We will work with either the field of real numbers ℝ or the complex numbers ℂ. To avoid repetition, we use K to denote either ℝ or ℂ.
Recall, see Defn. 3, a norm on a vector space V is a map ||·||:V→[0,∞) such that
Note, that the second and third conditions imply that linear operations—multiplication by a scalar and addition of vectors respectively—are continuous in the topology defined by the norm.
A norm induces a metric, see Defn. 1, on V by setting d(u,v)=||u−v||. When V is complete, see Defn. 6, for this metric, we say that V is a Banach space.
We will use the following simple inequality:
⎪ ⎪ | ab | ⎪ ⎪ | ≤ |
| + |
| , (62) |
S1= |
| xp−1 d x= |
| , S2= |
| yq−1 d y= |
| . |
| ⎪ ⎪ | uj vj | ⎪ ⎪ | ≤ | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uj | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | vj | ⎪ ⎪ | q | ⎞ ⎟ ⎟ ⎠ |
| . |
ai= |
| and bi= |
| . |
⎪ ⎪ | ai bi | ⎪ ⎪ | ≤ |
| + |
| , |
Using Hölder inequality we can derive the following one:
⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uj+vj | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| ≤ | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uj | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| + | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | vj | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| . |
| ⎪ ⎪ | uk+vk | ⎪ ⎪ | p = |
| ⎪ ⎪ | uk | ⎪ ⎪ | ⎪ ⎪ | uk+vk | ⎪ ⎪ | p−1 + |
| ⎪ ⎪ | vk | ⎪ ⎪ | ⎪ ⎪ | uk+vk | ⎪ ⎪ | p−1. (63) |
| ⎪ ⎪ | uk | ⎪ ⎪ | ⎪ ⎪ | uk+vk | ⎪ ⎪ | p−1 ≤ | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uk | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uk+vk | ⎪ ⎪ | q(p−1) | ⎞ ⎟ ⎟ ⎠ |
| . |
Minkowski’s inequality shows that for 1≤ p<∞ (the case p=1 is easy) we can define a norm ||·||p on Kn by
⎪⎪ ⎪⎪ | u | ⎪⎪ ⎪⎪ | p = | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uj | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| ( u =(u1,⋯,un)∈Kn ). |
See, Figure 2 for illustration of various norms of this type defined in ℝ2.
We can define an infinite analogue of this. Let 1≤ p<∞, let lp be the space of all scalar sequences (xn) with ∑n | xn |p < ∞. A careful use of Minkowski’s inequality shows that lp is a vector space. Then lp becomes a normed space for the ||·||p norm. Note also, that l2 is the Hilbert space introduced before in Example 2.
Recall that a Cauchy sequence, see Defn. 5, in a normed space is bounded: if (xn) is Cauchy then we can find N with ||xn−xm||<1 for all n,m≥ N. Then ||xn|| ≤ ||xn−xN|| + ||xN|| < ||xN||+1 for n≥ N, so in particular, ||xn|| ≤ max( ||x1||,||x2||,⋯,||xN−1||,||xN||+1).
For each n, x(n)∈lp so is a sequence of scalars, say (xk(n))k=1∞. As (x(n)) is Cauchy, for each є>0 there exists Nє so that ||x(n) − x(m)||p ≤ є for n,m≥ Nє.
For k fixed,
⎪ ⎪ | xk(n) − xk(m) | ⎪ ⎪ | ≤ | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | xj(n) − xj(m) | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| = | ⎪⎪ ⎪⎪ | x(n) − x(m) | ⎪⎪ ⎪⎪ | p ≤ є, |
when n,m≥ Nє. Thus the scalar sequence (xk(n))n=1∞ is Cauchy in K and hence converges, to xk say. Let x=(xk), so that x is a candidate for the limit of (x(n)).
