivalence if we replaced ≤ r by < r, as we only have a one way implication: if |h(x)| < r for all x ∈ [a,b] then sup|h(x)|a≤x≤b ≤ r.
5.11 Theorem: (Limits in B[a,b] and Uniform Convergence) Let (fn)n≥1 be a sequence in B[a,b], and let g ∈ B[a,b]. Then fn → g in B[a,b],d∞ if and only if fn → g uniformly on [a, b].
Proof: This follows immediately from the above two remarks. Indeed we have fn→ginB[a,b] ⇐⇒ ∀ε>0∃m∈Z+ ∀n∈Z+n≥m=⇒∥fn−g∥∞≤ε
⇐⇒ ∀ε>0∃m∈Z+ ∀n∈Z+n≥m=⇒fn(x)−g(x)≤εforallx∈[a,b]
⇐⇒ fn → g uniformly on [a, b].
5.12 Remark: In Chapter 3, we discussed pointwise convergence and uniform convergence of sequences of functions. In this Chapter, we are discussing convergence in a metric space. For a metric space X whose elements are functions, such as B[a,b] or C[a,b], a sequence in X is a sequence of functions, so we now have several different notions of convergence for sequences of functions. The above theorem shows that convergence in the metric space B[a,b] hence also in C[a,b] using the supremum metric d∞, is the same thing as uniform convergence. One might ask whether convergence in C[a,b] using the metrics d1 or d2 implies, or is implied by, pointwise convergence. The answer is negative, as the following exercises illustrate.
5.13Exercise: Definefn :[0,1]→Rbyfn(x)=1−nxfor0≤x≤ 1 andfn(x)=0for n
1 ≤x≤1. Showthatfn →0inC[0,1]usingeitherofthemetricsd1 ord2,butfn ̸→0 n
pointwise on [0, 1].
5.14Exercise: Definefn :[0,1]→Rbyfn(x)=n2x−n3x2 for0≤x≤ 1 andfn(x)=0
n
for 1 ≤x≤1. Showthatfn →0pointwiseon[0,1]butfn ̸→0inC[0,1]usingeitherof n
the metrics d1 or d2.
5.15 Exercise: Define fn : [0,1] → R by fn(x) = √nxn. Show that (fn)n≥1 converges
in C[0, 1], d1 but diverges in C[0, 1], d2.
3
Limits and Closed Sets
5.16 Theorem: (The Sequential Characterization of Limit Points and Closed Sets) Let X beametricspace,leta∈X,andletA⊆X.
(1)a∈A′ ifandonlyifthereexistsasequence(xn)inA\{a}with lim xn =ainX. n→∞
(2)a∈Aifandonlyifthereexistsasequence(xn)inAwith lim xn =ainX. n→∞
(3) A is closed in X if and only if for every sequence (xn) in A which converges in X, we have lim xn ∈ A.
n→∞
Proof: We prove Parts 1 and 3 and leave the proof of Part 2 as an exercise. Suppose that
a ∈ A′ (which means that for every r > 0 we have B∗(a,r)∩A ̸= ∅). For each n ∈ Z+,
choose xn ∈ B∗a, 1∩A, that is choose xn ∈ A\{a} with d(xn,a) < 1. Then (xn)n≥1 nn
isasequenceinA\{a}with lim xn =a. n→∞
Suppose, conversely, that (xn)n≥1 is a sequence in A \ {a} with lim xn = a. Let n→∞
r>0. Choosem∈Z+ suchthatd(xn,a)
B(a, r) = B(a, r) (so the closed ball is equal to the closure of the open ball).
Solution: We saw, in Example 4.32, that B(a,r) is closed. Since B(a,r) is closed and B(a,r) ⊆ B(a,r), it follows that B(a,r) ⊆ B(a,r). Let b ∈ B(a,r), that is let b ∈ U with ∥b−a∥ ≤ r. If ∥b−a∥ < r then we have b ∈ B(a,r) ⊆ B(a,r). Suppose that ∥b−a∥ = r. For n ∈ Z+, let xn = a + 1 − 1 (b − a) ∈ U. Note that
n
∥xn −a∥=1− 1(b−a)=1− 1∥b−a∥=1− 1r
∞
∥xn −a∥1 = |xn,k −ak|= |ak|≤ |ak|<ε. k=1 k>n k>m
It follows, from Part 2 of Theorem 5.16, that a ∈ R∞, as required.
