As mentioned above, it makes absolutely no difference which norm you decide to use. This holds in general
finite dimensional normed spaces and is shown here.
Definition 11.5.1Let
(V,∥⋅∥)
be a normed linear space and let a basis be
{v1,⋅⋅⋅,vn}
. For x ∈ V, let itscomponent vector in F^{n}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^{n}. It is also linear and its inverse θ^{−1}satisfies allthe same algebraic properties.
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 →
∥x ∥
is continuous as a map from V to ℝ.
Lemma 11.5.2There exists δ > 0 and Δ ≥ δ such that
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 11.5.3Let
(V,||⋅||)
be a finite dimensional normed linear space. Then the compact 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 achieves its maximum and minimum on K.
Proof:First note that the inequalities 11.13 and 11.14 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 11.4.7, it converges to some β∈ F^{n}. 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 11.13 and 11.14. Thus there is a subsequence still denoted with k
such that θw_{k}→β∈ F^{n}. 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 11.1.7, 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 11.4.7. 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.
Theorem 11.5.4Let
||⋅||
,
|||⋅|||
be two norms on V a finite dimensional vector space. Then they areequivalent, which means there are constants 0 < a < b such that for all v,
It follows right away that the closed and open sets are the same with two different norms. Also, all
considerations involving limits are unchanged from one norm to another.
Corollary 11.5.5Consider the metric spaces
(V,∥⋅∥1)
,
(V,∥⋅∥2)
where V has dimension n. Thena set is closed or open in one of these if and only if it is respectively closed or open in the other. Inother words, the two metric spaces have exactly the same open and closed sets. Also, a set is boundedin one metric space if and only if it is bounded in the other.
Proof: This follows from Theorem 11.1.27, the theorem about the equivalent formulations of continuity.
Using this theorem, it follows from Theorem 11.5.4 that the identity map I
(x)
≡ x is continuous.
The reason for this is that the inequality of this theorem implies that if
∥vm − v∥
_{1}→ 0 then
∥Ivm − Iv ∥
_{2} =
∥I(vm − v)∥
_{2}→ 0 and the same holds on switching 1 and 2 in what was just
written.
Therefore, the identity map takes open sets to open sets and closed sets to closed sets. In other words,
the two metric spaces have the same open sets and the same closed sets.
Suppose S is bounded in
(V,∥⋅∥ )
1
. This means it is contained in B
(0,r)
_{1} where the subscript of 1
indicates the norm is