Firstly, we check that x−x(n)∈lp for some n. Indeed, for a given є>0 find n0 such that ||x(n)−x(m)||<є for all n,m>n0. For any K and m:
| ⎪ ⎪ | xk(n)−xk(m) | ⎪ ⎪ | p ≤ | ⎪⎪ ⎪⎪ | x(n)−x(m) | ⎪⎪ ⎪⎪ | p<єp. |
Let m→ ∞ then ∑k=1K
| xk(n)−xk |p ≤ єp.
Let K→ ∞ then ∑k=1∞| xk(n)−xk |p ≤ єp. Thus
x(n)−x∈lp and because lp is a
linear space then x = x(n)−(x(n)−x) is also in
lp.
Finally, we saw above that for any є >0 there is n0 such that ||x(n)−x||<є for all n>n0. Thus x(n)→ x. □
For p=∞, there are two analogies to the lp spaces. First, we define l∞ to be the vector space of all bounded scalar sequences, with the sup-norm (||·||∞-norm):
⎪⎪ ⎪⎪ | (xn) | ⎪⎪ ⎪⎪ | ∞ = |
| ⎪ ⎪ | xn | ⎪ ⎪ | ( (xn)∈ l∞ ). (64) |
Second, we define c0 to be the space of all scalar sequences (xn) which converge to 0. We equip c0 with the sup norm (64). This is defined, as if xn→0, then (xn) is bounded. Hence c0 is a subspace of l∞, and we can check (exercise!) that c0 is closed.
Let m≥ N, so that for any k, we have that
⎪ ⎪ | xk − xk(m) | ⎪ ⎪ | = |
| ⎪ ⎪ | xk(n) − xk(m) | ⎪ ⎪ | ≤ є. |
As k was arbitrary, we see that supk | xk−xk(m) | ≤ є. So, firstly, this shows that (x−x(m))∈l∞, and so also x = (x−x(m)) + x(m) ∈ l∞. Secondly, we have shown that ||x−x(m)||∞ ≤ є when m≥ N, so x(m)→ x in norm. □
⎪⎪ ⎪⎪ | f | ⎪⎪ ⎪⎪ | p= | ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ | ∫ |
| ⎪ ⎪ | f(t) | ⎪ ⎪ | p d t | ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ |
| . |
Recall what a linear map is, see Defn. 1. A linear map is often called an operator. A linear map T:E→ F between normed spaces is bounded if there exists M>0 such that ||T(x)|| ≤ M ||x|| for x∈ E, see Defn. 3. We write B(E,F) for the set of operators from E to F. For the natural operations, B(E,F) is a vector space. We norm B(E,F) by setting
⎪⎪ ⎪⎪ | T | ⎪⎪ ⎪⎪ | = sup | ⎧ ⎪ ⎨ ⎪ ⎩ |
| : x∈ E, x≠0 | ⎫ ⎪ ⎬ ⎪ ⎭ | . (65) |
We write B(E) for B(E,E). For normed spaces E, F and G, and for T∈B(E,F) and S∈B(F,G), we have that ST=S∘ T∈B(E,G) with ||ST|| ≤ ||S|| ||T||.
For T∈B(E,F), if there exists S∈B(F,E) with ST=IE, the identity of E, and TS=IF, then T is said to be invertible, and write T=S−1. In this case, we say that E and F are isomorphic spaces, and that T is an isomorphism.
If ||T(x)||=||x|| for each x∈ E, we say that T is an isometry. If additionally T is an isomorphism, then T is an isometric isomorphism, and we say that E and F are isometrically isomorphic.
Let E be a normed vector space, and let E* (also written E′) be B(E,K), the space of bounded linear maps from E to K, which we call functionals, or more correctly, bounded linear functionals, see Defn. 1. Notice that as K is complete, the above theorem shows that E* is always a Banach space.