5.21 Exercise: Let
R[a, b] = f ∈ B[a, b] f is Riemann integrable,
P[a, b] = f ∈ B[a, b] f is a polynomial,
C1[a, b] = f ∈ B[a, b] f is continuously differentiable.
Note that
Determine which of the above spaces are closed in the metric space B[a,b], using the
supremum metric d∞ (we deal with the space C[a,b] in the following example).
5.22 Example: Show that C[a, b] is closed in the metric space B[a, b], d∞.
Solution: Let (fn) be a sequence in C[a, b] which converges in B[a, b] (using the metric d∞). Let g = lim fn in B[a,b]. By Theorem 5.11, we know that fn → g uniformly on [a,b].
n→∞
Since each function fn is continuous, it follows from Theorem 3.9 (Uniform Convergence
and Continuity) that g is continuous, that is g ∈ C[a, b]. By the Sequential Characterization of Closed Sets (Part 3 of Theorem 5.16), it follows that C[a,b] is closed in B[a,b].
P[a,b] ⊆ C1[a,b] ⊆ C[a,b] ⊆ R[a,b] ⊆ B[a,b].
5
Limits of Functions and Continuity in Metric Spaces
5.23 Definition: Let (X,dX) and (Y,dY ) be metric spaces. Let A ⊆ X, let f : A → Y,
leta∈A′,andletb∈Y. Wesaythatthelimitoff(x)asxtendstoaisequaltob,and
we write lim f(x) = b, when x→a
∀ε>0 ∃δ>0 ∀x∈A0 < dX(x,a) < δ =⇒ dY f(x),b < ε.
5.24 Theorem: (The Sequential Characterization of Limits) Let X and Y be metric
spaces, let A ⊆ X, let f : A → Y , let a ∈ A′ ⊆ X, and let b ∈ Y . Then lim f(x) = b if x→a
andonlyifforeverysequence(xn)inA\{a}withxn →awehave lim f(xn)=b. n→∞
Proof: Suppose that lim f(x) = b. Let (xn) be a sequence in A \ {a} with xn → a. Let x→A
ε>0. Since limf(x)=bwecanchooseδ>0suchthat0
existsx∈Asuchthat0
for all x ∈ X, if dX(x,a) < δ then dY f(x),f(a) < ε. We say that f is continuous (on
X) when f is continuous at every point a ∈ X. We say that f is uniformly continuous
(on X) when for every ε > 0 there exists δ > 0 such that for all x,y ∈ X, if dX(x,y) < δ
then dY f(x),f(y) < ε. We say that f is Lipschitz continuous (on X) when there is
a constant l ≥ 0, called a Lipschitz constant for f , such that for all x, y ∈ X we. have
df (x), f (y) ≤ l · d(x, y). Note that if f is Lipschitz continuous then f is also uniformly
continuous (indeed we can take δ = ε in the definition of uniform continuity). A bijective l −1
map f : X → Y such that both f and f are continuous is called a homeomorphism. 5.26 Note: Let X and Y be metric spaces and let a ∈ X. If a is a limit point of X then
f is continuous at a if and only if lim f(x) = f(a). If a is an isolated point of X then f x→a
is necessarily continuous at a, vacuously.
5.27 Theorem: (The Sequential Characterization of Continuity) Let X and Y be metric
spacesusingmetricsdX anddY,letf:X→Y,andleta∈X.Thenfiscontinuousata
if and only if for every sequence (xn) in X with xn → a we have lim f(xn) = f(a). n→∞
Proof: The proof is left as an exercise.
5.28Exercise: LetX,Y andZbemetricspaces,letf:X→Y,letg:Y →Z. Show that if f is continuous at the point a ∈ X and g is continuous at the point f(a) ∈ Y then the composite function g ◦ f is continuous at a.
6
5.29 Theorem: (The Topological Characterization of Continuity) Let X and Y be metric spaces and let f : X → Y . Then f is continuous (on X) if an only if f−1(V ) is open in X foreveryopensetV inY.
