Not surprisingly all of the above holds for a finite dimensional normed linear space. First here
is an easy lemma which follows right away from Theorem 2.6.2, the theorem about equivalent
formulations of continuity.
Lemma 3.7.1Let
(V,∥⋅∥V)
and
(W,∥⋅∥W )
be two normed linear spaces. Then alinear map f : V → W is continuous if and only if it takes bounded sets to bounded sets.(f is bounded)If V is finite dimensional, then f must be continuous.
Proof:
=⇒
Consider f
(B(0,1))
. If this is not bounded, then there exists
∥vm∥
_{V }≤ 1
but
∥f (vm )∥
_{W}≥ m. Then it follows that
∥ ( m)∥
∥f vm--∥
_{W}≥ 1 which is impossible for all m since
∥ m∥
∥vm-∥
≤
1m-
and so continuity requires that lim_{m→∞}f
( m )
vm-
= 0 (Theorem 2.6.2). Thus there
exists M such that
∥f (v)∥
≤ M whenever v ∈ B
(0,1)
. In general, let S be a bounded set.
Then S ⊆ B
(0,r)
for large enough r. Hence, for v ∈ B, it follows that v∕2r ∈ B
(0,1)
. It
follows that
∥f (v∕2r)∥
_{W}≤ M and so
∥f (v)∥
_{W}≤ 2rM. Thus f takes bounded sets to
bounded sets.
⇐= Suppose f is bounded and not continuous. Then by Theorem 2.6.2 again, there is a
sequence v_{n}→ v but f
(vn )
fails to converge to f
(v)
. Then there exists ε > 0 and a
subsequence, still denoted as v_{n} such that
∥f (vn)− f (v)∥
=
∥f (vn − v)∥
≥ ε.
Then
∥∥ ( vn − v )∥∥ 1
∥∥f ∥v--−-v∥-∥∥ ≥ ε∥v-−-v∥-
n n
The right side is unbounded, but the left is bounded, a contradiction.
Consider the last claim about continuity. Let
{v1,⋅⋅⋅,vn}
be a basis for V . By Lemma
3.4.6, if y^{m}→ 0, in V for
n
ym = ∑ ym vk,
k=1 k
then it follows that lim_{m→∞}y_{k}^{m} = 0 and consequently, f
(ym )
→ f
(0)
= 0. In general, if
y^{m}→ y, then
(ym − y)
→ 0 and so f
(ym − y)
= f
(ym)
−f
(y)
→ 0. That is, f
(ym )
→ f
(y)
.
■
Definition 3.7.2For f :
(V,∥⋅∥V )
→
(W, ∥⋅∥W )
continuous, it was just shown thatthere exists M such that
∥f (v)∥ ≤ M, v ∈ B (0,1).
It follows that, since
v
2∥v∥-
∈ B
(0,1)
, then
∥f (v)∥
≤ 2M
∥v∥
. Therefore, letting
∥f∥ ≡ sup ∥f (v)∥
∥v∥≤1
it follows that for all v ∈ V,
∥f (v)∥ ≤ ∥f∥∥v∥
Thus a linear map is bounded if and only if
∥f ∥
< ∞ if and only if f is continuous.
The number
∥f∥
is called the operator norm. You can show that for ℒ
(V,W )
the space of
bounded linear maps from V to W, ℒ
(V,W )
becomes a normed linear space with this
definition. This is true whether V,W are finite or infinite dimensional. You can also show that
if W is complete then so is ℒ
(V,W )
. This is left as an exercise. However, we will use
this notation. Also, when the vector spaces are finite dimensional, it was shown
in the above lemma that any linear function f is automatically bounded, hence
continuous, hence
∥f∥
exists. Here is an interesting observation about the operator
norm.
Lemma 3.7.3Let f ∈ℒ
(V,W )
and let h ∈ℒ
(W,Z )
where X,Y,Z are normed vectorspaces. Then
∥h ∘f∥ ≤ ∥h∥∥f∥
Proof:This follows right away from the definition. If
be any linear map which is one toone and onto. Then both f and f^{−1}are continuous. Also the compact sets of
(V,∥⋅∥V )
are exactly those which are closed and bounded.
Proof:Define another norm
∥⋅∥
_{1} on F^{n} as follows.
∥x∥1 ≡ ∥f (x )∥V
Since f is one to one and onto and linear, this is indeed a norm. The details are left as an
exercise. Then from the theorem on the equivalence of norms, there are positive constants δ,Δ
such that
δ∥x∥ ≤ ∥f (x)∥V ≤ Δ ∥x ∥
Since f is one to one and onto, this implies
∥ −1 ∥ ∥ −1 ∥
δ∥f (v)∥ ≤ ∥v∥V ≤ Δ ∥f (v )∥
The first of these above inequalities implies f is continuous. The second says
∥ − 1 ∥ 1
∥f (v)∥ ≤ δ ∥v∥V
and so f^{−1} is continuous. Thus, from the above theorems, both f and f^{−1} map closed sets to
closed sets, compact sets to compact sets, open sets to open sets and bounded sets to
bounded sets.
Now let K ⊆ V be closed and bounded. Then from the above observations, f^{−1}
(K )
is also
closed and bounded. Therefore, it is compact. Now f
( −1 )
f (K )
= K must be compact
because the continuous image of a compact set is compact, Theorem 2.7.1. Conversely, if
K ⊆ V is compact, then by the theorem just mentioned, f^{−1}
(K)
is compact and
so it is closed and bounded. Hence f
( − 1 )
f (K)
= K is also closed and bounded.
■
This is a remarkable theorem. It says that an algebraic isomorphism is also
a homeomorphism which is what it means to say that the map takes open sets
to open sets and the inverse does the same. In other words, there really isn’t any
algebraic or topological distinction between a finite dimensional normed vector
space of dimension n and F^{n}. Of course when one considers geometry, this is not
so.
Here is another interesting theorem about coordinate maps. It follows right away from
earlier theorems.
Theorem 3.7.5Let f :
(V,∥⋅∥V)
→
(W, ∥⋅∥W )
be a continuous functionwhere here
(V,∥⋅∥V)
is a normed linear space and
(W, ∥⋅∥W )
is a finite dimensionalnormed linear space with basis
{w1,⋅⋅⋅,wn}
. Thus f
(v)
= ∑_{k=1}^{n}f_{k}
(v)
w_{k}. Then f iscontinuous if and only if each f_{k}is a continuous F valued map.
Proof:
= ⇒
Why is the coordinate function f_{k} continuous? First note that
each of these coordinate maps is linear with values in F. From Lemma 3.7.1, it
suffices to verify that f_{k} is bounded. If this is not so, there exists v_{m},
∥vm ∥
_{V }≤ 1
but
|fk(vm )|
_{W}≥ m. It follows that
| ( )|
|fk vmm--|
≥ 1. Since f is continuous, and
v_{m}∕m → 0, it follows that f