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Theorem
...
If
F : X → Y is a linear map, then F is continuous
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The case X = {0} is trivial
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, am } be a basis for X
...
We choose kn,j , kj ∈ K, 1 ≤ j ≤ m, n ∈ N such that xn =
kn,j aj
j=1

and x =

m
P

kj aj
...
, m
...

(k
−
k
)F
(a
)
kF (xn ) − F (x)k = kF (xn − x)k =
n,j
j
j

j=1
j=1
Thus, F is continuous
...
If X is infinite dimensional and Y 6= 0, then there exists a linear map F : X → Y which
is not continuous
...
Let {an : n ∈ N} be a linearly independent set in X
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If L = {xn : n ∈ N},
then we can find a basis B of X such that L ⊂ B
...
Since xn → 0 as n → ∞ and F (xn ) = y0 6= 0,
∀n ∈ N, F (xn ) 6→ F (0) = 0 as n → ∞
...

Theorem
...
Then, the
following conditions are equivalent
...

(ii) F is continuous at 0
...

(iv) F is uniformly continuous on X
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(vi) The zero space Z(F ) of F is closed subspace of X and the linear map F˜ : X/Z(F ) → Y defined
by F˜ (x + Z(F )) = F (x), ∀x ∈ X is continuous
...

(i) ⇒(v) Let kF (x)k ≤ β, ∀x ∈ U (0, r), for some r > 0 and β > 0
...

x


kxk
r
Clearly, the above inequality holds for x = 0
...
Given  > 0, we choose δ =


α,

so that

x, y ∈ X with kx − yk < δ ⇒ kF (x) − F (y)k = kF (x − y)k ≤ αkx − yk < α δ = α


= 
...

(iv) ⇒(iii) Clearly, F is uniformly continuous on X ⇒ F is continuous on X
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(ii) ⇒(i) Assume that F is continuous at 0
...

If we take r = 2δ , then U (0, r) ⊂ U (0, δ) and hence F is bounded on U (0, r)
...
Since Z(F ) = F −1 ({0}) and {0} is closed in Y , we get
that Z(F ) is closed in X
...

For arbitrary x ∈ X and arbitrary z ∈ Z(F ), we have
kF˜ (x + Z(F ))k = kF˜ (x + z + Z(F ))k = kF (x + z)k ≤ αkx + zk
...

α
Therefore, using (v), we get that F˜ is continuous
...
Then, using (v), there exists α > 0 such that
kF˜ (x + Z(F ))k ≤ α 9 x + Z(F )9, ∀x ∈ X
...

Hence the theorem follows
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A function f : X → Y is a homeomorphism if f is one-to-one, f is continuous, and
f −1 : f (X) → X is continuous
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Let F : X → Y be a linear transformation
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(ii) If F is a homeomorphism from X onto Y , then X is complete iff Y is complete
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(i) Every bijective linear map from X to Y is a homeomorphism
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2

(iii) There exists a linear homeomorphism from K n onto X and X is complete with respect
to each norm
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(a)(i) Assume that F is a homeomorphism
...
Then, by previous theorem, there exists α > 0 and γ > 0 such that
kF (x)k ≤ α kxk, ∀x ∈ X and kF −1 (F (x))k ≤ γkF (x)k, ∀x ∈ X
...

Conversely, assume that there exist positive reals α, β such that β kxk ≤ kF (x)k ≤ α kxk, ∀x ∈
X
...

If x1 , x2 ∈ X such that F (x1 ) = F (x2 ), then we have
β kx1 − x2 k ≤ kF (x1 ) − F (x2 )k = 0 ⇒ kx1 − x2 k = 0 ⇒ x1 = x2
...
Since
β kxk ≤ kF (x)k, ∀x ∈ X ⇒ kF −1 (F (x))k ≤ β −1 kF (x)k, ∀x ∈ X,
again using the previous theorem, we get that F −1 is continuous on F (X)
...

(a)(ii) Let F be a homeomorphism from X onto Y
...
If (yn ) is a Cauchy
sequence from Y , using that F : X → Y is onto, we find xn ∈ X such that F (xn ) = yn ,
∀n ∈ N
...

Therefore,
kxn − xm k = kF −1 (yn ) − F −1 (ym )k = kF −1 (yn − ym )k ≤ β kxn − xm k → 0 as m, n → ∞
...
Since X is complete, xn → x as n → ∞ for some
x ∈ X
...
Thus, Y is complete
...

(b)(i) If F : X → Y is a bijective linear map, then dim X = n = dim Y
...
So, by a theorem, we get that F and F −1
are continuous and hence F is a homeomorphism
...
Let k · k and
k · k0 be two norms on X
...
Then,
using (a)(i), there exists α > 0 and β > 0 such that β kxk ≤ kI(x)k0 = kxk0 ≤ α kxk, ∀x ∈ X
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3

(b)(iii) Since dim X = n, there exists a basis {x1 , x2 ,
...
If {e1 , e2 ,
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Since K n is complete, X is also complete with respect to all norms
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Let X and Y be normed spaces and F : X → Y be a linear map such that the range
R(F ) of F is finite dimensional
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Proof
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If R(F ) 6= {0}, then
let {y1 , y2 ,
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We choose xj ∈ X such that F (xj )yj , ∀j = 1, 2,
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We
claim that Span {x1 + Z(F ), x2 + Z(F ),
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Let x + Z(F ) ∈ X/Z(F ) be
arbitrary
...

j=1

Therefore, x + Z(F ) =

j=1
m
P

j=1

!
kj xj

j=1

+ Z(F ) =

m
P

j=1

(kj xj + Z(F )), hence the claim follows
...

If Z(F ) is closed in X, then X/Z(F ) is finite dimensional normed space
...
Therefore, by a theorem, we get Z(F ) is
closed in X iff F is continuous on X
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A linear functional f on X is continuous iff Z(f ) is closed in X
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Since the range of a linear functional f is at most 1, by previous corollary, we get that f is
continuous iff Z(f ) is closed in X

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