Proof: Suppose that f is continuous in X. Let V be open in Y . Let a ∈ f−1(V ) and let f(a) ∈ V. Since V is open, we can choose ε > 0 such that Bf(a),ε ⊆ V. Since f is continuous at a we can choose δ > 0 such that for all x ∈ X with d(x,a) < δ we have df(x),f(a) < ε. Then we have fB(a,δ) ⊆ Bf(a),ε ⊆ V and so B(a,δ) ⊆ f−1(V). Thus f−1(V ) is open in X, as required.
Suppose, conversely, that f−1(V ) is open in X for every open set V in Y . Let a ∈ X and let ε > 0. Taking V = Bf(a),ε, which is open in Y, we see that f−1Bf(a),ε is open in X. Since a ∈ f−1B(f(a),ε and f−1Bf(a),ε is open in X, we can choose δ > 0 such that B(a, δ) ⊆ f−1Bf(a), ε. Then we have fB(a, δ) ⊆ Bf(a), ε or, in other words, for all x ∈ X, if d(x,a) < δ then df(x),f(a) < ε. Thus f is continuous at a hence, since a was arbitrary, f is continuous on X.
5.30 Definition: Let X and Y be topological spaces and let f : X → Y . We say that f is continuous (on X) when f−1(V ) is open in X for every open set V in Y . A bijective map f : X → Y such that both f and f−1 are continuous is called a homeomorphism.
5.31 Theorem: (Composition of Continuous Functions) Let X, Y and Z be metric spaces (or topological spaces), let f : X → Y , and let g : Y → Z. If f and g are continuous then the composite function g ◦ f : X → Z is continuous.
Proof:Leth=g◦f:X→Z.IfW⊆ZisopeninZ,theng−1(W)isopeninY (sinceg is continuous), hence h−1(W) = f−1g−1(W) is open in X (since f is continuous). Thus h is continuous, by Theorem 5.29 (or by Definition 5.30)
5.32 Example: Let A = (x,y) ∈ R2 y < x2. Show that A is open in R2.
Solution: We remark that it is surprisingly difficult to show that A is open directly from the definition of an open set (as mentioned in Remark 4.34). But we can make use of the Topological Characterization of Open Sets to give a quick proof. Define f : R2 → R by f(x, y) = y − x2. Note that f is continuous (polynomial functions, and indeed all elementary functions, are continuous) and we have A = (x,y)f(x,y) < 0 = f−1(B) where B is the open interval (−∞, 0). Since B is open in R and f is continuous, it follows that A = f−1(B) is open in R2.
5.33 Example: Recall from Example 4.41 that every set U ⊆ C[a, b] which is open using the metric d1 is also open using the metric d∞, but not vice versa. It follows (from Theorem 5.29) that the identity map I : C → C[a,b] given by I(f) = f is continuous as a map from the metric space C[a, b], d∞ to the metric space C[a, b], d1, but not vice versa.
7
Continuity of Linear Maps
5.34 Note: If U and V are inner product spaces and L : U → V is an inner product space isomorphism, then L and its inverse preserve distance so they are both continuous (we can take δ = ε in the definition of continuity), hence L is a homeomorphism.
If U and V are finite-dimensional inner product spaces with say dimU = n and dimV =m,andifφ:Rn →U andψ:Rm →V areinnerproductspaceisomorphisms (obtained by choosing orthonormal bases for U and V ) then a map F : U → V is continuous if and only if the composite map ψ−1Fφ : Rn → Rm is continuous. In particular, if F is linear then F is continuous (since ψ−1F φ : Rn → Rm is linear, hence continuous).
We shall see below (in Corollary 5.39) that the same is true for finite dimensional normed linear spaces: every linear map between finite dimensional normed linear spaces is continuous. But this is not always true (see Example 5.33) for infinite dimensional spaces.
5.35 Theorem: Let U and V be normed linear spaces and let F : U → V be a linear map. Then the following are equivalent:
(1) F is Lipschitz continuous on U,
(2) F is continuous at some point a ∈ U, (3) F is continuous at 0, and
(4) F B(0, 1) is bounded.
In this case, if m ≥ 0 with FB(0,1) ⊆ B(0,m) then m is a Lipschitz constant for F.
