all norms on finite-dimensional vector spaces are equivalent

all norms on finite-dimensional vector spaces are equivalent

Theorem

.

All norms on

finite-dimensional

vector spaces

over

R

or

C

are

equivalent

(

http://planetmath.org/EquivalentNorms

).

A

consequence

of this is that there is only one norm induced topology on a finite dimensional vector space. This means that on such a vector space, we
need not worry about what norm we use when we talk about convergence of a

sequence

of vectors in norm. So a standard use of this

theorem

is in

continuity

arguments

over finite dimensional vector spaces, and it allows you to pick the most convenient norm for your argument (the

Euclidean norm

is not always very convenient).

This obviously is not true for infinite dimensional spaces, for example see the different

L

p

spaces (

http://planetmath.org/LpSpace

). Note that the reason all this works is because a

unit sphere

is

compact

in a finite dimensional vector space, while that is not true in an infinite dimensional one.

Proof.

Any such finite-dimensional space is really just the same as

ℝ

n

so
we can talk about just those spaces. That is, any finite-dimensional vector
space over

ℝ

or

ℂ

is

isomorphic

to

ℝ

n

for some

n

(note that

ℂ

is just isomorphic to

ℝ

2

as a vector space over

ℝ

).
To see this, just write any element of the space in of
the basis and then define the

isomorphism

to take that basis to the

standard
basis

in

ℝ

n

and then extend linearly.

First let’s show that if two norms are equivalent on the unit sphere
(all

x

→

such that

∥

x

→

∥

=

1

with respect to some norm,
for example the standard Euclidean norm) then they are equivalent everywhere.
We can write any

x

→

∈

ℝ

n

as a multiple of some scalar

γ

≥

0

and a vector on the unit sphere, say

x

0

→

, that is

x

→

=

γ

⁢

x

0

→

.
Then when suppose we have two

equivalent
norms

, say

∥

⋅

∥

a

and

∥

⋅

∥

b

, on the unit sphere

α

⁢

∥

x

0

→

∥

a

≤

∥

x

0

→

∥

b

≤

β

⁢

∥

x

0

→

∥

a

γ

⁢

α

⁢

∥

x

0

→

∥

a

≤

γ

⁢

∥

x

0

→

∥

b

≤

γ

⁢

β

⁢

∥

x

0

→

∥

a

α

⁢

∥

γ

⁢

x

0

→

∥

a

≤

∥

γ

⁢

x

0

→

∥

b

≤

β

⁢

∥

γ

⁢

x

0

→

∥

a

α

⁢

∥

x

→

∥

a

≤

∥

x

→

∥

b

≤

β

⁢

∥

x

→

∥

a

.

So the norms are equivalent everywhere.

Suppose we are working with the 2-norm. Now we want to show that any other
norm is a

continuous function

with respect to the 2-norm.
Take an arbitrary finite-dimensional space

X

and
an arbitrary norm

∥

⋅

∥

.
Also suppose that

{

b

i

→

}

1

n

is a basis of

X

and so an

element

x

→

∈

X

may be written as

x

→

=

∑

1

n

x

i

⁢

b

i

→

.
Now given an

ϵ

>

0

, choose

δ

>

0

such that

∥

x

→

-

y

→

∥

2

<

δ

(the

Euclidean distance

is less then

δ

) implies that

max

⁡

{

|

x

i

-

y

i

|

}

<

ϵ

∑

i

=

1

n

∥

b

i

→

∥

In fact we can just choose

δ

to be the right side of
the above

inequality

.
Now we note that the

triangle inequality

immediately also yields the
inequality

|

∥

x

→

∥

-

∥

y

→

∥

|

≤

∥

x

→

-

y

→

∥

. So

|

∥

x

→

∥

-

∥

y

→

∥

|

≤

∥

x

→

-

y

→

∥

=

∥

∑

i

=

1

n

x

i

⁢

b

i

→

-

∑

i

=

1

n

y

i

⁢

b

i

→

∥

=

∥

∑

i

=

1

n

(

x

i

-

y

i

)

⁢

b

i

→

∥

≤

∑

i

=

1

n

|

x

i

-

y

i

|

⁢

∥

b

i

→

∥

≤

(

max

i

⁡

|

x

i

-

y

i

|

)

⁢

∑

i

=

1

n

∥

b

i

→

∥

<

ϵ

∑

i

=

1

n

∥

b

i

→

∥

⁢

∑

i

=

1

n

∥

b

i

→

∥

=

ϵ

.

And so

∥

⋅

∥

is a continuous function.

Suppose we are given two norms

∥

⋅

∥

a

and

∥

⋅

∥

b

, we know
that they are both continuous functions with respect to the 2-norm.
And so the

function

defined as

f

⁢

(

x

→

)

:=

∥

x

→

∥

a

∥

x

→

∥

b

is a continuous function on the unit sphere (with respect to the 2-norm).
This function is

continuous

except perhaps at 0, but we don’t
care about the value at zero. On the unit sphere however

f

⁢

(

x

→

)

is continuous and thus achieves a

maximum

and a

minimum

since the unit sphere
is compact. Let’s call the minimum and maximum,

α

and

β

respectively.
Then for any

x

→

on the unit sphere we have

α

≤

f

⁢

(

x

→

)

≤

β

α

≤

∥

x

→

∥

a

∥

x

→

∥

b

≤

β

α

⁢

∥

x

→

∥

b

≤

∥

x

→

∥

a

≤

β

⁢

∥

x

→

∥

b

.

And so the norms are equivalent on the unit sphere and thus as we shown
above, everywhere.
∎

Title

all norms on finite-dimensional vector spaces are equivalent

Canonical

name

AllNormsOnFinitedimensionalVectorSpacesAreEquivalent

Date of creation

2013-03-22 14:08:54

Last modified on

2013-03-22 14:08:54

Owner

jirka (4157)

Last modified by

jirka (4157)

Numerical id

15

Author

jirka (4157)

Entry type

Theorem

Classification

msc 46B99

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