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Finite Dimensional Spaces
Let X be a vector space of dimension d < ∞ over F, and let {en}dn=1 ⊂ X be a basis, which is to say that for any x ∈ X, there are unique xn ∈ F such that
x =
∑^ d
n=
xn en.
We define a map T : X → F by T (x) = (x 1 , ..., xd), which gives a one-to-one correspondance between X and F. This map, called the coordinate mapping, is easily seen to be linear, so we have a vector space isomorphism between the spaces. Consequently, X and F have the same vector space structure. That is, given a dimension d, there is only one vector space structure of the given dimension (for the field F). Define for 1 ≤ p ≤ ∞ the map ‖ · ‖`p : Fd^ → [0, ∞) by
‖x‖`p =
( (^) ∑d
n=
|xn|p
) 1 /p for p < ∞ ,
‖x‖`∞ = max n=1,...,d
|xn| for p = ∞.
It is easy to veryfy that ‖x‖1 and ‖x‖∞ are norms on Fd. In fact, the zero and scaling properties of a norm are easily verified for ‖ · ‖p , but we need a few facts before we can verify the triangle inequality. Once done, note that then also ‖T (·)‖p is a norm on X. The following simple inequality turns out to be quite useful in practice. Lemma 1. Let 1 < p < ∞ and let q denote the conjugate exponent to p defined by 1 p
q
If a and b are nonnegative real numbers, then
ab ≤
ap p
bq q
with equality if and only if ap/bq^ = 1. Moreover, for any > 0 , then there is C = C(p, ) > 0 such that
ab ≤ ap^ + Cbq^. Proof. The function u : [0, ∞) → R given by
u(t) =
tp p
q
− t
has minimum value 0, attained only with t = 1. Apply this fact to t = ab−q/p^ to obtain main result. Replace ab by [(p)^1 /pa][(p)−^1 /pb] to obtain the final result.
This leads us immediately to H¨older’s Inequality. When p = 2, the inequality is also called the Cauchy-Schwarz Inequality.
Theorem 2 (H¨older’s Inequality). Let 1 ≤ p ≤ ∞ and let q denote the conjugate exponent (i.e., 1 /p + 1/q = 1, with the convention that q = ∞ if p = 1 and q = 1 if p = ∞). If x, y ∈ Fd, then ∣ ∣∣ ∣
∑^ d
n=
xnyn
∣ ≤ ‖x‖p^ ‖y‖q^.
1
2
Proof. The result is trivial if either p or q is infinity. Otherwise, simply apply (1) to a = |xn|/‖x‖p and b = |yn|/‖y‖q and sum on n to see that
∑^ d
n=
|xn| ‖x‖`p
|yn| ‖y‖`q
∑^ d
n=
|xn|p p‖x‖p`p
∑^ d
n=
|yn|q q‖y‖q`q
p
q
Thus ∣∣ ∣∣
∑^ d
n=
xnyn
∑^ d
n=
|xn| |yn| ≤ ‖x‖p ‖y‖q.
We can now finish our proof that each ‖ · ‖`p is in fact a norm on Fd; it remains only to show
the triangle inequality for 1 < p < ∞. For x, y ∈ Fd, simply apply H¨older’s inequality twice as follows:
‖x + y‖p`p =
∑^ d
n=
|xn + yn|p^ ≤
∑^ d
n=
|xn + yn|p−^1 (|xn| + |yn|)
( (^) ∑d
n=
|xn + yn|(p−1)q
) 1 /q (^) ( ‖x‖p + ‖y‖p
Since (p − 1)q = p and 1 − 1 /q = 1/p, we have the triangle inequality
‖x + y‖p ≤ ‖x‖p + ‖y‖`p
as desired.
Proposition 3. Let 1 ≤ p ≤ ∞. For any x ∈ Fd, ‖x‖∞ ≤ ‖x‖p ≤ d^1 /p‖x‖`∞ ,
with equality possible.
This result is trivial, and shows that all the p-norms ‖x‖p , 1 ≤ p ≤ ∞, are equivalent on
Fd, and it gives the optimal bounding constants with respect to the `∞-norm. A fundamental difference between finite and infinite dimensional spaces NLS’s is that in finite dimensions, a closed and bounded set is always compact, but this statement turns out to be untrue in infinite dimensions. This is closely related to another fundamental difference: in finite dimensions, all norms are equivalent, and so there is in fact only one norm topology.
Proposition 4. Let X be a finite dimensional NLS. All norms on X are equivalent. More- over, a subset of X is compact if and only if it is closed and bounded.
Proof. Let d be the dimension of X, and let {en}dn=1 be a basis. We defined earlier the coordinate mapping T : X → Fd. Let ‖ · ‖ denote any norm of X, and let
‖x‖ 1 = ‖T (x)‖` 1
be a second norm. We will show that these two norms are equivalent, which then implies that any pair are equivalent. The space (X, ‖ · ‖ 1 ) is essentially Fd, on which we assume the norm ‖ · ‖` 1. In fact, by definition of the norm ‖ · ‖ 1 , the coordinate map T : (X, ‖ · ‖ 1 ) → Fd^ is bounded, i.e., continuous, as is its inverse, which is also a linear function. Thus, (X, ‖ · ‖ 1 ) and Fd^ are homeomorphic as topological spaces, and also isomorphic as vector spaces. The Heine-Borel Theorem states that