Proof: It is clear that if F is Lipschitz continuous on U then F is continuous at some point
a ∈ U (indeed F is continuous at every point a ∈ U). Let us show that if F is continuous
at some point a ∈ U then F is continuous at 0. Suppose that F is continuous at a ∈ U.
Letε>0. SinceF iscontinuousata∈U,wecanchooseδ1 >0suchthatforallu∈U we
have ∥u−a∥ ≤ δ1 =⇒ F(u)−F(a) ≤ 1. Choose δ = δ1ε. Let x ∈ U with ∥x−0∥ < δ.
If x = 0 then F(x)−F(0) = ∥0∥ = 0. Suppose that x ̸= 0. Then for u = a+ δ1x we ∥x∥
have ∥u−a∥ = δ1x = δ1 and so F(u−a) = F(u)−F(a) ≤ 1, that is Fδ1x ≤ 1 ∥x∥ ∥x∥
hence, by the linearity of F and the scaling property of the norm, we have F(x)−F(0)=F(x)= ∥x∥Fδ1x≤ ∥x∥ < δ1ε =ε.
δ1 ∥x∥ δ1 δ1
Thus F is continuous at 0, as required
Next we show that if F is continuous at 0 then F B(0, 1) is bounded. Suppose that
F iscontinuousat0. Chooseδ>0sothatforallu∈U wehave∥u∥≤δ=⇒∥F(u)∥≤1. Letm= 1. Forx∈U,whenx=0wehave∥F(x)∥=0≤mandwhen0<∥x∥≤1we
have
δ
∥x∥ δx ∥x∥ δx ∥x∥
∥F(x)∥= δ F ∥x∥ = δ F ∥x∥ ≤ δ =m∥x∥≤m.
Thus F B(0, 1) is bounded, as required.
Finally we show that if F B(0, 1) is bounded then F is Lipschitz continuous. Suppose
that FB(0,1) is bounded. Choose m > 0 so that ∥F(u)∥ ≤ m for all u ∈ U with ∥u∥ ≤ 1. Let x,y ∈ U. If x = y then F(x)−F(y) = 0. Suppose that x ̸= y. Then we have
x−y =1sothatF x−y ≤mandso ∥ x − y ∥ ∥ x − y ∥
F(x) − F(y) = F(x − y) = ∥x − y∥F x−y ≤ m∥x − y∥. ∥x−y∥
Thus F is Lipschitz continuous with Lipschitz constant m, as required. 8
5.36 Example: Define L : C[a,b],d∞ → C[a,b],d∞ by L(f) = L is Lipschitz continuous.
x a
f(t)dt. Show that Solution: Let f ∈ C[a,b] with ∥f∥∞ ≤ 1, that is with max |f(x)| ≤ 1. Then
a≤x≤b
F(f) = max f(t)dt≤ max 1dt= max |x−a|=|b−a|.
5.37 Example: Define D : C1[0, 1], d∞ → C[0, 1], d∞ by D(f) = f′. Show that D is not continuous.
Solution: For n ∈ Z+, define fn : [0,1] → R by fn(x) = xn. Then fn ∈ C1[a,b], and ∥fn∥∞ = max |xn| = 1 so that fn ∈ B(0, 1), and D(fn)∞ = max n xn−1 = n. Thus
0≤x≤1 0≤x≤1
D B(0, 1) is not bounded, so D is not continuous at any point g ∈ C[0, 1].
5.38 Theorem: Let U be an n-dimensional normed linear space over R. Let {u1, · · · , un} be any basis for U and let φ : Rn → U be the associated vector space isomorphism given
n
by φ(t) = tkuk. Then both φ and φ−1 are Lipschitz continuous.
k=1 Proof:LetM=
n k=1
2 1 / 2 n .Fort∈R wehave
∥uk∥ n n
xx
∞ a≤x≤ba a≤x≤ba a≤x≤b Thus F B(0, 1) is bounded and so F is uniformly continuous.
t u ≤ |t |∥u ∥ , by the Triangle Inequality, k=1
φ(t) =
kk kk
k=1 n
21/2 n k=1
21/2
∥uk∥ , by the Cauchy-Schwarz Inequality,
≤
= M∥t∥.
k=1
tk
For all s, t ∈ Rn , φ(s) − φ(t) = φ(s − t) ≤ M ∥s − t∥, so φ is Lipschitz continuous.
