As mentioned above, it makes absolutely no difference which norm you decide to use. This
holds in general finite dimensional normed spaces. First are some simple lemmas featuring one
dimensional considerations. In this case, the distance is given by d
(x,y)
=
|x − y|
and so the
open balls are sets of the form
(x − δ,x + δ)
. First recall the nested interval lemma.
Lemma 3.4.1Let
[a ,b ]
k k
⊇
[a ,b ]
k+1 k+1
for all k = 1,2,3,
⋅⋅⋅
. Then there exists apoint p in ∩_{k=1}^{∞}
[a ,b ]
k k
.
Proof: We note that for any k,l,a_{k}≤ b_{l}. Here is why. If k ≤ l, then
ak ≤ al ≤ bl
If k > l, then
bl ≥ bk ≥ ak
It follows that for each l,
sup a ≤ b
k k l
Hence sup_{k}a_{k} is a lower bound to the set of all b_{l} and so it is no larger than the greatest
lower bound. It follows that
supa ≤ infb
k k l l
Pick x ∈
[supkak,inflbl]
. Then for every k,a_{k}≤ x ≤ b_{k}. Hence x ∈∩_{k=1}^{∞}
[ak,bk]
.■
Lemma 3.4.2The closed interval
[a,b]
is compact.
Proof: Let C be an open cover of
[a,b]
and suppose it fails to admit a finite
subcover. Then this must be the case for one of the two intervals
[ a+b]
a, 2
and
[a+b ]
-2 ,b
. Let I_{1} be the one for which this is so. Then split it into two equal pieces
like what was just done and let I_{2} be a half for which there is no finite subcover
of sets of C. Continue this way. This yields a nested sequence of closed intervals
I_{1}⊇ I_{2}⊇
⋅⋅⋅
and by the above lemma, there exists a point x in all of these intervals.
There exists U ∈C such that x ∈ U. Now since U is open, there exists an open ball
(x − δ,x + δ)
⊆ U. However, for all n large enough, the length of I_{n} is less than δ. Hence I_{n} is
actually contained in a single set of C contrary to the construction. Hence
[a,b]
is compact. By Theorem 2.5.5, the closed interval is also sequentially compact.
■
The next corollary gives the definition of a closed disk and shows that, like a closed
interval, a closed disk is compact.
Corollary 3.4.3In ℂ, let D
(z,r)
≡
{w ∈ ℂ : |z − w | ≤ r}
. Then D
(z,r)
iscompact.
Proof:Let
{xn + iyn}
_{n=1}^{∞}⊆ D
(z,r)
where z = a + ib. Then in particular,
{xn}
_{n=1}^{∞}⊆
[a− r,a+ r]
,
{yn}
_{n=1}^{∞}⊆
[b − r,b+ r]
. It follows from Lemma 3.4.2 that
there is a subsequence obtained by taking a succession of two subsequences, and still denoted
with the index n such that x_{n}→ x ∈
[a− r,a+ r]
,y_{n}→ y ∈
[b − r,b+ r]
. Thus
x_{n} + iy_{n}→ x + iy. However,
∘ -------2---------2
(xn − a) +(yn − b) ≤ r
and so, on passing to a limit,
∘ ------2--------2
(x − a) +(y − b) ≤ r
so x + iy ∈ D
(z,r)
. ■
A set K in F^{n} is said to be bounded if it is contained in some ball B
(0,r)
.
Theorem 3.4.4A set K ⊆ F^{n}is compact if it is closed and bounded. Iff : K → ℝ, then f achieves its maximum and its minimum on K.
Proof: Let
{zk}
_{k=1}^{∞} be contained in K. Then the sequence of i^{th} components
{zk}
i
_{k=1}^{∞} is also bounded. Thus there exists r large enough that
n
K ⊆ ∏ D (0,r)
j=1
Taking a succession of n subsequences, there is a subsequence, still denoted with
the superscript k such that for each i, lim_{k→∞}z_{i}^{k}→ z_{i}∈ D
(0,r)
. It follows
z ≡
(z1,⋅⋅⋅,zn)
^{T}∈∏_{j=1}^{n}D
(0,r)
. Actually, z ∈ K because K is closed. Thus K is
sequentially compact and by Theorem 2.5.5, K is compact also. Now consider the last claim.
Let M ≡ sup
{f (x) : x ∈ K }
. Let
{ k}
x
_{k=1}^{∞} be a maximizing sequence which means
lim_{k→∞}f
( k)
x
= M. By compactness, there is a subsequence, still denoted by superscript k
such that x^{k}→ x ∈ K. Then by continuity, M = f
(x)
= lim_{k→∞}f
( k)
x
. The case of
the minimum is exactly similar. You just use a minimizing sequence followed by
compactness. You could also get it from what was just done by consideration of −f.
■
Definition 3.4.5Let
{v1,⋅⋅⋅,vn }
be a basis for V where
(V,||⋅||)
is a finitedimensional normed vector space with field of scalars equal to either ℝ or ℂ. Defineθ : V → F^{n}as follows.
( )
∑n
θ ( αjvj) ≡ α ≡ (α1,⋅⋅⋅,αn)T
j=1
Thus θ maps a vector to its coordinates taken with respect to a given basis.
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 defined on F^{n}. 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 3.4.6There 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 . ■
Note that these inequalities yield the fact that convergence of the coordinates with respect
to a given basis is equivalent to convergence of the vectors. More precisely, to say that
lim_{k→∞}v^{k} = v is the same as saying that lim_{k→∞}θv^{k} = θv. Indeed,
δ|θvn − θv| ≤ ||vn − v|| ≤ Δ |θvn − θv|
Now we can draw several conclusions about
(V,||⋅||)
for V finite dimensional.
Theorem 3.4.7Let
(V, ||⋅||)
be a finite dimensional normed linear space. Thenthe 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 3.21 and 3.22 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 3.4.4, 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 3.21 and 3.22. 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 2.1.7, 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 3.4.4. 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 3.4.8Let
||⋅||
,
|||⋅|||
be two norms on V a finite dimensional vector space.Then they are equivalent, which means there are constants 0 < a < b such that for allv,
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 3.4.9Consider the metric spaces
(V,∥⋅∥1)
,
(V,∥⋅∥2)
where V hasdimension n. Then a set is closed or open in one of these if and only if it is respectivelyclosed or open in the other. In other words, the two metric spaces have exactly the sameopen and closed sets. Also, a set is bounded in one metric space if and only if it isbounded in the other.
Proof: This follows from Theorem 2.6.2, the theorem about the equivalent formulations of
continuity. Using this theorem, it follows from Theorem 3.4.8 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
∥⋅∥
_{1}. Let δ
∥⋅∥
_{1}≤
∥⋅∥
_{2}≤ Δ
∥⋅∥
_{1} as described above.
Then
S ⊆ B(0,r)1 ⊆ B (0,Δr)2
so S is also bounded in
(V,∥⋅∥ )
2
. Similarly, if S is bounded in
∥⋅∥
_{2} then it is bounded in
∥⋅∥
_{1}. ■
One can show that in the case of ℝ where it makes sense to consider sup and inf,
convergence of Cauchy sequences can be shown to imply the other definition of completeness
involving sup, and inf.