φu(x) = |
| uj xj | ⎛ ⎝ | x=(xj)∈lp | ⎞ ⎠ | . |
⎪ ⎪ | φu(x) | ⎪ ⎪ | ≤ |
| ⎪ ⎪ | uj | ⎪ ⎪ | ⎪ ⎪ | xj | ⎪ ⎪ | ≤ | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uj | ⎪ ⎪ | q | ⎞ ⎟ ⎟ ⎠ |
| ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | xj | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| = | ⎪⎪ ⎪⎪ | u | ⎪⎪ ⎪⎪ | q | ⎪⎪ ⎪⎪ | x | ⎪⎪ ⎪⎪ | p. |
Now let φ∈(lp)*. For each n, let en = (0,⋯,0,1,0,⋯) with the 1 in the nth position. Then, for x=(xn)∈lp,
⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ | x − |
| xk ek | ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ | p = | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | xk | ⎪ ⎪ | p | ⎞ ⎟ ⎟ ⎠ |
| → 0, |
as n→∞. As φ is continuous, we see that
φ(x) = |
|
| φ(xkek) = |
| xk φ(ek). |
Let uk=φ(ek) for each k. If u=(uk)∈lq then we would have that φ=φu.
Let us fix N∈ℕ, and define
xk = |
|
Then we see that
| ⎪ ⎪ | xk | ⎪ ⎪ | p = |
| ⎪ ⎪ | uk | ⎪ ⎪ | p(q−1) = |
| ⎪ ⎪ | uk | ⎪ ⎪ | q, |
as p(q−1) = q. Then, by the previous paragraph,
φ(x) = |
| xk uk = |
| ⎪ ⎪ | uk | ⎪ ⎪ | q. |
Hence
⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ≥ |
| = | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uk | ⎪ ⎪ | q | ⎞ ⎟ ⎟ ⎠ |
| = | ⎛ ⎜ ⎜ ⎝ |
| ⎪ ⎪ | uk | ⎪ ⎪ | q | ⎞ ⎟ ⎟ ⎠ |
| . |
By letting N→∞, it follows that u∈lq with ||u||q ≤ ||φ||. So φ=φu and ||φ|| = ||φu|| ≤ ||u||q. Hence every element of (lp)* arises as φu for some u, and also ||φu|| = ||u||q. □
Loosely speaking, we say that lq = (lp)*, although we should always be careful to keep in mind the exact map which gives this.
Similarly, we can show that c0*=l1 and that (l1)*=l∞ (the implementing isometric isomorphism is giving by the same summation formula).
Mathematical induction is a well known method to prove statements depending from a natural number. The mathematical induction is based on the following property of natural numbers: any subset of ℕ has the least element. This observation can be generalised to the transfinite induction described as follows.
A poset is a set X with a relation ≼ such that a≼ a for all a∈ X, if a≼ b and b≼ a then a=b, and if a≼ b and b≼ c, then a≼ c. We say that (X,≼) is total if for every a,b∈ X, either a≼ b or b≼ a. For a subset S⊆ X, an element a∈ X is an upper bound for S if s≼ a for every s∈ S. An element a∈ X is maximal if whenever b∈ X is such that a≼ b, then also b≼ a.
Then Zorn’s Lemma tells us that if X is a non-empty poset such that every total subset has an upper bound, then X has a maximal element. Really this is an axiom which we have to assume, in addition to the usual axioms of set-theory. Zorn’s Lemma is equivalent to the axiom of choice and Zermelo’s theorem.
