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. Of course the main
interest here is where the normed linear space is
p
(ℝ ,∥⋅∥)
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.
Definition B.2.30Let
(V,∥⋅∥)
be a normed linear space and let a basis be
{v ,⋅⋅⋅,v }
1 n
. For x ∈ V, let its component vector in F^{p}be
(α ,⋅⋅⋅,α )
1 n
so that x = ∑_{i}α_{i}v_{i}.Then define
( )
θx ≡ α = α1 ⋅⋅⋅ αn T
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 all the same algebraic properties as θ. In particular,
(V,∥⋅∥)
could be
(ℝp,∥⋅∥)
where
∥⋅∥
is some norm on ℝ^{p}.
The following fundamental lemma comes from the extreme value theorem for continuous
functions defined on a compact set. Let
||||∑ |||| || ||
f (α ) ≡ |||| αivi|||| ≡ ||θ−1α||
|| 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 B.2.31There exists δ > 0 and Δ ≥ δ such that
δ = min {f (α) : |α | = 1},Δ = max {f (α) : |α | = 1}
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 B.2.32Let
(V,||⋅||)
be a finite dimensional normed linear space.Then the compact sets are exactly those which are closed and bounded. Also
(V,||⋅||)
iscomplete. If K is a closed and bounded set in
(V,||⋅||)
and f : K → ℝ, then f achievesits maximum and minimum on K.
Proof:First note that the inequalities 2.8 and 2.9 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 B.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
{w }
k
⊆ K. Then
{θw }
k
⊆ θK which is also
a closed and bounded set thanks to the inequalities 2.8 and 2.9. 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 B.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 B.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 B.2.33Let
||⋅||
,
|||⋅|||
be two norms on V a finite dimensional vectorspace. Then they are equivalent, which means there are constants 0 < a < b such that for allv,