Note that the map N : U → R given by N(x) = ∥x∥ is (uniformly) continuous, indeed
we can take δ = ε in the definition of continuity. Since φ and N are both continuous, so is
the composite G = N ◦ φ : Rn → R, which given by G(t) = φ(t). By the Extreme Value
Theorem, the map G attains its minimum value on the unit sphere t ∈ Rn∥t∥ = 1,
which is compact. Let m = min G(t) = min φ(t). Note that m > 0 because when ∥t∥=1 ∥t∥=1
t ̸= 0 we have φ(t) ̸= 0 (since φ is a bijective linear map) and hence ∥φ(t)∥ ≠ 0. For t∈Rn,if∥t∥>1thenwehave t =1so,bythechoiceofm,
∥t∥
t
∥t∥
φ(t) =∥t∥φ ≥∥t∥·m>m.
It follows that for all t ∈ Rn, if φ(t) ≤ m then ∥t∥ ≤ 1. Since φ is bijective, it follows
thatforx∈U,if∥x∥≤mthenφ−1(x)≤1. Thusforallx∈U,ifx=0then
∥φ−1(x)∥=0=∥x∥ andifx̸=0thensincemx=mwehave m ∥x∥
φ−1(x) = ∥x∥ φ−1 mx ≤ ∥x∥ . m ∥x∥ m
For all x,y ∈ U, we have φ−1(x) − φ−1(y) = φ−1(x − y) ≤ 1 ∥x − y∥, so φ−1 is m
Lipschitz continuous.
9
5.39 Corollary: When U and V are finite-dimensional normed linear spaces, every linear map F : U → V is Lipschitz continuous.
Proof: Let U and V be finite-dimensional vector spaces and let F : U → V be linear. Let {u1,···,un}and{v1,···,vm}bebasesforU andV,andletφ:Rn →U andψ:Rm →V
nm
be the vector space isomorphisms given by φ(t) = tkuk and ψ(s) = skvk. Since ψ−1
k=1 k=1
and φ are both linear, the composite G = ψ−1Fφ : Rn → Rm is linear, hence continuous
(linear maps from Rn to Rm, using the standard metric, are continuous). By the above theorem, we know that ψ and φ−1 are continuous, and so the composite map F = ψ Gφ−1 is continuous, hence also Lipschitz continuous, by Theorem 5.35.
5.40 Corollary: Any two norms on a finite-dimensional vector space U induce the same topology on U.
Proof: Let U have two norms ∥ ∥1 and ∥ ∥2, inducing two metrics d1 and d2, determining two topologies on U. Let I : (U,d1) → (U,d2) be the identity map (given by I(x) = x), and let J = I−1 : (U, d2) → (U, d1) (so J is also the identity map). By the above corollary, I and J are continuous. Let A ⊆ U. If A is open in (U,d1) then, since J is continuous, J−1(A) is open in (U,d2), but J−1(A) = I(A) = A and so A is open in (U,d2). Similarly, if A is open in (U,d2) then A = J(A) = I−1(A) is open in (U,d1).
5.41 Corollary: Let U be a finite-dimensional vector space. Let ∥ ∥1 and ∥ ∥2 be two norms on U inducing the two metric d1 and d2 on U. Let (xn)n≥1 be a sequence in U, and leta∈U. Thenxn →ain(U,d1)ifandonlyifxn →ain(U,d2).
Proof: Let I : (U, d1) → (U, d2) be the identity map (given by I(x) = x). By Corollary 5.39,
I is Lipschitz continuous. Let l ≥ 0 be a Lipschitz constant for I. Suppose that xn → a in
(U,d1). Letε>0. Choosem∈Z+ suchthatwhenn≥mwehaved1(xn,a)< ε . Then l+1
when n ≥ m we have d2(xn,a) = d2I(xn),I(a) ≤ l·d1(xn,a) < l· ε < ε. Thus xn → a l+1
in (U, d2 ). Similarly, since the identity map J : (U, d2 ) → (U, d1 ) is Lipschitz continuous, it follows that if xntoa in (U,d2) then xn → a in (U,d1). We remark that I and J might have different Lipschits constants (even though I and J are both the identity map from U to itself).
10