Let x∉G, so an extension φ1 of φ to the linear span of G and x must have the form
φ1(x′+ax) = φ(x) + a α (x′∈ G, a∈ℝ), |
for some α∈ℝ. Under this, φ1 is linear and extends φ, but we also need to ensure that ||φ1||≤||φ||. That is, we need
⎪ ⎪ | φ(x′) + aα | ⎪ ⎪ | ≤ | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | x′+ax | ⎪⎪ ⎪⎪ | (x′∈ G, a∈ℝ). (66) |
It is straightforward for a=0, otherwise to simplify proof put −a y=x′ in (66) and divide both sides of the identity by a. Thus we need to show that there exist such α that
⎪ ⎪ | α−φ(y) | ⎪ ⎪ | ≤ | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | x−y | ⎪⎪ ⎪⎪ | for all y∈ G, a∈ℝ, |
or
φ(y)− | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | x−y | ⎪⎪ ⎪⎪ | ≤ α ≤ φ(y)+ | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | x−y | ⎪⎪ ⎪⎪ | . |
For any y1 and y2 in G we have:
φ(y1)−φ(y2)≤ | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | y1−y2 | ⎪⎪ ⎪⎪ | ≤ | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ( | ⎪⎪ ⎪⎪ | x−y2 | ⎪⎪ ⎪⎪ | + | ⎪⎪ ⎪⎪ | x−y1 | ⎪⎪ ⎪⎪ | ). |
Thus
φ(y1)− | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | x−y1 | ⎪⎪ ⎪⎪ | ≤ φ(y2)+ | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | x−y2 | ⎪⎪ ⎪⎪ | . |
As y1 and y2 were arbitrary,
| (φ(y) − | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | y+x | ⎪⎪ ⎪⎪ | ) ≤ |
| (φ(y) + | ⎪⎪ ⎪⎪ | φ | ⎪⎪ ⎪⎪ | ⎪⎪ ⎪⎪ | y+x | ⎪⎪ ⎪⎪ | ). |
Hence we can choose α between the inf and the sup.
The complex case follows by “complexification”. □
The Hahn-Banach theorem tells us that a functional from a subspace can be extended to the whole space without increasing the norm. In particular, extending a functional on a one-dimensional subspace yields the following.
Another useful result which can be proved by Hahn-Banach is the following.
So, x∉F. Define ψ:{F,x}→K by
ψ(y+tx) = t (y∈ F, t∈K). |
This is well-defined, for y, y′∈ F if y+tx=y′+t′x then either t=t′, or otherwise x = (t−t′)−1(y′−y) ∈ F which is a contradiction. The map ψ is obviously linear, so we need to show that it is bounded. Towards a contradiction, suppose that ψ is not bounded, so we can find a sequence (yn+tnx) with ||yn+tnx||≤1 for each n, and yet | ψ(yn+tnx) |=| tn |→∞. Then || tn−1 yn + x || ≤ 1/| tn | → 0, so that the sequence (−tn−1yn), which is in F, converges to x. So x is in the closure of F, a contradiction. So ψ is bounded. By Hahn-Banach theorem, we can find some φ∈ E* extending ψ. For y∈ F, we have φ(y)=ψ(y)=0, while φ(x)=ψ(x)=1, so 2 doesn’t hold, as required. □
We define E** = (E*)* to be the bidual of E, and define J:E→ E** as follows. For x∈ E, J(x) should be in E**, that is, a map E*→K. We define this to be the map φ↦φ(x) for φ∈ E*. We write this as
J(x)(φ) = φ(x) (x∈ E, φ∈ E*). |
The Corollary 16 shows that J is an isometry; when J is surjective (that is, when J is an isomorphism), we say that E is reflexive. For example, lp is reflexive for 1<p<∞. On the other hand c0 is not reflexive.
This section is not examinable. Standard facts about topology will be used in later sections of the course.
All our topological spaces are assumed Hausdorff. Let X be a compact space, and let CK(X) be the space of continuous functions from X to K, with pointwise operations, so that CK(X) is a vector space. We norm CK(X) by setting
⎪⎪ ⎪⎪ | f | ⎪⎪ ⎪⎪ | ∞ = |
| ⎪ ⎪ | f(x) | ⎪ ⎪ | (f∈ CK(X)). |
Let E be a vector space, and let ||·||(1) and ||·||(2) be norms on E. These norms are equivalent if there exists m>0 with
m−1 | ⎪⎪ ⎪⎪ | x | ⎪⎪ ⎪⎪ | (2) ≤ | ⎪⎪ ⎪⎪ | x | ⎪⎪ ⎪⎪ | (1) ≤ m | ⎪⎪ ⎪⎪ | x | ⎪⎪ ⎪⎪ | (2) (x∈ E). |
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