As mentioned above, it makes absolutely no difference which norm you decide to use on ℝp.
This holds in general finite dimensional normed spaces and is shown here. Of course the main
interest here is where the normed linear space is
(ℝp,∥⋅∥)
but it is no harder to present the more
general result where you have a finite dimensional vector space V which has a norm. If you have
not seen such things, just let V be ℝp in what follows or consider the problems at end of the
chapter.
Definition 2.5.1Let
(V,∥⋅∥)
be a normed linear space and let a basis be
{v1,⋅⋅⋅,vn}
. For x ∈ V, let itscomponent vector in Fpbe
(α1,⋅⋅⋅,αn)
so that x = ∑iαivi. Then define
( )T
θx ≡ α = α1 ⋅⋅⋅ αn
Thus θ is well defined, one to one and onto from V to Fp. It is also linear and its inverse θ−1satisfies allthe same algebraic properties as θ. In particular,
(V,∥⋅∥)
could be
p
(ℝ ,∥⋅∥)
where
∥⋅∥
is some norm onℝp.
The following fundamental lemma comes from the extreme value theorem for continuous functions
defined on a compact set. Let
∥ ∥
∥∥∑ ∥∥ |||| −1 ||||
f (α ) ≡ ∥∥ αivi∥∥ ≡ θ α
i
Then it is clear that f is a continuous function. This is because α→∑iαivi is a continuous map into V
and from the triangle inequality x →
Proof:These numbers exist thanks to the extreme value theorem, Theorem 2.2.25. It
cannot be that δ = 0 because if it were, you would have
|α|
= 1 but ∑j=1nαkvj = 0 which is
impossible since
{v1,⋅⋅⋅,vn}
is linearly independent. The first of the above inequalities follows
from
∥∥ −1 α ∥∥ ( α )
δ ≤ ∥∥θ |α|∥∥ = f |α-| ≤ Δ
the second follows from observing that θ−1α is a generic vector v in V . ■
Now we can draw several conclusions about
(V,∥⋅∥)
for V finite dimensional.
Theorem 2.5.3Let
(V,||⋅||)
be a finite dimensionalnormed linear space. Then thecompact setsare exactly those which are closed and bounded. Also
(V,||⋅||)
is complete. If K is a closed andbounded set in
(V,||⋅||)
and f : K → ℝ, then f achievesits maximum and minimum on K.
Proof:First note that the inequalities 2.15 and 2.16 show that both θ−1 and θ are continuous. Thus
these take convergent sequences to convergent sequences.
k=1∞ is a Cauchy sequence. Thanks to
Theorem 2.2.25, it converges to some β∈ Fp. It follows that limk→∞θ−1θwk = limk→∞wk = θ−1β∈ V .
This shows completeness.
Next let K be a closed and bounded set. Let
{wk }
⊆ K. Then
{θwk }
⊆ θK which is also a closed and
bounded set thanks to the inequalities 2.15 and 2.16. Thus there is a subsequence still denoted with k
such that θwk→β∈ Fp. Then as just done, wk→ θ−1β. Since K is closed, it follows that
θ−1β ∈ K.
Finally, why are the only compact sets those which are closed and bounded? Let K be compact. If it is
not bounded, then there is a sequence of points of K,
{km}
m=1∞ such that
∥km ∥
≥ m. It
follows that it cannot have a convergent subsequence because the points are further apart from
each other than 1/2. Hence K is not sequentially compact and consequently it is not compact.
It follows that K is bounded. If K is not closed, then there exists a limit point k which is
not in K. (Recall that closed means it has all its limit points.) By Theorem 2.1.9, there is a
sequence of distinct points having no repeats and none equal to k denoted as
{km }
m=1∞
such that km→ k. Then this sequence
{km}
fails to have a subsequence which converges to a
point of K. Hence K is not sequentially compact. Thus, if K is compact then it is closed and
bounded.
The last part is identical to the proof in Theorem 2.2.25. You just take a convergent subsequence of a
minimizing (maximizing) sequence and exploit continuity. ■
Next is the theorem which states that any two norms on a finite dimensional vector space are
equivalent. In particular, any two norms on ℝp are equivalent.
Theorem 2.5.4Let
∥⋅∥
,
∥|⋅∥|
be two norms on V a finite dimensional vector space. Then they areequivalent, which means there are constants 0 < a < b suchthat for all v,