As mentioned above, it makes absolutely no difference which norm you decide to use on ℝ^{p}.
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 ℝ^{p} 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 F^{p}be
(α1,⋅⋅⋅,αn)
so that x = ∑_{i}α_{i}v_{i}. Then define
( )T
θx ≡ α = α1 ⋅⋅⋅ αn
Thus θ is well defined, one to one and onto from V to F^{p}. It is also linear and its inverse θ^{−1}satisfies allthe same algebraic properties as θ. In particular,
(V,∥⋅∥)
could be
p
(ℝ ,∥⋅∥)
where
∥⋅∥
is some norm onℝ^{p}.
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→∑_{i}α_{i}v_{i} 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=1}^{n}α_{k}v_{j} = 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^{−1}α 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∈ F^{p}. It follows that lim_{k→∞}θ^{−1}θw_{k} = lim_{k→∞}w_{k} = θ^{−1}β∈ 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 θw_{k}→β∈ F^{p}. Then as just done, w_{k}→ θ^{−1}β. Since K is closed, it follows that
θ^{−1}β ∈ 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 k^{m}→ 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 ℝ^{p} 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,